Latihan Soal 8 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

 $\begin{array}{ll}\\ 71.&\textrm{Grafik fungsi trigonometri pada gambar}\\ &\textrm{berikut adalah}\: .... \end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{a}.\quad \displaystyle y=2\cos x \\\\ &\textrm{b}.\quad \displaystyle y=\cos 2x \\\\ &\textrm{c}.\quad \displaystyle y=\cos \displaystyle \frac{1}{2}x \\\\ &\textrm{d}.\quad \displaystyle \color{red}y=2\cos 2x \\\\ &\textrm{e}.\quad \displaystyle y=2\cos \displaystyle \frac{1}{2}x \\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Dari grafik tampak jelas bahwa}\\ &\textrm{gambar di atas adalah garfik}\\ &\textrm{fungsi cosinus di geser ke}\\ & \textbf{atas dan ke bawah}\\ &\textrm{dengan}\: \: \textbf{amplitudo}\: \: 2\\ &\textrm{dan}\: \: \textbf{periodenya}\: \: \displaystyle \frac{360^{\circ}}{2}=180^{\circ}=\pi ,\\ & \textrm{maka}\\ &\textrm{bentuk persamaan}\: \: \textbf{grafik fungsinya}\\ &y=\color{red}2\cos 2x \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 72.&\textrm{Grafik fungsi trigonometri pada gambar}\\ &\textrm{berikut adalah}\: .... \end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{a}.\quad \displaystyle y=2\sin \left (2x+\pi \right ) \\\\ &\textrm{b}.\quad \displaystyle y=\sin \left (2x-\displaystyle \frac{1}{2}\pi \right ) \\\\ &\textrm{c}.\quad \displaystyle y=2\sin \left (2x-\displaystyle \frac{1}{2}\pi \right ) \\\\ &\textrm{d}.\quad \displaystyle y=\sin \left (2x+\displaystyle \frac{1}{2}\pi \right ) \\\\ &\textrm{e}.\quad \displaystyle \color{red}y=2\sin \displaystyle \left (x+\displaystyle \frac{1}{2}\pi \right ) \\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Dari grafik tampak jelas bahwa}\\ &\textrm{gambar di atas adalah garfik}\\ &\textrm{fungsi cosinus di geser ke}\: \: \textbf{kiri}\\ &\textrm{dengan}\: \: \textbf{amplitudo}\: \: 2\\ &\textrm{dan}\: \: \textbf{periodenya}\: \: \displaystyle \frac{360^{\circ}}{1}=360^{\circ}=2\pi ,\\ &\textrm{maka}\\ &\textrm{bentuk persamaan}\: \: \textbf{grafik fungsinya}\\ &y=\color{red}2\sin \left ( x+\displaystyle kx \right )\\ &\textrm{dengan}\: \: +k\: \: \textrm{adalah}\\ &\textbf{besar geseran ke kiri}\: \: \displaystyle \frac{1}{2}\pi \: \: \textrm{atau}\: \: 90^{\circ}\\ &\textrm{Jadi},\: \: y=\color{red}2\sin (x+\displaystyle \frac{1}{2}\pi )^{\circ} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 73.&\textrm{Himpunan penyelesaian dari persamaan}\\ &\sin x=\sin \displaystyle \frac{2}{10}\pi \: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \displaystyle \left \{ \displaystyle \frac{22}{10}\pi ,\displaystyle \frac{8}{10}\pi \right \} \\\\ &\textrm{b}.\quad \displaystyle \left \{ \displaystyle \frac{2}{10}\pi ,\displaystyle \frac{28}{10}\pi \right \} \\\\ &\textrm{c}.\quad \displaystyle \color{red}\left \{ \displaystyle \frac{2}{10}\pi ,\displaystyle \frac{8}{10}\pi \right \} \\\\ &\textrm{d}.\quad \displaystyle \left \{ \displaystyle \frac{22}{10}\pi ,\displaystyle \frac{28}{10}\pi \right \} \\\\ &\textrm{e}.\quad \displaystyle \left \{ \displaystyle \frac{12}{10}\pi ,\displaystyle \frac{8}{10}\pi \right \} \\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\begin{aligned} & \sin x=\sin \displaystyle \frac{2}{10}\pi \\ &\Leftrightarrow \: \: x_{1}=\displaystyle \frac{2}{10}\pi+k.2\pi \: \: \: \: \color{blue}\textrm{atau}\\ &\Leftrightarrow \quad x_{2} =\left (\pi -\displaystyle \frac{2}{10}\pi \right )+k.2\pi=\displaystyle \frac{8}{10}\pi +k.2\pi \\ &k=0\Rightarrow x_{1}=\displaystyle \frac{2}{10}\pi\: \: (\color{blue}\textrm{mm})\: \: \color{black}\textrm{atau}\\ &\qquad\qquad x_{2}=\displaystyle \frac{8}{10}\pi\: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x_{1,2}=....+2\pi \quad (\color{red}\textrm{tidak memenuhi})\\ \end{aligned} \\ &\textbf{HP}=\color{red}\left \{\displaystyle \frac{2}{10}\pi,\: \displaystyle \frac{8}{10}\pi \right \} \end{array} \end{array}$.

$\begin{array}{ll}\\ 74.&\textrm{Himpunan penyelesaian dari persamaan}\\ &\tan \left ( 2x-\displaystyle \frac{1}{4}\pi \right )=\tan \displaystyle \frac{1}{4}\pi \: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \displaystyle \left \{ \displaystyle \frac{1}{3}\pi ,\pi ,\displaystyle \frac{5}{3}\pi ,\displaystyle \frac{7}{3}\pi \right \} \\\\ &\textrm{b}.\quad \displaystyle \left \{ \displaystyle \frac{1}{4}\pi ,\displaystyle \frac{3}{5}\pi ,\frac{5}{4}\pi ,\frac{8}{5}\pi \right \} \\\\ &\textrm{c}.\quad \displaystyle \left \{ \displaystyle \frac{1}{4}\pi ,\displaystyle \frac{3}{4}\pi ,\frac{6}{4}\pi ,\displaystyle \frac{7}{4}\pi \right \} \\\\ &\textrm{d}.\quad \displaystyle \left \{ \displaystyle \frac{2}{4}\pi ,\displaystyle \frac{3}{4}\pi ,\pi ,\displaystyle \frac{7}{4}\pi \right \} \\\\ &\textrm{e}.\quad \displaystyle \color{red}\left \{ \displaystyle \frac{1}{4}\pi ,\frac{3}{4}\pi ,\frac{5}{4}\pi ,\displaystyle \frac{7}{4}\pi \right \} \\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\begin{aligned} & \tan \left ( 2x-\displaystyle \frac{1}{4}\pi \right )=\tan \displaystyle \frac{1}{4}\pi \\ &\Leftrightarrow \: \: 2x-\displaystyle \frac{1}{4}\pi=\displaystyle \frac{1}{4}\pi+k.\pi\\ &\Leftrightarrow \quad 2x =\displaystyle \frac{2}{4}\pi +k.\pi \\ &\Leftrightarrow \quad x =\displaystyle \frac{1}{4}\pi +k.\frac{\pi}{2} \\ &k=0\Rightarrow x=\displaystyle \frac{1}{4}\pi\: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x=\displaystyle \frac{1}{4}\pi+\frac{\pi}{2}=\displaystyle \frac{3}{4}\pi \: \: (\color{blue}\textrm{mm}) \\ &k=2\Rightarrow x=\displaystyle \frac{1}{4}\pi+\pi =\displaystyle \frac{5}{4}\pi \: \: (\color{blue}\textrm{mm}) \\ &k=3\Rightarrow x=\displaystyle \frac{1}{4}\pi+\frac{3\pi}{2}=\displaystyle \frac{7}{4}\pi \: \: (\color{blue}\textrm{mm}) \\ &k=4\Rightarrow x=\displaystyle \frac{1}{4}\pi+2\pi =\displaystyle \frac{9}{4}\pi \: \: (\color{red}\textrm{tidak memenuhi}) \\ \end{aligned} \\ &\textbf{HP}=\color{red}\color{red}\left \{\displaystyle \frac{1}{4}\pi,\: \displaystyle \frac{3}{4}\pi ,\frac{5}{4}\pi ,\frac{7}{4}\pi \right \} \end{array} \end{array}$.

$\begin{array}{ll}\\ 75.&\textrm{Himpunan penyelesaian dari persamaan}\\ &\cos 2x-2\cos x=-1\: \: \textrm{untuk}\: \: 0< x< 2\pi \\ &\textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \color{red}\displaystyle \left \{ 0,\displaystyle \frac{1}{2}\pi ,\displaystyle \frac{3}{2}\pi, 2\pi \right \} \\\\ &\textrm{b}.\quad \displaystyle \left \{ 0,\displaystyle \frac{1}{2}\pi ,\displaystyle \frac{2}{3}\pi, 2\pi \right \} \\\\ &\textrm{c}.\quad \displaystyle \left \{ 0,\displaystyle \frac{1}{2}\pi ,\pi ,\displaystyle \frac{3}{2}\pi \right \} \\\\ &\textrm{d}.\quad \displaystyle \left \{ 0,\displaystyle \frac{1}{2}\pi ,\displaystyle \frac{2}{3}\pi \right \} \\\\ &\textrm{e}.\quad \displaystyle \left \{ 0,\displaystyle \frac{1}{2}\pi ,\pi \right \} \\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\begin{aligned} & \cos 2x-2\cos x=-1\\ &\Leftrightarrow \cos 2x-2\cos x+1=0\\ &\Leftrightarrow \left ( 2\cos ^{2}x-1 \right )-2\cos x+1=0 \\ &\Leftrightarrow 2\cos x\left ( \cos x-1 \right )=0\\ &\Leftrightarrow \cos x=0\: \: \textrm{atau}\: \: \cos x=1\\ &\Leftrightarrow \cos x=\cos \displaystyle \frac{1}{2}\pi \: \: \textrm{atau}\: \: \cos x=\cos 0\\ &\Leftrightarrow x_{1,2}=\pm \displaystyle \frac{1}{2}\pi +k.2\pi \: \: \textrm{atau}\: \: x_{3}=k.2\pi\\ &\textrm{maka}\\ &k=0\Rightarrow x_{1}=-\displaystyle \frac{1}{2}\pi\: \: (\color{red}\textrm{tm})\: \: \color{black}\textrm{atau}\\ &\qquad\qquad x_{2}=\displaystyle \frac{1}{2}\pi\: \: (\color{blue}\textrm{mm})\\ &\qquad\qquad x_{3}=\displaystyle 0\: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x_{1}=\displaystyle \frac{3}{2}\pi \quad (\color{blue}\textrm{mm})\\ &\qquad\qquad x_{2}=\displaystyle \frac{5}{2}\pi\: \: (\color{red}\textrm{tm})\\ &\qquad\qquad x_{3}=\displaystyle 2\pi\: \: (\color{blue}\textrm{mm})\\ \end{aligned} \end{array} \end{array}$.

$\begin{array}{ll}\\ 76.&\textrm{Himpunan penyelesaian dari persamaan}\\ &\qquad\quad 3\tan \left (2x-\displaystyle \frac{1}{3}\pi \right )=-\sqrt{3} \\ & \textrm{untuk}\: \: 0\leq x\leq \pi \: \: \: \textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \color{red}\displaystyle \left \{ \displaystyle \frac{1}{12}\pi ,\displaystyle \frac{7}{12}\pi \right \} \\\\ &\textrm{b}.\quad \displaystyle \left \{ \displaystyle \frac{2}{12}\pi ,\displaystyle \frac{9}{12}\pi \right \} \\\\ &\textrm{c}.\quad \displaystyle \left \{ \displaystyle \frac{3}{12}\pi ,\displaystyle \frac{7}{12}\pi \right \} \\\\ &\textrm{d}.\quad \displaystyle \left \{ \displaystyle \frac{3}{12}\pi ,\displaystyle \frac{9}{12}\pi \right \} \\\\ &\textrm{e}.\quad \displaystyle \left \{ \displaystyle \frac{5}{12}\pi ,\displaystyle \frac{7}{12}\pi \right \} \\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\begin{aligned} & 3\tan \left (2x-\displaystyle \frac{1}{3}\pi \right )=-\sqrt{3}\\ &\tan \left (2x-\displaystyle \frac{1}{3}\pi \right )=-\frac{\sqrt{3}}{3}\\ &(\textrm{kuadran IV, karena Y negatif, X positif})\\ &\tan \left (2x-\displaystyle \frac{1}{3}\pi \right )=-\tan \displaystyle \frac{1}{6}\pi ,\: \: \textbf{menjadi}\\ &\tan \left (2x-\displaystyle \frac{1}{3}\pi \right )=\tan \left ( 2\pi -\displaystyle \frac{1}{6}\pi \right )=\tan \displaystyle \frac{11}{6}\pi \\ &\left (2x-\displaystyle \frac{1}{3}\pi \right )=\displaystyle \frac{11}{6}\pi\\ &\Leftrightarrow \: \: 2x=\displaystyle \frac{1}{3}\pi+\displaystyle \frac{11}{6}\pi +k.\pi =\displaystyle \frac{13}{6}\pi +k.\pi \\ &\Leftrightarrow \: \: x=\displaystyle \frac{13}{12}\pi +\displaystyle \frac{k.\pi}{2} \\ &k=0\Rightarrow x=\displaystyle \frac{13}{12}\pi=\displaystyle \frac{1}{12}\pi \: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x=\displaystyle \frac{13}{12}\pi +\displaystyle \frac{1}{2}\pi =\displaystyle \frac{19}{12}\pi=\displaystyle \frac{7}{12}\pi \quad (\color{blue}\textrm{mm})\\ &k=2\Rightarrow x=\displaystyle \frac{13}{12}\pi +\pi \quad \color{red}\textrm{tidak memenuhi} \end{aligned} \\ &\textbf{HP}=\color{red}\left \{\displaystyle \frac{1}{12}\pi,\: \displaystyle \frac{7}{12}\pi \right \} \end{array} \end{array}$.

$\begin{array}{ll}\\ 77.&\textrm{Salah satu nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &\textrm{persamaan}\: \: \cos x+\sin x=\displaystyle \frac{1}{2}\sqrt{6}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \frac{1}{24}\pi &&&\textrm{d}.&\displaystyle \frac{1}{8}\pi \\\\ \textrm{b}.&\displaystyle \frac{1}{15}\pi&\textrm{c}.&\displaystyle \color{red}\frac{1}{12}\pi&\textrm{e}.&\displaystyle \frac{1}{6}\pi \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\textrm{Diketahui bahwa}\\ &\sin x+\cos x=\displaystyle \frac{1}{2}\sqrt{6}\quad \left (\textbf{ingat}:a=1,\: b=1 \right )\\ &\sin x+\cos x=k\cos \left ( x-\theta \right )=\displaystyle \frac{1}{2}\sqrt{6}\\ &\begin{cases} k & =\sqrt{1^{2}+1^{2}}=\sqrt{2} \\ \tan & \theta =\displaystyle \frac{a}{b}=\displaystyle \frac{1}{1}=1\Rightarrow \theta =45^{\circ}=\displaystyle \frac{1}{4}\pi \end{cases}\\ &\qquad\quad\textrm{sudut}\: \: \theta \: \: \textrm{di kuadran I, karena}\: \: a,b>0\\ &\textrm{selanjutnya}\\ &\begin{aligned}&\sin x+\cos x=k\cos \left ( x-\theta \right )=\displaystyle \frac{1}{2}\sqrt{6}\\ &\Leftrightarrow \sqrt{2}\cos\left ( x-\displaystyle \frac{1}{4}\pi \right )=\displaystyle \frac{1}{2}\sqrt{6}\\ &\Leftrightarrow \: \: \, \cos \left ( x-\displaystyle \frac{1}{4}\pi \right )=\displaystyle \frac{\displaystyle \frac{1}{2}\sqrt{6}}{\sqrt{2}}=\displaystyle \frac{1}{2}\sqrt{3}\\ &\Leftrightarrow \: \: \,\cos \left ( x-\displaystyle \frac{1}{4}\pi \right )=\cos \displaystyle \frac{1}{6}\pi \\ &\Leftrightarrow \quad x-\displaystyle \frac{1}{4}\pi =\pm \displaystyle \frac{1}{6}\pi +k.2\pi \\ &\Leftrightarrow \quad x=\displaystyle \frac{1}{4}\pi\pm \displaystyle \frac{1}{6}\pi+k.2\pi \\ &\Leftrightarrow \quad x_{1}=\displaystyle \frac{5}{12}\pi+k.2\pi \: \: \textbf{atau}\\ &\: \: \: \quad\quad x_{2}=\displaystyle \frac{1}{12}\pi +k.2\pi \\ &k=0\Rightarrow x_{1}=\displaystyle \frac{5}{12}\pi\qquad (\color{blue}\textrm{memenuhi})\\ &\: \: \qquad\Rightarrow x_{2}=\displaystyle \color{red}\frac{1}{12}\pi\qquad \color{black}(\color{blue}\textrm{memenuhi})\\ &\textrm{Langkah berikutnya tidak diperlukan}\\ &\textrm{karena jawaban sudah kita dapatkan}\\ &\textrm{yaitu}:\: \: \color{red}\displaystyle \frac{1}{12}\pi \end{aligned} \end{array} \end{array}$.

$\begin{array}{ll}\\ 78.&\textrm{Himpunan penyelesaian persamaan}\\ &\qquad\: \: \cos x^{\circ}-\sqrt{3}\sin x^{\circ}=1\\ &\textrm{untuk}\: \: 0\leq x< 360\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \color{red}\left \{ 0,240 \right \} &&&\textrm{d}.&\displaystyle \left \{ 180,240 \right \} \\\\ \textrm{b}.&\displaystyle \left \{ 150,270 \right \}&\textrm{c}.&\displaystyle \left \{ 180,300 \right \}&\textrm{e}.&\displaystyle \left \{ 210,270 \right \} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\textrm{Diketahui dari soal bahwa}\\ &\cos x^{\circ}-\sqrt{3}\sin x^{\circ}=1,\\ &\textrm{lalu kita ubah posisinya menjadi}\\\\ &-\sqrt{3}\sin x+\cos x=1\: \: \left (\textbf{ingat}:a=-\sqrt{3},\: b=1 \right )\\ &-\sqrt{3}\sin x+\cos x=k\cos \left ( x-\theta \right )=1\\ &\begin{cases} k & =\sqrt{\left ( -\sqrt{3} \right )^{2}+\left ( 1 \right )^{2}}=\sqrt{4}=2 \\ \tan & \theta =\displaystyle \frac{a}{b}=\displaystyle \frac{-\sqrt{3}}{1}=-\sqrt{3}\Rightarrow \theta =300^{\circ} \end{cases}\\ &\quad\quad\textrm{sudut}\: \: \theta \: \: \textrm{di kuadran IV, karena}\: \: a<0,\: b>0\\ &\textrm{selanjutnya}\\ &\begin{aligned}&-\sqrt{3}\sin x+\cos x=k\cos \left ( x-\theta \right )=1\\ &\Leftrightarrow 2\cos\left ( x-300^{\circ} \right )=1\\ &\Leftrightarrow \: \: \, \cos \left ( x-300^{\circ} \right )=\displaystyle \frac{1}{2}\\ &\Leftrightarrow \: \: \,\cos \left ( x-300^{\circ} \right )=\cos 60^{\circ}\\ &\Leftrightarrow \quad x-300^{\circ} =\pm 60^{\circ}+k.360^{\circ}\\ &\Leftrightarrow \quad x=300^{\circ}\pm 60^{\circ}+k.360^{\circ}\\ &k=0\Rightarrow x_{1}=300^{\circ}+60^{\circ}=360^{\circ}=0^{\circ}\: \: (\color{blue}\textrm{mm})\\ & \qquad\qquad\quad\color{black}\textrm{atau}\\ &\qquad\qquad x_{2}=300^{\circ}-60^{\circ}=240^{\circ}\: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x=300^{\circ}\pm 60^{\circ}+360^{\circ}\quad (\color{red}\textrm{tm})\\ \end{aligned} \\ &\textbf{HP}=\left \{0^{\circ},240^{\circ} \right \} \end{array} \end{array}$

$\begin{array}{ll}\\ 79.&\textrm{Diketahui}\: \: \alpha -\beta =\displaystyle \frac{\pi }{3}\: \: \textrm{dan}\: \: \sin \alpha \sin \beta =\displaystyle \frac{1}{4}\\ &\textrm{dengan}\: \: \alpha \: \: \textrm{dan}\: \: \beta \: \: \textrm{adalah sudut} \: \: \textbf{lancip}\\ &\textrm{Nilai dari}\: \: \cos \left ( \alpha +\beta \right )\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&1&&&\textrm{d}.&\displaystyle \frac{1}{4}\\ \textrm{b}.&\displaystyle \frac{3}{4}&\textrm{c}.&\displaystyle \frac{1}{2}&\textrm{e}.&\color{red}\displaystyle 0 \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &\bullet \quad\alpha -\beta =\displaystyle \frac{\pi }{3}\: \: \textrm{dan}\: \: \sin \alpha \sin \beta =\displaystyle \frac{1}{4}\\ &\bullet \quad\textrm{dengan}\: \: \alpha \: \: \textrm{dan}\: \: \beta \: \: \textrm{sudut} \: \: \textbf{lancip}\\ &\qquad \textrm{akibatnya semua sudut dikuadran I}\\ &\qquad \textrm{sehingga}\color{purple}\begin{cases} \sin & =+ \\ \cos & =+ \\ \tan & =+ \end{cases}\\ &\textrm{ditanya}\: \: \cos \left ( \alpha +\beta \right ),\: \: \textrm{maka}\\ &\textrm{sebagai langkah awal kita adalah}:\\ &\cos \left ( \alpha -\beta \right )=\cos \left ( \displaystyle \frac{\pi }{3} \right )\\ &\Leftrightarrow \: \cos \alpha \cos \beta +\sin \alpha \sin \beta =\displaystyle \frac{1}{2}\\ &\Leftrightarrow \: \cos \alpha \cos \beta+\displaystyle \frac{1}{4} =\displaystyle \frac{1}{2}\\ &\Leftrightarrow \: \cos \alpha \cos \beta =\displaystyle \frac{1}{2}-\displaystyle \frac{1}{4}=\displaystyle \frac{1}{4}\\ &\textbf{Selanjutnya nilai dari}\\ &\cos \left ( \alpha +\beta \right )\\ &=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ &=\displaystyle \frac{1}{4}-\displaystyle \frac{1}{4}=0\\ \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 80.&\textrm{Nilai}\: \: \sin 75^{\circ}-\sin 165^{\circ}\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \frac{1}{4}\sqrt{2}&&&\textrm{d}.&\color{red}\displaystyle \frac{1}{2}\sqrt{2}\\\\ \textrm{b}.&\displaystyle \frac{1}{4}\sqrt{3}&\textrm{c}.&\displaystyle \frac{1}{4}\sqrt{6}&\textrm{e}.&\displaystyle \frac{1}{2}\sqrt{6} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\sin 75^{\circ}-\sin 165^{\circ}\\ &=2\cos \left ( \displaystyle \frac{75^{\circ}+165^{\circ}}{2} \right )\sin \left ( \displaystyle \frac{75^{\circ}-165^{\circ}}{2} \right )\\ &=2\cos \displaystyle \frac{240^{\circ}}{2}\sin \left (-\displaystyle \frac{90^{\circ}}{2} \right ) \\ &=2\cos 120^{\circ}\sin \left ( -45^{\circ} \right )\\ &=2\left (-\cos 60^{\circ} \right )\left (- \sin 45^{\circ} \right )\\ &=2\left ( -\displaystyle \frac{1}{2} \right )\left ( -\displaystyle \frac{1}{2}\sqrt{2} \right )\\ &=\color{red}\displaystyle \frac{1}{2}\sqrt{2} \end{aligned} \end{array}$.

Latihan Soal 7 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

 $\begin{array}{ll}\\ 61.&\textrm{Nilai}\: \: 105^{\circ}\: \: \textrm{jika dinyatakan ke radian}\\ &\textrm{adalah}\: \: ....\: \: \textrm{radian}\\\\ &\textrm{a}.\quad \displaystyle \frac{1}{3}\pi \\\\ &\textrm{b}.\quad \displaystyle \frac{5}{6}\pi \\\\ &\textrm{c}.\quad \displaystyle \frac{5}{12}\pi \\\\ &\textrm{d}.\quad \color{red}\displaystyle \frac{7}{12}\pi \\\\ &\textrm{e}.\quad \displaystyle \frac{9}{12}\pi \\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{Diketah}&\textrm{ui bahwa}\\ 180^{\circ}&=\pi \: \: \: radian\\ 1^{\circ}&=\displaystyle \frac{\pi }{180}\: \: \: radian\\ 105\times 1^{\circ}&=105\times \displaystyle \frac{\pi }{180}\: \: \: radian\\ 105^{\circ}&=\displaystyle \frac{7}{12}\pi \: \: \: radian \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 62.&\textrm{Nilai}\: \: \tan 240^{\circ} \: \: \: \textrm{adalah}\: \: ....\\\\ &\textrm{a}.\quad \displaystyle \color{red}\sqrt{3} \\\\ &\textrm{b}.\quad \displaystyle \frac{1}{3}\sqrt{3} \\\\ &\textrm{c}.\quad \displaystyle -\frac{1}{3}\sqrt{3} \\\\ &\textrm{d}.\quad \displaystyle \frac{1}{2}\sqrt{3} \\\\ &\textrm{e}.\quad \displaystyle -\sqrt{3} \\\\ &\textbf{Jawab}:\\ &\begin{aligned}\tan 240^{\circ}&=\tan \left ( 180^{\circ}+60^{\circ} \right )\\ &=\tan 60^{\circ}\\ &=\color{red}\sqrt{3}\\ \textbf{catatan}&: \textrm{ingat sudut berelasi} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 63.&\textrm{Perhatikanlah gambar berikut}\\ \end{array}$.

$.\qquad\begin{array}{ll}\\ &\textrm{Pada gambar di atas perbandingan}\\ &\sin \theta \: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad \sqrt{\displaystyle \frac{a^{2}-d^{2}}{f^{2}+g^{2}}} \\ &\textrm{b}.\quad \sqrt{\displaystyle \frac{a^{2}+b^{2}}{f^{2}+g^{2}}} \\ &\textrm{c}.\quad \sqrt{\displaystyle \frac{a^{2}-b^{2}}{f^{2}-g^{2}}} \\ &\textrm{d}.\quad \sqrt{\displaystyle \frac{a^{2}+b^{2}}{f^{2}-g^{2}}}\\ &\textrm{e}.\quad \color{red}\sqrt{\displaystyle \frac{a^{2}-b^{2}}{f^{2}+g^{2}}} \\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{Dari so}&\textrm{al diketahui bahwa}\\ \sin \theta &=\displaystyle \frac{c}{e}=\displaystyle \frac{\sqrt{a^{2}-b^{2}}}{\sqrt{f^{2}+g^{2}}}\\ &=\color{red}\sqrt{\displaystyle \frac{a^{2}-b^{2}}{f^{2}+g^{2}}} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 64.&\textrm{Nilai dari}\: \: \left ( \cos ^{2}17^{\circ}-\sin ^{2}73^{\circ} \right ) \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\color{red}0&&&\textrm{d}.&1\\ \textrm{b}.&\displaystyle \frac{1}{3}&\textrm{c}.&\displaystyle \frac{2}{\sqrt{3}}&\textrm{e}.&\displaystyle \frac{1}{2}\sqrt{3} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\left ( \cos ^{2}17^{\circ}-\sin ^{2}73^{\circ} \right )\\ &=\left ( \cos ^{2}17^{\circ}-\left (\sin 73^{\circ} \right )^{2} \right )\\ &=\left ( \cos ^{2}17^{\circ}-\left (\sin \left ( 90^{\circ}-17^{\circ} \right ) \right )^{2} \right )\\ &=\left ( \cos ^{2}17^{\circ}-\cos ^{2}17^{\circ} \right )\\ &=0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 65.&\textrm{Jika diketahui}\\ & \displaystyle \frac{x\csc ^{2}30^{\circ}\sec ^{2}45^{\circ}}{8\cos ^{2}45^{\circ}\sin ^{2}60^{\circ}}=\tan ^{2}60^{\circ}-\tan ^{2}30^{\circ},\\ & \textrm{maka nilai}\: \: x\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&-2&&&\textrm{d}.&\color{red}1\\ \textrm{b}.&\displaystyle -1&\textrm{c}.&0&\textrm{e}.&\displaystyle 2 \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{x\csc ^{2}30^{\circ}\sec ^{2}45^{\circ}}{8\cos ^{2}45^{\circ}\sin ^{2}60^{\circ}}=\tan ^{2}60^{\circ}-\tan ^{2}30^{\circ}\\ &\displaystyle \frac{x\left ( 4 \right )\left ( \displaystyle \frac{4}{2} \right )}{8\left ( \displaystyle \frac{2}{4} \right )\left ( \displaystyle \frac{3}{4} \right )}=3-\left ( \displaystyle \frac{1}{3} \right )\\ &\displaystyle \frac{8x}{3}=\displaystyle \frac{8}{3}\\ &x=1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 66.&\textrm{Nilai dari}\: \: \displaystyle \frac{\sin 49^{\circ}}{\cos 41^{\circ}}-\displaystyle \frac{\cos 17^{\circ}}{\sin 73^{\circ}}\\ &\textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&-1&&&\textrm{d}.&0,143\\ \textrm{b}.&\displaystyle -0,321&\textrm{c}.&\color{red}0&\textrm{e}.&\displaystyle 0,321 \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\sin 49^{\circ}}{\cos 41^{\circ}}-\displaystyle \frac{\cos 17^{\circ}}{\sin 73^{\circ}}\\ &=\displaystyle \frac{\sin 49^{\circ}}{\cos \left (90^{\circ}-49^{\circ} \right )}-\displaystyle \frac{\cos 17^{\circ}}{\sin \left (90^{\circ}-17^{\circ} \right )}\\ &=\displaystyle \frac{\sin 49^{\circ}}{\sin 49^{\circ}}-\displaystyle \frac{\cos 17^{\circ}}{\cos 17^{\circ}}\\ &=1-1\\ &=0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 67.&\textrm{Nilai dari}\\ &p=r\sin \alpha \cos \beta \\ &q=r\sin \alpha \sin \beta \\ &s=r\cos \alpha \\ &\textrm{maka pernyataan berikut yang}\\ &\textrm{tepat adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\color{red}p^{2}+t^{2}+s^{2}=r^{2}&&&\\ \textrm{b}.&p^{2}-t^{2}+s^{2}=r^{2} \\ \textrm{c}.&p^{2}+t^{2}-s^{2}=r^{2}&\\ \textrm{d}.&-p^{2}+t^{2}+s^{2}=r^{2}&\\ \textrm{e}.&-p^{2}-t^{2}+s^{2}=r^{2} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Saat}\\ &p^{2}+q^{2}\: \: \textrm{maka hasilnya adalah}\\ &\color{purple}\begin{array}{lll}\\ p^{2}&=r^{2}\sin^{2} \alpha \cos^{2} \beta&\\ q^{2}&=r^{2}\sin^{2} \alpha \sin^{2} \beta&+\\\hline &=r^{2}\sin ^{2}\alpha \left ( \cos ^{2}\beta +\sin ^{2}\beta \right )\\ &=r^{2}\sin ^{2}\alpha (1)\\ &=r^{2}\sin ^{2}\alpha \end{array}\\ &\textrm{Dan saat}\\ &p^{2}+q^{2}+s^{2}\: \: \textrm{akan diperoleh hasil}\\ &\color{purple}\begin{array}{lll}\\ p^{2}+q^{2}&=r^{2}\sin ^{2}\alpha &\\ \qquad s^{2}&=r^{2}\cos ^{2}\alpha &+\\\hline &=r^{2}\sin ^{2}\alpha+r^{2}\cos ^{2}\alpha\\ &=r^{2}\left ( \sin ^{2}\alpha+\cos ^{2}\alpha \right )\\ &=r^{2}(1)\\ &=r^{2} \end{array}\\ \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 68.&\textrm{Nilai dari}\\ & \displaystyle \frac{\cos \left ( 90^{\circ}+\theta \right )\sec \left ( 2\pi -\theta \right )\tan \left ( \pi -\theta \right )}{\sec \left ( \theta -2\pi \right )\sin \left ( 540^{\circ}+\theta \right )\cot \left ( \theta -90^{\circ} \right )}\\ &\textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&-1&&&\textrm{d}.&-\tan \theta \\ \textrm{b}.&\displaystyle 0&\textrm{c}.&\color{red}1&\textrm{e}.&\displaystyle \tan \theta \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\color{purple}\textrm{Ingat kembali sudut-sudut}\\ &\color{purple}\textrm{yang berelasi dari kudran selain I}\\ &\color{purple}\textrm{ke kuadran I beserta tandanya}\\ &\displaystyle \frac{\cos \left ( 90^{\circ}+\theta \right )\sec \left ( 2\pi -\theta \right )\tan \left ( \pi -\theta \right )}{\sec \left ( \theta -2\pi \right )\sin \left ( 540^{\circ}+\theta \right )\cot \left ( \theta -90^{\circ} \right )}\\ &=\displaystyle \frac{\left (-\sin \theta \right ) .\sec \theta .\left (-\tan \theta \right )}{\sec \theta .\left (-\sin \theta \right ). \left (-\tan \theta \right )}\\ &=1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 69.&\textrm{Diketahui bahwa}\\ &\sin \theta +\cos \theta =\displaystyle \frac{1}{2} \\ &\textrm{maka nilai dari}\\ &\sin ^{3}\theta +\cos ^{3}\theta \: \: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \frac{1}{2}&&&\textrm{d}.&\displaystyle \frac{5}{8} \\\\ \textrm{b}.& \displaystyle \frac{3}{4}&\textrm{c}.&\displaystyle \frac{9}{15}&\textrm{e}.&\color{red}\displaystyle \frac{11}{16} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &\sin \theta +\cos \theta =\displaystyle \frac{1}{2}\\ &\Leftrightarrow \: \left (\sin \theta +\cos \theta \right )^{2} =\displaystyle \frac{1}{4}\\ &\Leftrightarrow \: \sin ^{2}\theta +\cos ^{2}\theta +2\sin \theta \cos \theta =\displaystyle \frac{1}{4}\\ &\Leftrightarrow \: 1+2\sin \theta \cos \theta =\displaystyle \frac{1}{4}\\ &\Leftrightarrow \: 2\sin \theta \cos \theta =-\displaystyle \frac{3}{4}\\ &\Leftrightarrow \: \sin \theta \cos \theta =-\displaystyle \frac{3}{8}\\ &\textbf{Selanjutnya}\\ &\color{purple}\sin ^{3}\theta +\cos ^{3}\theta\\ &=\color{purple}\left ( \sin \theta +\cos \theta \right )\left ( \sin ^{2}\theta -\sin \theta \cos \theta +\cos ^{2}\theta \right )\\ &=\color{purple}\left ( \displaystyle \frac{1}{2} \right )\left ( 1-\left ( -\displaystyle \frac{3}{8} \right ) \right ) \\ &=\color{purple}\displaystyle \frac{1}{2}\times \displaystyle \frac{11}{8}\\ &=\color{red}\displaystyle \frac{11}{16} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 70.&\textrm{Jika diketahui}\: \: \: \displaystyle \frac{3}{2}\pi <x<2\pi \\ &\textrm{dan}\: \: \: \tan x=m,\\ &\textrm{maka nilai dari}\: \: \sin x \cos x \: \: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle -\frac{1}{m^{2}+1}&&&\textrm{d}.&\displaystyle -\frac{m}{m^{2}-1} \\\\ \textrm{b}.& \color{red}\displaystyle -\frac{m}{m^{2}+1}&\textrm{c}.&\displaystyle \frac{m}{m^{2}+1}&\textrm{e}.&\displaystyle \frac{m}{m^{2}-1} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\: \: \displaystyle \frac{3}{2}\pi <x<2\pi\\ &\textrm{ini daerah Kwadran IV, akibatnya adalah nilai}\\ &\begin{cases} \sin x & = -\\ \cos x & =+ \\ \tan x & =- \end{cases}\\ &\textbf{Selanjutnya ada pernyataan}\: \: \tan x=m\\ &\textrm{ini artinya}\: \: \tan x=\displaystyle \frac{m}{1}\\ &\textbf{Perhatikanlah ilustrasi gambar berikut} \end{aligned} \end{array}$.

$.\qquad\begin{aligned} &\textrm{maka nilai dari}\\ &\sin x\cos x\: \: \left (\textrm{ingat yang diminta di Kwadran IV} \right )\\ &=\left (-\displaystyle \frac{m}{\sqrt{m^{2}+1}} \right )\times \left (+\displaystyle \frac{1}{\sqrt{m^{2}+1}} \right )\\ &=\color{red}-\displaystyle \frac{m}{m^{2}+1} \end{aligned}$

Latihan Soal 6 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

 $\begin{array}{ll}\\ 51.&\textrm{Nilai dari}\: \: \displaystyle \frac{1}{\sec ^{2}A}+\frac{1}{\csc ^{2}A}=....\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\ \textrm{b}.&\displaystyle -1\\ \textrm{c}.&\displaystyle 0\\ \color{red}\textrm{d}.&\displaystyle 1\\ {e}.&\displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\displaystyle \frac{1}{\sec ^{2}A}+\frac{1}{\csc ^{2}A}\\ &=\cos ^{2}A+\sin ^{2}=\color{red}1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 52.&\textrm{Nilai dari}\: \: \displaystyle \frac{\tan B+\tan C}{\cot B+\cot C}=....\\ &\begin{array}{llll}\\ \textrm{a}.&\cot B\times \cot C \\ \color{red}\textrm{b}.&\displaystyle \tan B\times \tan C\\ \textrm{c}.&\displaystyle \sec B\times \csc C\\ \textrm{d}.&\displaystyle \tan B\times \cot C\\ {e}.&\displaystyle \tan B\times \csc C \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\displaystyle \frac{\tan B+\tan C}{\cot B+\cot C}\\ &=\displaystyle \frac{\tan B+\tan C}{\displaystyle \frac{1}{\tan B}+\frac{1}{\tan C}}\\ &=\displaystyle \frac{\tan B+\tan C}{\left ( \displaystyle \frac{\tan B+\tan C}{\tan B\times \tan C} \right )}\\ &=\color{red}\tan B\times \tan C \end{aligned} \end{array}$

$\begin{array}{ll}\\ 53.&\textrm{Nilai dari}\\ &\displaystyle \frac{\tan A}{\sec A-1}+\frac{\tan A}{\sec A+1}=....\\ &\begin{array}{llll}\\ \textrm{a}.&2\tan A \\ \textrm{b}.&2\cot A\\ \textrm{c}.&\displaystyle 2\sec A\\ \color{red}\textrm{d}.&\displaystyle 2\csc A\\ {e}.&\displaystyle 2\tan A.\sec A \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\displaystyle \frac{\tan A}{\sec A-1}+\frac{\tan A}{\sec A+1}\\ &=\tan A\left (\displaystyle \frac{1}{\displaystyle \frac{1}{\cos A}-1}+\frac{1}{\displaystyle \frac{1}{\cos A}+1} \right )\\ &=\displaystyle \frac{\sin A}{\cos A}\left ( \displaystyle \frac{\cos A}{1-\cos A}+\frac{\cos A}{1+\cos A} \right )\\ &=\displaystyle \frac{\sin A}{1-\cos A}+\frac{\sin A}{1+\cos A}\\ &=\displaystyle \frac{\sin A(1+\cos A)+\sin A(1-\cos A)}{(1-\cos A)(1+\cos A)}\\ &=\displaystyle \frac{2\sin A}{1-\cos ^{2}}\\ &=\displaystyle \frac{2\sin A}{\sin ^{2}A}\\ &=\displaystyle \frac{2}{\sin A}\\ &=\color{red}2\csc A \end{aligned}\\\\ &\textrm{Sebagai catatanya}\\ &\textrm{Anda bisa gunakan cara yang lain} \end{array}$

$\begin{array}{ll}\\ 54.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &\sin \left ( 2x-20^{\circ} \right )=-\cos \left ( 3x+50^{\circ} \right )\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-30^{\circ}\\ \textrm{b}.&-25^{\circ}\\ \color{red}\textrm{c}.&20^{\circ}\\ \textrm{d}.&25^{\circ}\\ {e}.&30^{\circ} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\sin \left ( 2x-20^{\circ} \right )&=-\cos \left ( 3x+50^{\circ} \right )\\ \sin \left ( 20^{\circ}-2x \right )&=\cos \left ( 3x+50^{\circ} \right )\\ \sin A&=\cos B,\: \: \color{black}\textrm{artinya}\\ A+B&=90^{\circ},\: \: \color{magenta}\textrm{maka}\\ \left ( 20^{\circ}-2x \right )+\left ( 3x+50^{\circ} \right )&=90^{\circ}\\ x+70^{\circ}&=90^{\circ}\\ x&=90^{\circ}-70^{\circ}\\ &=\color{red}20^{\circ} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 55.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &\tan \left ( 2x+60^{\circ} \right )=\cot \left ( 90^{\circ}-3x \right )\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&20^{\circ}\\ \textrm{b}.&30^{\circ}\\ \textrm{c}.&40^{\circ}\\ \textrm{d}.&50^{\circ}\\ \color{red}\textrm{e}.&60^{\circ} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}\tan \left ( 2x+60^{\circ} \right )&=\cot \left ( 90^{\circ}-3x \right )\\ \tan (2x+60^{\circ})&=\tan 3x\\ 2x+60^{\circ}&=3x\\ 2x-3x&=-60^{\circ}\\ -x&=-60^{\circ}\\ x&=\color{red}60^{\circ} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 56.&\textrm{Jika nilai}\: \: \cot A+\cos A=x\\ &\textrm{dan}\: \: \cot A-\cos A=y,\\ &\textrm{maka nilai}\: \: \left ( x^{2}-y^{2} \right )=\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&4\sqrt{xy}\\ \textrm{b}.&2\sqrt{xy}\\ \textrm{c}.&xy\\ \textrm{d}.&2\\ \textrm{e}.&4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}xy&=\left (\cot A+\cos A \right )\left ( \cot A-\cos A \right )\\ &=\cot ^{2}A-\cos ^{2}A\\ &=\displaystyle \frac{\cos^{2}A }{\sin ^{2}A}-\cos ^{2}A\\ &=\displaystyle \frac{\cos^{2}A }{\sin ^{2}A}-\cos ^{2}A\times \frac{\sin ^{2}A}{\sin ^{2}A}\\ &=\displaystyle \frac{\cos ^{2}A}{\sin ^{2}A}\left ( 1-\sin ^{2}A \right )\\ &=\displaystyle \frac{\cos ^{4}A}{\sin ^{2}A}\\ \sqrt{xy}&=\displaystyle \frac{\cos A}{\sin A}\times \cos A\\ \color{black}\textrm{Se}&\color{black}\textrm{lanjutnya}\\ x^{2}-y^{2}&=\left (\cot A+\cos A \right )^{2}-\left ( \cot A-\cos A \right )^{2}\\ (x+y)&(x-y)=(\cot A+\cos A+\cot A-\cos A)\\ &\qquad\times (\cot A+\cos A-(\cot A-\cos A))\\ x^{2}-y^{2}&=2\cot A\times 2\cos A\\ &=4\times \displaystyle \frac{\cos A}{\sin A}\times \cos A\\ &=\color{red}4\sqrt{xy} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 57.&\textrm{Jika nilai}\: \: \cos A+\sin A=\sqrt{2}\cos A\\ &\textrm{maka nilai}\: \: \left ( \cos A-\sin A \right )=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-\sqrt{2}\cos A\\ \textrm{b}.&-\sqrt{2}\sin A\\ \color{red}\textrm{c}.&\sqrt{2}\sin A\\ \textrm{d}.&\displaystyle \frac{1}{\sqrt{2}}\sec A\\ \textrm{e}.&\displaystyle \frac{1}{\sqrt{2}}\csc A \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\cos A+\sin A&=\sqrt{2}\cos A\\ \left (\cos A+\sin A \right )^{2}&=\left (\sqrt{2}\cos A \right )^{2}\\ 1+2\sin A\cos A&=2\cos ^{2}A\\ 2\sin A\cos A&=2\cos ^{2}A-1\\ \textrm{maka}&\\ \left (\cos A-\sin A \right )^{2}&=1-2\sin A\cos A\\ &=1-\left ( 2\cos ^{2}A-1 \right )\\ &=2-2\cos ^{2}A\\ &=2\left ( 1-\cos ^{2}A \right )\\ &=2\sin ^{2}A\\ \cos A-\sin A&=\sqrt{2\sin ^{2}A}\\ &=\sin A\sqrt{2}\\ &=\color{red}\sqrt{2}.\sin A \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 58.&\textrm{Nilai}\: \: \cos \gamma \left ( \csc \gamma +\tan \gamma \right ) \\ & \textrm{adalah ekivalen dengan}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\color{red}\cot \gamma +\sin \gamma \\ \textrm{b}.&\tan \gamma +\cos \gamma \\ \textrm{c}.&\cot \gamma -\sin \gamma \\ \textrm{d}.&\tan \gamma -\cos \gamma \\ \textrm{e}.&\cot \gamma -\tan \gamma \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\cos \gamma \left ( \csc \gamma +\tan \gamma \right )&=\cos \gamma \left ( \displaystyle \frac{1}{\sin \gamma } +\displaystyle \frac{\sin \gamma }{\cos \gamma } \right )\\ &=\cos \gamma \left ( \displaystyle \frac{\cos \gamma +\sin ^{2}\gamma }{\sin \gamma \cos \gamma } \right )\\ &=\displaystyle \frac{\cos \gamma +\sin ^{2}\gamma }{\sin \gamma }\\ &=\displaystyle \frac{\cos \gamma }{\sin \gamma }+\displaystyle \frac{\sin ^{2}\gamma }{\sin \gamma }\\ &=\color{red}\cot \gamma +\sin \gamma \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 59.&\textrm{Jika diketahui}\: \: \sin \theta \cos \theta =\displaystyle \frac{3}{8},\\ &\textrm{maka nilai}\: \: \displaystyle \frac{1}{\sin \theta }-\displaystyle \frac{1}{\cos \theta }\: \: \textrm{adalah}=....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\displaystyle \frac{1}{4}\\ \textrm{b}.&\displaystyle \frac{1}{2}\\ \textrm{c}.&\displaystyle \frac{3}{4} \\ \textrm{d}.&\color{red}\displaystyle \frac{4}{3}\\ \textrm{e}.&\displaystyle 4 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\displaystyle \frac{1}{\sin \theta }-\displaystyle \frac{1}{\cos \theta }\\ &=\sqrt{\left ( \displaystyle \frac{1}{\sin \theta }-\displaystyle \frac{1}{\cos \theta } \right )^{2}}\\ &=\sqrt{\displaystyle \frac{1}{\sin ^{2}\theta }-\displaystyle \frac{2}{\sin \theta \cos \theta }+\displaystyle \frac{1}{\cos ^{2}\theta }}\\ &=\sqrt{\displaystyle \frac{\cos ^{2}\theta +\sin ^{2}\theta }{\sin ^{2}\theta \cos ^{2}\theta }-\displaystyle \frac{2}{\sin \theta \cos \theta }}\\ &=\sqrt{\displaystyle \frac{1}{\left ( \displaystyle \frac{3}{8} \right )^{2}}-\displaystyle \frac{2}{\left ( \displaystyle \frac{3}{8} \right )}}=\sqrt{\displaystyle \frac{1}{\displaystyle \frac{9}{64}}-\displaystyle \frac{2}{\displaystyle \frac{3}{8}}}=\sqrt{\displaystyle \frac{64}{9}-\displaystyle \frac{16}{3}}\\ &=\sqrt{\displaystyle \frac{64}{9}-\displaystyle \frac{48}{9}}=\sqrt{\displaystyle \frac{16}{9}}=\color{red}\displaystyle \frac{4}{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 60.&\textrm{Bentuk}\: \: \displaystyle \frac{\left ( \sin \alpha +\sin \beta \right )\left ( \sin \alpha -\sin \beta \right )}{\left ( \cos \alpha \cos \beta \right )^{2}}\\ & \textrm{ekivalen dengan}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\tan \alpha -\tan \beta \\ \textrm{b}.&\tan \alpha +\tan \beta \\ \textrm{c}.&\color{red}\tan ^{2}\alpha -\tan ^{2}\beta \\ \textrm{d}.&\tan ^{2}\alpha +\tan ^{2}\beta \\ \textrm{e}.&\tan ^{3}\alpha -\tan ^{3}\beta \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\displaystyle \frac{\left ( \sin \alpha +\sin \beta \right )\left ( \sin \alpha -\sin \beta \right )}{\left ( \cos \alpha \cos \beta \right )^{2}}\\ &=\displaystyle \frac{\sin ^{2}\alpha -\sin ^{2}\beta }{\left ( \cos \alpha \cos \beta \right )^{2}}\\ &=\displaystyle \frac{\sin ^{2}\alpha -\sin ^{2}\alpha \sin ^{2}\beta -\sin ^{2}\beta +\sin ^{2}\alpha \sin ^{2}\beta}{\left ( \cos \alpha \cos \beta \right )^{2}}\\ &=\displaystyle \frac{\sin ^{2}\alpha\left ( 1-\sin ^{2}\beta \right ) -\sin ^{2}\beta \left ( 1-\sin ^{2}\alpha \right )}{\left ( \cos \alpha \cos \beta \right )^{2}}\\ &=\displaystyle \frac{\sin ^{2}\alpha \cos ^{2}\beta -\sin ^{2}\beta \cos ^{2}\alpha }{\left ( \cos \alpha \cos \beta \right )^{2}}\\ &=\displaystyle \frac{\sin ^{2}\alpha \cos ^{2}\beta }{\left ( \cos \alpha \cos \beta \right )^{2}}-\displaystyle \frac{\sin ^{2}\beta \cos ^{2}\alpha }{\left ( \cos \alpha \cos \beta \right )^{2}}\\ &=\displaystyle \frac{\sin ^{2}\alpha }{\cos ^{2}\alpha }-\displaystyle \frac{\sin ^{2}\beta }{\cos ^{2}\beta }\\ &=\color{red}\tan ^{2}\alpha -\tan ^{2}\beta \end{aligned} \end{array}$.

DAFTAR PUSTAKA

1. Sembiring, S., Zulkifli, M., Marsito, dan Rusdi, I. 2017. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SEWU.
2. Sukino. 2016. Matematika untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA

Latihan Soal 5 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

 $\begin{array}{ll}\\ 41.&\textrm{Jika}\: \: \tan^{2} x +\sec x =5 \: \: \textrm{untuk}\\ &0\leq x\leq \displaystyle \frac{\pi }{2}\: \: \textrm{maka nilai} \: \: \cos x=....\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \color{red}\textrm{b}.&\displaystyle \frac{1}{2}\\ \textrm{c}.&\displaystyle \frac{1}{3}\\ \textrm{d}.&\displaystyle \frac{1}{\sqrt{2}}\\ \textrm{e}.&\displaystyle \frac{1}{2}\sqrt{3} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\textrm{Ingat bahwa}&\: \: 0\leq x\leq \displaystyle \frac{\pi }{2}\\ \textrm{berarti sudu}&\textrm{t}\: \: x\: \: \textrm{berada di kuadran I}\\ \textrm{sehingga ak}&\textrm{an menyebabkan nilai}\\ & \color{magenta}\cos x=+\\ \color{black}\textrm{Selanjutnya}&\\ \tan^{2} x +\sec x &=5\\ \sec ^{2}x-1+\sec x&=5\\ \sec ^{2}x+\sec x-6&=0\\ (\sec x+3)(\sec x-2)&=0\\ \sec x=-3\: \: \textrm{atau}&\sec x=2\\ \textrm{untuk}\: \: \sec x&=-3\: \: (\color{red}\textrm{tidak memenuhi})\\ \textrm{untuk}\: \: \sec x&=2\: \: (\color{magenta}\textrm{memenuhi})\\ \color{black}\textrm{Selanjutnya}&\: \color{black}\textrm{lagi}\\ \sec x&=2\\ \displaystyle \frac{1}{\cos x}&=2\\ \cos x&=\color{red}\displaystyle \frac{1}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 42.&\textrm{Himpunan penyelesaian persamaan}\\ &3\cos 2x+5\sin x+1=0\: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 2\pi \\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\left \{ \displaystyle \frac{7}{6}\pi ,\frac{11}{6}\pi \right \}\\ \textrm{b}.&\left \{ \displaystyle \frac{5}{6}\pi ,\frac{11}{6}\pi \right \}\\ \textrm{c}.&\left \{ \displaystyle \frac{1}{6}\pi ,\frac{7}{6}\pi \right \}\\ \textrm{d}.&\left \{ \displaystyle \frac{1}{5}\pi ,\frac{5}{6}\pi \right \}\\ \textrm{e}.&\left \{ \displaystyle \frac{5}{6}\pi ,\frac{7}{6}\pi \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}3\cos 2x+5\sin x+1&=0\\ 3\left ( 1-2\sin ^{2}x \right )+5\sin x+1&=0\\ -6\sin ^{2}+5\sin x+4&=0\\ 6\sin ^{2}x-5\sin x-4&=0\\ \left ( 3\sin x-4 \right )\left ( 2\sin x+1 \right )&=0\\ \sin x=\displaystyle \frac{4}{3}\: \: \textrm{atau}\: \: \sin x&=-\frac{1}{2}\\ \sin x&=\sin 150^{\circ}=\frac{5}{6}\pi \\ x&=\begin{cases} \displaystyle \frac{7}{6}\pi &+k.2\pi \\ \pi -\displaystyle \frac{7}{6}\pi & +k.2\pi \end{cases}\\ \textrm{saat}\: \: k&=0\\ x_{1}&=\displaystyle \frac{7}{6}\pi \\ x_{2}&=-\displaystyle \frac{1}{6}\pi \\ \textrm{saat}\: \: k&=1\\ x_{1}&=\color{red}\displaystyle \frac{7}{6}\pi +2\pi \\ x_{2}&=-\displaystyle \frac{1}{6}\pi +2\pi =\color{red}\frac{11}{6}\pi \end{aligned} \end{array}$

$\begin{array}{ll}\\ 43.&\textrm{Himpunan penyelesaian dari}\\ &\sqrt{3}\sin 2x+2\cos ^{2}x=-1\: \: \textrm{untuk}\\ &0^{\circ}\leq x\leq 360^{\circ}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ 240^{\circ},300^{\circ} \right \}\\ \textrm{b}.&\left \{ 30^{\circ},60^{\circ} \right \}\\ \textrm{c}.&\left \{ 150^{\circ},315^{\circ} \right \}\\ \color{red}\textrm{d}.&\left \{ 120^{\circ},300^{\circ} \right \}\\ \textrm{e}.&\left \{ 60^{\circ},150^{\circ} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\sqrt{3}\sin 2x+2\cos ^{2}x&=-1\\ \sqrt{3}\sin 2x+1+\cos 2x&=-1\\ \sqrt{3}\sin 2x+\cos 2x&=-2\\ \sqrt{\sqrt{3}^{2}+1^{2}}\cos \left ( 2x-\alpha \right )&=-2\\ a=x=1,\: \: b&=y=\sqrt{3}\\ \alpha &=\arctan \displaystyle \frac{b}{a}=\arctan \frac{\sqrt{3}}{1}\\ \alpha &=60^{\circ}\\ \textrm{maka persamaan akan}&\: \textrm{menjadi}\\ 2\cos \left ( 2x-60^{\circ} \right )&=-2\\ \cos \left ( 2x-60^{\circ} \right )&=-1\\ \cos \left ( 2x-60^{\circ} \right )&=\cos 180^{\circ}\\ \left ( 2x-60^{\circ} \right )&=\pm 180^{\circ}+k.360^{\circ}\\ 2x&=60^{\circ}\pm 180^{\circ}+k.360^{\circ}\\ x&=30^{\circ}\pm 90^{\circ}+k.180^{\circ}\\ \textrm{saat}\: \: k&=0\\ x_{1}&=120^{\circ}\\ x_{2}&=-60^{\circ}\\ \textrm{saat}\: \: k&=1\\ x_{3}&=120^{\circ}+360^{\circ}=....\\ x_{2}&=-60^{\circ}+360^{\circ}=\color{red}300^{\circ} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 44.&\textrm{Jika}\: \: 3\sin \theta +4\cos \theta =5 \: \: \textrm{maka}\\ &\textrm{nilai dari} \: \: \sin \theta \: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&0,3\\ \color{red}\textrm{b}.&0,60\\ \textrm{c}.&0,75\\ \textrm{d}.&0,80\\ \textrm{e}.&1,20 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}3\sin \theta +4\cos \theta &=5\\ k\cos \left ( \theta -\alpha \right )&=5\\ a=x=4,\: &b=y=3\\ \theta &=\arctan \displaystyle \frac{b}{a}\\ \theta &=\arctan \displaystyle \frac{3}{4}\\ \color{black}\textrm{atau}\: \: &\\ \tan \theta &=\displaystyle \frac{3}{4}\\ \color{black}\textrm{maka}\: \: &\\ \sin \theta &=\displaystyle \frac{3}{\sqrt{3^{2}+4^{2}}}\\ &=\displaystyle \frac{3}{5}\\ &=\color{red}0,6 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 45.&\textrm{Jika}\: \: \tan \theta +\sec \theta =x \: \: \textrm{maka}\\ &\textrm{nilai dari} \: \: \tan \theta \: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{2x}{x^{2}-1}\\ \textrm{b}.&\displaystyle \frac{2x}{x^{2}+1}\\ \textrm{c}.&\displaystyle \frac{x^{2}+1}{2x}\\ \color{red}\textrm{d}.&\displaystyle \frac{x^{2}-1}{2x}\\ \textrm{e}.&\displaystyle \frac{x^{2}-1}{x^{2}+1} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\color{black}\textrm{Langkah}&\: \: 1\\ \tan \theta +\sec \theta &=x\\ \displaystyle \frac{\sin \theta }{\cos \theta } +\frac{1}{\cos \theta }&=x\\ \displaystyle \frac{\sin \theta +1}{\cos \theta }&=x\\ \sin \theta +1&=x\cos \theta ...........1\\ \color{black}\textrm{Langkah}&\: \: 2\\ \left (\tan \theta +\sec \theta \right )^{2}&=x^{2}\\ \tan ^{2}\theta +2\tan \theta \sec \theta +\sec ^{2}\theta &=x^{2}\\ \sec ^{2}\theta -1+2\tan \theta \sec \theta &+\sec ^{2}\theta =x^{2}\\ 2\sec ^{2}\theta +2\tan \theta \sec \theta &=x^{2}+1\\ \displaystyle \frac{2}{\cos ^{2}\theta }+\frac{2\sin \theta }{\cos ^{2}\theta }&=x^{2}+1\\ 1+\sin \theta &=\left (\displaystyle \frac{x^{2}+1}{2} \right )\cos ^{2}\theta.....2\\ \color{black}\textrm{Langkah}&\: \: 3\\ \left (\displaystyle \frac{x^{2}+1}{2} \right )\cos ^{2}&=x\cos \theta \\ \cos \theta &=\displaystyle \frac{2x}{x^{2}+1}\\ \textrm{maka}\: \: &(\color{blue}\textbf{dengan sisi segitiga})\\ \tan \theta =\displaystyle \frac{\sqrt{\left ( x^{2}+1 \right )^{2}-(2x)^{2}}}{2x}&=\displaystyle \frac{\sqrt{x^{4}+2x^{2}+1-4x^{2}}}{2x}\\ \tan \theta &=\displaystyle \frac{\sqrt{x^{4}-2x^{2}+1}}{2x}\\ &=\displaystyle \frac{\sqrt{\left ( x^{2}-1 \right )^{2}}}{2x}\\ &=\color{red}\displaystyle \frac{x^{2}-1}{2x} \end{aligned} \end{array}$

$\begin{aligned}.\: \: \qquad \textbf{Sehingga}&\: \textbf{dari kasus di atas didapatkan}\\ \color{red}\sin \theta &=\displaystyle \frac{x^{2}-1}{x^{2}+1}\\ \color{red}\cos \theta &=\displaystyle \frac{2x}{x^{2}+1}\\ \color{red}\tan \theta &=\displaystyle \frac{x^{2}-1}{2x} \end{aligned}$

$\begin{array}{ll}\\ 46.&\textrm{Jika}\: \: \sec x +\tan x =\displaystyle \frac{3}{2} \: \: \textrm{untuk}\\ &0\leq x\leq \displaystyle \frac{\pi }{2}\: \: \textrm{maka nilai} \: \: \sin x=....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{5}{13}\\ \textrm{b}.&\displaystyle \frac{12}{13}\\ \textrm{c}.&\displaystyle 1\\ \textrm{d}.&\displaystyle \frac{2}{13}\\ \textrm{e}.&\displaystyle \frac{5}{12} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\sec x +\tan x &=\displaystyle \frac{3}{2}\\ \textrm{ingat saat}&\: \color{black}\textrm{mengerjakan soal no}.14\\ \sec \theta +\tan \theta &=x\\ \sin \theta &=\displaystyle \frac{x^{2}-1}{x^{2}+1},\: \: \color{black}\textrm{maka}\\ \sin x&=\color{red}\displaystyle \frac{\left ( \displaystyle \frac{3}{2} \right )^{2}-1}{\left (\displaystyle \frac{3}{2} \right )^{2}+1}\\ &=\displaystyle \frac{\displaystyle \frac{9}{4}-1}{\displaystyle \frac{9}{4}+1}\\ &=\frac{\displaystyle \frac{5}{4}}{\displaystyle \frac{13}{4}}\\ &=\color{red}\displaystyle \frac{5}{13} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 47.&\textrm{Nilai}\\ &\left ( \sin A+\cos A \right )^{2}+\left ( \sin A-\cos A \right )^{2}=....\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \color{red}\textrm{b}.&\displaystyle 2\\ \textrm{c}.&\displaystyle 3\\ \textrm{d}.&\displaystyle 3\cos A\\ {e}.&\displaystyle 4\sin A \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\left ( \sin A+\cos A \right )^{2}+\left ( \sin A-\cos A \right )^{2}\\ &=\sin ^{2}A+2\sin A\cos A+\cos ^{2}A\\ &\quad +\sin ^{2}A-2\sin A\cos A+\cos ^{2}A\\ &=1+1\\ &=\color{red}2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 48.&\textrm{Nilai}\: \: \sqrt{\displaystyle \frac{1+\sin A}{1-\sin A}}=....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\sec A+\tan A\\ \textrm{b}.&\sec ^{2}A+\tan ^{2}A\\ \textrm{c}.&\sec ^{2}A-\tan ^{2}A\\ \textrm{d}.&\tan ^{2}A-\sec ^{2}A\\ {e}.&\sec A\times \tan A \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\sqrt{\displaystyle \frac{1+\sin A}{1-\sin A}}\\ &=\sqrt{\displaystyle \frac{1+\sin A}{1-\sin A}\times \frac{1+\sin A}{1+\sin A}}\\ &=\sqrt{\displaystyle \frac{\left ( 1+\sin A \right )^{2}}{1-\sin ^{2}A}}\\ &=\sqrt{\displaystyle \frac{\left ( 1+\sin A \right )^{2}}{\cos ^{2}A}}\\ &=\displaystyle \frac{1+\sin A}{\cos A}\\ &=\displaystyle \frac{1}{\cos A}+\frac{\sin A}{\cos A}\\ &=\color{red}\sec A+\tan A \end{aligned} \end{array}$

$\begin{array}{ll}\\ 49.&\textrm{Jika}\: \: 0^{\circ}\leq \theta \leqslant 90^{\circ},\: \textrm{maka nilai}\\ &\left ( \displaystyle \frac{5\cos \theta -4}{3-5\sin \theta }-\frac{3+5\sin \theta }{4+5\cos \theta } \right )=....\\ &\begin{array}{llll}\\ \textrm{a}.&-1\\ \color{red}\textrm{b}.&\displaystyle 0\\ \textrm{c}.&\displaystyle \frac{1}{4}\\ \textrm{d}.&\displaystyle \frac{1}{2}\\ {e}.&\displaystyle 1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\left ( \displaystyle \frac{5\cos \theta -4}{3-5\sin \theta }-\frac{3+5\sin \theta }{4+5\cos \theta } \right )\\ &=\left ( \displaystyle \frac{5\cos \theta -4}{3-5\sin \theta }\times \frac{4+5\cos \theta }{4+5\cos \theta } \right )\\ &\qquad-\left ( \displaystyle \frac{3+5\sin \theta }{4+5\cos \theta }\times \frac{3-5\sin \theta }{3-5\sin \theta } \right )\\ &=\displaystyle \frac{25\cos ^{2}-16-\left ( 9-25\sin ^{2}\theta \right )}{(3-5\sin \theta )(4+5\cos \theta )}\\ &=\displaystyle \frac{25\left ( \sin ^{2}\theta +\cos ^{2}\theta \right )-25}{(3-5\sin \theta )(4+5\cos \theta )}\\ &=\displaystyle \frac{0}{(3-5\sin \theta )(4+5\cos \theta )}\\ &=\color{red}0 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 50.&\textrm{Nilai}\\ &\left ( 1+\cot \theta -\csc \theta \right )\left ( 1+\tan \theta +\sec \theta \right )=....\\ &\begin{array}{llll}\\ \textrm{a}.&-2\\ \textrm{b}.&\displaystyle -1\\ \textrm{c}.&\displaystyle 0\\ \textrm{d}.&\displaystyle 1\\ \color{red}\textrm{e}.&\displaystyle 2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\left ( 1+\cot \theta -\csc \theta \right )\left ( 1+\tan \theta +\sec \theta \right )\\ &=\left ( 1+\displaystyle \frac{\cos \theta }{\sin \theta } -\frac{1}{\sin \theta } \right )\left ( 1+\frac{\sin \theta }{\cos \theta } +\frac{1}{\cos \theta } \right )\\ &=\left ( \displaystyle \frac{\sin \theta +\cos \theta -1}{\sin \theta } \right )\left ( \displaystyle \frac{\cos \theta +\sin \theta +1}{\cos \theta } \right )\\ &=\displaystyle \frac{\left (\sin \theta +\cos \theta \right )^{2}-1}{\sin \theta \cos \theta }\\ &=\displaystyle \frac{1+2\sin \theta \cos \theta -1}{\sin \theta \cos \theta }\\ &=\color{red}2 \end{aligned} \end{array}$

Latihan Soal 4 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

 $\begin{array}{ll}\\ 31.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &2\cos ^{2}x+\cos x-1=0\: \: \textrm{untuk}\: \: 0\leq x\leq \pi \\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{1}{3}\pi \: \: \textrm{dan}\: \: \pi \\ \textrm{b}.&\displaystyle \frac{1}{3}\pi \: \: \textrm{dan}\: \: \frac{2}{3}\pi \\ \textrm{c}.&\displaystyle \frac{1}{3}\pi \: \: \textrm{dan}\: \: \frac{3}{4}\pi \\ \textrm{d}.&\displaystyle \frac{1}{4}\pi \: \: \textrm{dan}\: \: \frac{3}{4}\pi \\ \textrm{e}.&\displaystyle \frac{1}{4}\pi \: \: \textrm{dan}\: \: \frac{2}{3}\pi \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}2\cos ^{2}x+\cos x-1&=0\\ \left (2\cos x-1 \right )\left (\cos x+1 \right )&=0\\ \cos x=\displaystyle \frac{1}{2}\: \: \color{black}\textrm{atau}\: \: &\cos x=-1\\ \cos x=\color{red}\cos 60^{\circ}=\cos \frac{1}{3}\pi \: \: &\\ \color{black}\textrm{atau}\: \: \cos x=\color{red}\cos 180^{\circ}&=\cos \pi \\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 32.&\textrm{Untuk}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &\tan ^{2}x-\tan x-6=0\: \: \textrm{pada}\: \: 0\leq x\leq \pi ,\\ &\textrm{maka himpunan nilai}\: \: \sin x\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\left \{ \displaystyle \frac{3\sqrt{10}}{10},\frac{2\sqrt{5}}{5} \right \}\\ \textrm{b}.&\left \{ \displaystyle \frac{3\sqrt{10}}{10},-\frac{2\sqrt{5}}{5} \right \} \\ \textrm{c}.&\left \{ -\displaystyle \frac{3\sqrt{10}}{10},\frac{2\sqrt{5}}{5} \right \} \\ \textrm{d}.&\left \{ \displaystyle \frac{\sqrt{10}}{10},\frac{\sqrt{5}}{5} \right \} \\ \textrm{e}.&\left \{ \displaystyle \frac{\sqrt{10}}{10},\frac{2\sqrt{5}}{5} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\tan ^{2}x-\tan x-6&=0\\ \left (\tan x-3 \right )\left (\tan x+2 \right )&=0\\ \tan x=3\: \: \textrm{atau}\: \: &\tan x=-2\\ \tan x=\displaystyle \frac{3}{1}\: \: \textrm{atau}\: \: &\tan x=\frac{-2}{1}\\ \sin x=\displaystyle \frac{3}{\sqrt{1^{2}+3^{2}}}\: \: \textrm{atau}\: \: &\sin x=\frac{2}{\sqrt{1^{2}+2^{2}}}\\ \sin x=\displaystyle \frac{3}{\sqrt{10}}\: \: \textrm{atau}\: \: &\sin x=\frac{2}{\sqrt{5}}\\ \sin x=\color{red}\displaystyle \frac{3}{10}\sqrt{10}\: \: \color{black}\textrm{atau}\: \: &\sin x=\color{red}\frac{2}{5}\sqrt{5} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 33.&\textrm{Jika}\quad 2\sin ^{2}x+3\cos x=0\\ &\textrm{dan}\quad 0^{0}\leq x\leq 180^{0},\\ &\textrm{maka nilai \textit{x} adalah} ....\\ &\begin{array}{llllll}\\ \textrm{a}.&30^{0}\\ \textrm{b}.&60^{0}\\ \textrm{c}.&\color{red}120^{0}\\ \textrm{d}.&150^{0}\\ \textrm{e}.&170^{0}\\ \end{array}\\\\ &\textbf{Jawab}:\quad\color{red}\textbf{c}\\ &\begin{aligned}2&\sin ^{2}x+3\cos x=0,\\ & \textrm{ingat identitas}\quad \sin ^{2}x+\cos ^{2}x=1\\ 2&\left ( 1-\cos ^{2}x \right )+3\cos x=0\\ 2&-2\cos ^{2}x+3\cos x=0,\\ & \textrm{(dikali dengan -1)} \\ 2&\cos ^{2}x-\cos x-2=0,\\ & \textrm{menjadi persamaan kuadrat dalam}\: \: \cos x\\ &\left ( \cos x-2 \right )\left ( 2\cos x+1 \right )=0,\qquad (\textrm{difaktorkan})\\ &\cos x-2=0\quad \textrm{atau}\quad 2\cos x+1=0\\ &\underset{\textrm{tidak memenuhi}}{\underbrace{\cos x=2}}\quad \textrm{atau}\quad \underset{\textrm{memenuhi}}{\underbrace{\cos x=-\displaystyle \frac{1}{2}}},\\ & \textrm{ingat rentang nilai cosinus adalah}:\left | \cos x \right |\leq 1\\ &\textrm{Selanjutnya pilih yang memenuhi, yaitu}\\ &\cos x=-\displaystyle \frac{1}{2}\\ &\cos x=\cos \left ( 180^{0}-60^{0} \right )\\ &\cos x=\cos 120^{0}\\ &\therefore \quad x=\color{red}120^{0} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 34.&\textrm{Akar-akar dari persamaan}\\ & 4\sin ^{2}x+4\cos x=1 \quad \textrm{pada selang}\\ & -\pi \leq x\leq \pi \quad \textrm{adalah} ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle \frac{3}{2}\: \: \textrm{atau}\: \: -\frac{1}{2}\\ \textrm{b}.&-\displaystyle \frac{3}{2}\: \: \textrm{atau}\: \: \frac{1}{2}\\ \textrm{c}.&\displaystyle \frac{3}{2}\pi \: \: \textrm{atau}\: \: -\frac{1}{2}\pi \\ \textrm{d}.&\displaystyle \frac{1}{3}\pi \: \: \textrm{atau}\: \: -\frac{1}{3}\pi \\ \textrm{e}.&\color{red}\displaystyle -\frac{2}{3}\pi \: \: \textrm{atau}\: \: \frac{2}{3}\pi \\ \end{array}\\\\ &\textbf{Jawab}:\quad\color{red}\textbf{e}\\ &\begin{aligned}4&\sin ^{2}x+4\cos x=1\\ 4&\left ( 1-\cos ^{2}x \right )+4\cos x-1=0\\ 4&-4\cos ^{2}x+4\cos x-1=0\\ 4&\cos ^{2}x-4\cos x-3=0\\ &\left ( 2\cos x-3 \right )\left ( 2\cos x+1 \right )=0\\ &\left ( 2\cos x-3 \right )=0\quad \textrm{atau}\quad \left ( 2\cos x+1 \right )=0\\ &\underset{\textrm{tidak memenuhi}}{\underbrace{\cos x=\displaystyle \frac{3}{2}}}\quad \textrm{atau} \quad\underset{\textrm{memenuhi}}{\underbrace{\cos x=-\displaystyle \frac{1}{2}}}\\ &\textrm{pilih yang memenuhi persamaan, yaitu}\\ &\cos x=-\displaystyle \frac{1}{2}\\ &\cos x=\cos \left ( 180^{0}-60^{0} \right )\\ &\cos x=\cos 120^{0}\\ &\therefore \quad x=120^{0}=\color{red}\displaystyle \frac{2}{3}\pi \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 35.&\textrm{Himpunan penyelesaian persamaan}\\ &\sin ^{2}2x+2\sin x\cos x -2=0\\ & \textrm{pada selang}\quad 0 \leq x\leq 360^{0} \quad \textrm{adalah} ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\left \{ 45^{0},135^{0} \right \}\\ \textrm{b}.&\color{red}\left \{ 45^{0},225^{0} \right \}\\ \textrm{c}.&\left \{ 135^{0},180^{0} \right \}\\ \textrm{d}.&\left \{ 135^{0},225^{0} \right \} \\ \textrm{e}.&\left \{ 135^{0},315^{0} \right \} \\ \end{array}\\\\ &\textbf{Jawab}:\quad\color{red}\textbf{b}\\ &\begin{aligned}&\sin ^{2}2x+2\sin x\cos x -2=0\\ &\sin ^{2}2x+\sin 2x-2=0,\\ & \textrm{(ingat bahwa)}:\: \sin 2x=2\sin x\cos x\\ &\left ( \sin 2x+2 \right )\left ( \sin 2x-1 \right )=0\\ &\sin 2x+2=0\quad \textrm{atau}\quad \sin 2x-1=0\\ &\underset{\textrm{tidak memenuhi}}{\underbrace{\sin 2x=-2}}\quad \textrm{atau}\quad \underset{\textrm{memenuhi}}{\underbrace{\sin 2x=1}}\\ &\textrm{pilih yang memenuhi persamaan, yaitu}\\ &\sin 2x=1\\ &\sin 2x=\sin 90^{0}\\ &\quad 2x=90^{0}+k.360^{0}\quad \textrm{atau}\\ &\quad 2x=\left (180^{0}-90^{0} \right )+k.360^{0}\\ & \textrm{(cukup ambil 1 persamaan saja, karena sama)}\\ &\qquad x=45^{0}+k.180^{0}\\ &k=0\Rightarrow x=45^{0}+0.180^{0}=45^{0}\\ &k=1\Rightarrow x=45^{0}+1.180^{0}=45^{0}+180^{0}=225^{0}\\ &k=2\Rightarrow x=45^{0}+2.180^{0}=45^{0}+360^{0}=405^{0}\\ & \textrm{(tidak memenuhi rentang)}\\ &\therefore \quad \textbf{HP}=\color{red}\left \{ 45^{0},225^{0} \right \} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 36.&\textrm{Bentuk}\: \: \sqrt{3}\cos x-\sin x\: \: \textrm{untuk}\: \: 0\leq x\leq 2\pi ,\\ &\textrm{dapat dinyatakan sebagai}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&2\cos \left ( x+\displaystyle \frac{\pi }{6} \right )\\ \textrm{b}.&2\cos \left ( x+\displaystyle \frac{7\pi }{6} \right ) \\ \color{red}\textrm{c}.&2\cos \left ( x-\displaystyle \frac{11\pi }{6} \right )\\ \textrm{d}.&2\cos \left ( x-\displaystyle \frac{7\pi }{6} \right )\\ \textrm{e}.&2\cos \left ( x-\displaystyle \frac{\pi }{6} \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textrm{c}\\ &\begin{aligned}\sqrt{3}\cos x-\sin x&=k\cos (x-\alpha )\\ (1)\qquad & (a,b)=\begin{cases} a &=\sqrt{3} \\ b &=-1 \end{cases}\\ \textrm{maka}\: &\textrm{titik ada dikadran IV}\\ (2)\: \quad k&=\sqrt{a^{2}+b^{2}}\\ &=\sqrt{\sqrt{3}^{2}+(-1)^{2}}\\ &=\sqrt{4}=2\\ (3)\quad \alpha &=\arctan \frac{b}{a}=\arctan \left (\frac{-1}{\sqrt{3}} \right )=-30^{\circ}\\ &=\left ( 360^{\circ}-30^{\circ} \right )=330^{\circ}=\displaystyle \frac{11}{6}\pi \\ \textrm{sehingga}&\\ \sqrt{3}\cos x-\sin x&=\color{red}2\cos \left ( x-\displaystyle \frac{11}{6}\pi \right )\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 37.&\textrm{Nilai-nilai}\: \: x\: \: \textrm{yang terletak pada}\: \: 0\leq x\leq 2\pi ,\\ &\textrm{yang memenuhi persamaan}\: \: \sqrt{3}\cos x+\sin x=\sqrt{2}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&75^{\circ}\: \: \textrm{atau}\: \: 285^{\circ}\\ \color{red}\textrm{b}.&75^{\circ}\: \: \textrm{atau}\: \: 345^{\circ}\\ \textrm{c}.&15^{\circ}\: \: \textrm{atau}\: \: 285^{\circ}\\ \textrm{d}.&15^{\circ}\: \: \textrm{atau}\: \: 345^{\circ}\\ \textrm{e}.&15^{\circ}\: \: \textrm{atau}\: \: 75^{\circ} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textrm{b}\\ &\begin{aligned}\sqrt{3}\cos x+\sin x&=\sqrt{2}\\ \sqrt{\sqrt{3}^{2}+1^{2}}&\left ( \cos \left ( \alpha -\arctan \displaystyle \frac{1}{\sqrt{3}} \right ) \right )=\sqrt{2}\\ 2\cos \left ( x-30^{\circ} \right )&=\sqrt{2}\\ \cos \left ( x-30^{\circ} \right )&=\displaystyle \frac{\sqrt{2}}{2}=\frac{1}{2}\sqrt{2}\\ \cos \left ( x-30^{\circ} \right )&=\cos 45^{\circ}\\ \left ( x-30^{\circ} \right )&=\pm 45^{\circ}+k.360^{\circ}\\ x&=30^{\circ}\pm 45^{\circ}+k.360^{\circ}\\ \textrm{saat}\: \: k&=0\\ x_{1}&=75^{\circ}\\ x_{2}&=\color{red}-15^{\circ}\\ \textrm{saat}\: \: k&=1\\ x_{3}&=75^{\circ}+360^{\circ}=\color{red}435^{\circ}\\ x_{4}&=-15^{\circ}+360^{\circ}=\color{red}345^{\circ} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 38.&\textrm{Diketahui fungsi trigonometri}\: \: f(x)=\displaystyle \frac{1}{2}\sin 3x\\ &\textrm{perhatikanlah pernyataan-pernyataan berikut}\\ &(1)\quad \textrm{hasil dari}\: \: f(0)+f\left ( \displaystyle \frac{\pi }{6} \right )=1\\ &(2)\quad \textrm{hasil dari}\: \: f\left ( \displaystyle \frac{\pi }{6} \right )+f\left ( \displaystyle \frac{\pi }{3} \right )=\displaystyle \frac{1}{2}\\ &(3)\quad \textrm{hasil dari}\: \: f\left ( \displaystyle \frac{\pi }{6} \right )-f\left ( \displaystyle \frac{\pi }{3} \right )=\displaystyle \frac{1}{2}\\ &(4)\quad \textrm{hasil dari}\: \: f\left ( \displaystyle \frac{\pi }{3} \right )-f\left ( \displaystyle \frac{\pi }{6} \right )=1\\ &\textrm{Pernyataan yang tepat adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&(1)\: \: \textrm{dan}\: \: (2)\\ \textrm{b}.&(1)\: \: \textrm{dan}\: \: (3)\\ \textrm{c}.&(1)\: \: \textrm{dan}\: \: (4)\\ \color{red}\textrm{d}.&(2)\: \: \textrm{dan}\: \: (3)\\ \textrm{e}.&(3)\: \: \textrm{dan}\: \: (4) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textrm{d}\\ &\begin{aligned}\textrm{Diketahui}&\\ f(x)&=\displaystyle \frac{1}{2}\sin 3x\\ \textrm{maka}\qquad&\\ f(0)&=\displaystyle \frac{1}{2}\sin 3(0^{\circ})=0\\ f\left ( \displaystyle \frac{\pi }{3} \right )&=\displaystyle \frac{1}{2}\sin 3\left ( \displaystyle \frac{\pi }{3} \right )=0\\ f\left ( \displaystyle \frac{\pi }{6} \right )&=\displaystyle \frac{1}{2}\sin 3\left ( \displaystyle \frac{\pi }{6} \right )=\color{red}\displaystyle \frac{1}{2}\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 39.&\textrm{Himpunan penyelesaian dari}\: \: \sin x-\sqrt{3}\cos x=-1\\ &\textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ 0^{\circ},120^{\circ} \right \}\\ \textrm{b}.&\left \{ 90^{\circ},330^{\circ} \right \}\\ \textrm{c}.&\left \{ 60^{\circ},180^{\circ} \right \}\\ \textrm{d}.&\left \{ 90^{\circ},120^{\circ} \right \}\\ \color{red}\textrm{e}.&\left \{ 30^{\circ},270^{\circ} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}\sin x-\sqrt{3}\cos x&=-1\\ \sqrt{1^{2}+\left ( -\sqrt{3} \right )^{2}}\cos \left ( x-\alpha \right )&=-1\\ a&=x=-\sqrt{3},\: \: b=y=1\\ &\textbf{kuadran II}\\ \alpha &=\arctan \left ( \displaystyle \frac{1}{-\sqrt{3}} \right )=-30^{\circ}\\ &=180^{\circ}-30^{\circ}=150^{\circ}\\ \textrm{maka persamaan akan}&\: \textrm{menjadi}\\ 2\cos \left ( x-150^{\circ} \right )&=-1\\ \cos \left ( x-150^{\circ} \right )&=\displaystyle \frac{-1}{2}\\ \cos \left ( x-150^{\circ} \right )&=\cos 120^{\circ}\\ \left ( x-150^{\circ} \right )&=\pm 120^{\circ}+k.360^{\circ}\\ x&=150^{\circ}\pm 120^{\circ}+k.360^{\circ}\\ \textrm{saat}\: \: &k=0\\ x_{1}&=270^{\circ}\\ x_{2}&=30^{\circ}\\ \textrm{saat}\: \: &k=1\\ x_{3}&=270^{\circ}+360^{\circ}=....\\ x_{4}&=\color{red}30^{\circ}+360^{\circ}=.... \end{aligned} \end{array}$

$\begin{array}{ll}\\ 40.&\textrm{Nilai}\: \: \tan x\: \: \textrm{yang memenuhi persamaan}\\ &\cos 2x+7\cos x-3=0\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\sqrt{3}\\ \textrm{b}.&\displaystyle \frac{1}{2}\sqrt{3}\\ \textrm{c}.&\displaystyle \frac{1}{3}\sqrt{3}\\ \textrm{d}.&\displaystyle \frac{1}{2}\\ \textrm{e}.&\displaystyle \frac{1}{5}\sqrt{5} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\cos 2x+7\cos x-3&=0\\ 2\cos ^{2}x-1+7\cos x-3&=0\\ 2\cos ^{2}x+7\cos x-4&=0\\ \left ( \cos x+4 \right )\left (2 \cos -1 \right )&=0\\ \color{red}\cos x=-4\: \: \color{black}\textrm{atau}\: \: \color{red}\cos x&=\displaystyle \frac{1}{2}=\cos 60^{\circ}\\ x&=60^{\circ},\\ \textrm{maka}&\\ \tan 60^{\circ}&=\sqrt{3}\\\\ \textrm{ dan ingat bahwa}&\: \: \cos x=-4\: \: \color{red}\textrm{tidak memenuhi} \end{aligned} \end{array}$

Latihan Soal 3 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

$\begin{array}{ll}\\ 21.&\textrm{Jika diketahui}\: \: \tan \left ( 3x+60^{0} \right )=\sqrt{3},\\ & \textrm{maka nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &\textrm{tersebut adalah}....\\ &\begin{array}{llllll}\\ \textrm{a}.&90^{0}\\ \textrm{b}.&110^{0}\\ \textrm{c}.&\color{red}120^{0}\\ \textrm{d}.&130^{0}\\ \textrm{e}.&230^{0} \end{array}\\\\ &\textbf{Jawab}:\qquad \color{red}\textbf{c}\\ &\begin{aligned}\tan \left ( 3x+60^{0} \right )&=\sqrt{3}\\ \tan \left ( 3x+60^{0} \right )&=\tan 60^{0}\\ \left ( 3x+60^{0} \right )&=60^{0}+k.180^{0}\\ 3x&=k.180^{0}\\ x&=k.60^{0}\\ k=0\rightarrow x&=0^{0}\\ k=1\rightarrow x&=60^{0}\\ k=2\rightarrow x&=120^{0}\\ k=3\rightarrow x&=180^{0}\\ k=4\rightarrow x&=240^{0}\\ k=5\rightarrow x&=300^{0}\\ k=\cdots \rightarrow x&=\cdots \\ \cdots \cdots \rightarrow &=....\textbf{dst} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 22.&\textrm{Himpunan penyelesaian dari }\\ & \sqrt{3}\tan x=-1,\quad 0^{0}\leq x\leq 360^{0}\: \: \textrm{adalah}....\\ &\begin{array}{llllll}\\ \textrm{a}.&\left \{ 30^{0},150^{0} \right \}\\ \textrm{b}.&\left \{ 150^{0},210^{0} \right \}\\ \textrm{c}.&\color{red}\left \{ 150^{0},330^{0} \right \}\\ \textrm{d}.&\left \{ 120^{0},300^{0} \right \}\\ \textrm{e}.&\left \{ 60^{0},240^{0} \right \} \end{array}\\\\ &\textbf{Jawab}:\qquad \color{red}\textbf{c}\\ &\begin{aligned}\sqrt{3}\tan x&=-1\\ \tan x&=-\displaystyle \frac{1}{\sqrt{3}}=-\frac{1}{3}\sqrt{3}\\ &\qquad \color{blue}\textrm{nilai di kuadran II}\\ \tan x&=\tan 150^{0}\\ x&=150^{0}+k.180^{0}\\ k=0,\Rightarrow x&=150^{0}\\ k=1,\Rightarrow x&=330^{0}\\ k=2,\Rightarrow x&=510^{0},\quad \textrm{tidak}\: \textrm{memenuhi}\: \textrm{(tm)}\\ k=3,\quad \textbf{dst}&=\textrm{(tm)}\\ \therefore \textbf{HP}&=\color{red}\left \{ 150^{0},330^{0} \right \} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 23.&\textrm{Himpunan penyelesaian dari }\\ & \tan 2x^{0}=-\displaystyle \frac{1}{3}\sqrt{3},\quad 0^{0}\leq x\leq 180^{0}\: \: \textrm{adalah}....\\ &\begin{array}{llllll}\\ \textrm{a}.&\color{red}\left \{ 75^{0},165^{0} \right \}\\ \textrm{b}.&\left \{ 60^{0},150^{0} \right \}\\ \textrm{d}.&\left \{ 30^{0},120^{0} \right \}\\ \textrm{c}.&\left \{ 45^{0},135^{0} \right \}\\ \textrm{e}.&\left \{ 15^{0},105^{0} \right \} \end{array}\\\\ &\textbf{Jawab}:\qquad \color{red}\textbf{a}\\ &\begin{aligned}\tan 2x^{0}&=-\displaystyle \frac{1}{3}\sqrt{3}\\ \tan 2x^{0}&=\tan \left ( 180^{0}-30^{0} \right )\\ &\textrm{nilai tan negatif paling kecil }\\ &\textrm{berada di kuadran II}\\ 2x^{0}&=150^{0}+k.180^{0}\\ x^{0}&=75^{0}+k.90^{0}\\ k=0,\Rightarrow x^{0}&=75^{0}\\ k=1,\Rightarrow x^{0}&=75^{0}+90^{0}=165^{0}\\ k=2,\Rightarrow x^{0}&=255^{0}\\ &\textrm{tidak memenuhi, karena }\\ &\textrm{berada di luar batas interval}\\ \therefore \textbf{HP}&=\color{red}\left \{ 75^{0},165^{0} \right \} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 24.&\textrm{Himpunan penyelesaian dari }\\ & \cos \left ( x-30^{0} \right )=-\sin 50^{0},\quad 0^{0}\leq x\leq 360^{0}\\ & \textrm{adalah}....\\ &\begin{array}{llllll}\\ \textrm{a}.&\left \{ 90^{0} \right \}\\ \textrm{b}.&\left \{ 140^{0},150^{0} \right \}\\ \textrm{c}.&\color{red}\left \{ 170^{0},250^{0} \right \}\\ \textrm{d}.&\left \{ 80^{0},280^{0} \right \}\\ \textrm{e}.&\left \{ 20^{0},340^{0} \right \} \end{array}\\\\ &\textbf{Jawab}:\qquad \color{red}\textbf{c}\\ &\begin{aligned}\cos \left ( x-30^{0} \right )&=-\sin 50^{0}\\ \cos \left ( x-30^{0} \right )&=\cos \left ( 90^{0}+50^{0} \right )\\ \cos \left ( x-30^{0} \right )&=\cos 140^{0}\\ x-30^{0}&=\pm 140^{0}+k.360^{0}\\ x&=30^{0}\pm 140^{0}+k.360^{0}\\ k=0,\Rightarrow x_{1}&=30^{0}+140^{0}=170^{0}\\ \Rightarrow x_{2}&=30^{0}-140^{0}=-110^{0}\: \: (\textrm{tm})\\ k=1,\Rightarrow x_{1}&=30^{0}+140^{0}+360^{0}=530^{0}\: \: (\textrm{tm})\\ \Rightarrow x_{2}&=30^{0}-140^{0}+360^{0}=250^{0}\\ \therefore \textbf{HP}&=\color{red}\left \{ 170^{0},250^{0} \right \} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 25.&\textrm{Himpunan penyelesaian dari}\\ &\sin 2x=\displaystyle \frac{1}{2}\sqrt{3}\: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ 30^{\circ},210^{\circ} \right \}\\ \textrm{b}.&\left \{ 60^{\circ},240^{\circ} \right \}\\ \textrm{c}.&\left \{ 30^{\circ},60^{\circ},210^{\circ} \right \}\\ \color{red}\textrm{d}.&\left \{ 30^{\circ},60^{\circ},210^{\circ},240^{\circ} \right \}\\ \textrm{e}.&\left \{ 30^{\circ},60^{\circ},210^{\circ},240^{\circ},270^{\circ} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\sin 2x&=\displaystyle \frac{1}{2}\sqrt{3}\\ \sin 2x&=\sin 60^{\circ}\\ 2x&=\begin{cases} 60^{\circ} & +k.360^{\circ} \\ \left ( 180^{\circ}-60^{\circ} \right ) & +k.360^{\circ} \end{cases}\\ x&=\begin{cases} 30^{\circ} & +k.180^{\circ}\\ 60^{\circ} & +k.180^{\circ} \end{cases}\\ \textrm{saat}&\: \: k=0\\ x&=\begin{cases} 30^{\circ} & \\ 60^{\circ} & \end{cases}\\ \textrm{saat}&\: \: k=1\\ x&=\begin{cases} 30^{\circ} & +1.180^{\circ}=210^{\circ}\\ 60^{\circ} & +1.180^{\circ}=240^{\circ} \end{cases}\\ \textrm{saat}&\: \: k=2\\ x&=\begin{cases} 30^{\circ} & +2.180^{\circ}=\color{red}390^{\circ}\\ 60^{\circ} & +2.180^{\circ}=\color{red}420^{\circ} \end{cases}\\ \color{red}\textrm{kedua}&\color{red}\textrm{nnya tidak memenuhi} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 26.&\textrm{Himpunan penyelesaian dari}\\ &\tan 2x-\sqrt{3}=0\: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ 15^{\circ},105^{\circ},195^{\circ},285^{\circ} \right \}\\ \color{red}\textrm{b}.&\left \{ 30^{\circ},120^{\circ},210^{\circ},300^{\circ} \right \}\\ \textrm{c}.&\left \{ 45^{\circ},135^{\circ},225^{\circ},315^{\circ} \right \}\\ \textrm{d}.&\left \{ 15^{\circ},105^{\circ},195^{\circ},285^{\circ} \right \}\\ \textrm{e}.&\left \{ 15^{\circ},30^{\circ},45^{\circ},60^{\circ},75^{\circ} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\tan 2x&-\sqrt{3}=0\\ \tan 2x&=\sqrt{3}\\ \tan 2x&=\tan 60^{\circ}\\ 2x&=60+k.180^{\circ}\\ x&=30^{\circ}+k.90^{\circ}\\ \textrm{saat}&\: \: k=0\\ x&=30^{\circ}\\ \textrm{saat}&\: \: k=1\\ x&=30^{\circ}+90^{\circ}=120^{\circ}\\ \textrm{saat}&\: \: k=2\\ x&=30^{\circ}+180^{\circ}=210^{\circ}\\ \textrm{saat}&\: \: k=3\\ x&=30^{\circ}+270^{\circ}=300^{\circ}\\ \textrm{saat}&\: \: k=4\\ x&=30^{\circ}+360^{\circ}=\color{red}390^{\circ}\\ \color{red}\textrm{tidak}&\: \color{red}\textrm{memenuhi} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 27.&\textrm{Himpunan penyelesaian dari}\\ &\cos 3x=-\displaystyle \frac{1}{2}\sqrt{3}\: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 180^{\circ}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ 40^{\circ},80^{\circ} \right \}\\ \textrm{b}.&\left \{ 50^{\circ},70^{\circ} \right \}\\ \textrm{c}.&\left \{ 40^{\circ},70^{\circ},80^{\circ} \right \}\\ \color{red}\textrm{d}.&\left \{ 50^{\circ},70^{\circ},170^{\circ} \right \}\\ \textrm{e}.&\left \{ 50^{\circ},80^{\circ},170^{\circ} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\cos 3x&=-\displaystyle \frac{1}{2}\sqrt{3}\\ \cos 3x&=-\cos 30^{\circ}\\ \cos 3x&=\cos \left (180^{\circ}-30^{\circ} \right )=\cos 150^{\circ}\\ 3x&=\pm 150^{\circ}+k.360^{\circ}\\ x&=\pm 50^{\circ}+k.120^{\circ}\\ \textrm{saat}&\: \: k=0\\ x&=\pm 50^{\circ}\: \rightarrow x=50^{\circ}\: \: (\textrm{mm})\\ \textrm{saat}&\: \: k=1\\ x&=\pm 50^{\circ}+120^{\circ}=\begin{cases} 170^{\circ} & (\textrm{mm}) \\ 70^{\circ} & (\textrm{mm}) \end{cases} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 28.&\textrm{Jika diketahui}\: \: \sin \beta -\tan \beta -2\cos \beta +2=0\\ & \textrm{dengan}\: \: 0<\beta <\displaystyle \frac{\pi }{2},\\ &\textrm{maka himpunan harga}\: \: \sin \beta =....\\ &\begin{array}{llllll}\\ \textrm{a}.&\color{red}\left \{ \displaystyle \frac{2}{5}\sqrt{5} \right \}\\ \textrm{b}.&\left \{ 0 \right \}\\ \textrm{c}.&\left \{ \displaystyle \frac{2}{5}\sqrt{5},0 \right \}\\ \textrm{d}.&\left \{ \displaystyle \frac{1}{5}\sqrt{5} \right \}\\ \textrm{e}.&\left \{ \displaystyle \frac{1}{5}\sqrt{5},0 \right \}\\\\ &&&&\textbf{(SIMAK UI 2009)} \end{array}\\ &\\ &\textbf{Jawab}:\quad\color{red}\textbf{a}\\ &\begin{aligned}\sin \beta -\tan \beta -2\cos \beta +2&=0\\ \sin \beta -\displaystyle \frac{\sin \beta }{\cos \beta } -2\cos \beta +2&=0\\ \sin \beta \cos \beta -\sin \beta -2\cos^{2} \beta +2\cos \beta &=0\\ \sin \beta \left ( \cos \beta -1 \right )-2\cos \beta \left ( \cos \beta -1 \right )&=0\\ \left ( \sin \beta -2\cos \beta \right )\left ( \cos \beta -1 \right )&=0\\ \left ( \sin \beta -2\cos \beta \right )=0\: \textbf{(mm)}&\\ \color{red}\textrm{atau}\: \: \color{black}\left ( \cos \beta -1 \right )&=0\: \textbf{(tm)}\\ \color{blue}\textrm{Selanjutnya adalah},\qquad\qquad\quad&\\ \left ( \sin \beta -2\cos \beta \right )&=0\\ \sin \beta =2\cos \beta&\\ \displaystyle \frac{\sin \beta }{\cos \beta }&=2\\ \tan \beta &=2=\displaystyle \frac{2}{1},\\ \textrm{buatlah ilustrasi}& \\ \textrm{dengan membuat segitiga siku-siku}&\\ \textrm{Sehingga akan didapatkan nilai}&\\ \sin \beta =\color{red}\displaystyle \frac{2}{5}\sqrt{5}& \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 29.&\textrm{Penyelesaian umum untuk nilai}\: \: \theta \\ & \textrm{yang memenuhi persamaan}\\ &\sin \theta =\displaystyle \frac{1}{2},\: \: \tan \theta =\displaystyle \frac{1}{\sqrt{3}} \: \: \textrm{adalah}\, ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\color{red}2n\pi +\displaystyle \frac{\pi }{6}\\\\ \textrm{b}.&2n\pi +\displaystyle \frac{\pi }{5}\\\\ \textrm{c}.&2n\pi +\displaystyle \frac{\pi }{4} \\\\ \textrm{d}.&2n\pi +\displaystyle \frac{\pi }{3}\\\\ \textrm{e}.&2n\pi +\displaystyle \frac{\pi }{2} \\ \end{array}\\ &\\ &\textbf{Jawab}:\quad\color{red}\textbf{a}\\ &\begin{aligned}\textrm{Perhatikan}&\: \textrm{bahwa}\, ;\\ &\begin{cases} \sin x & =\sin \theta \\ x& = \theta +k.2\pi \\ \tan x & =\tan \theta \\ x&=\theta +k.\pi \end{cases}\\ \textrm{Karena}\quad&\sin x=\sin \theta =\displaystyle \frac{1}{2}\Rightarrow \theta =30^{\circ}=\displaystyle \frac{\pi }{6}\\ \textrm{Demikian}&\: \textrm{juga untuk nilai}\: \: \tan \theta ,\: \: \textrm{yaitu}:\\ &\tan x=\tan \theta =\displaystyle \frac{1}{\sqrt{3}}\Rightarrow \theta =30^{\circ}=\displaystyle \frac{\pi }{6}\\ \textrm{maka dari}&\: \textrm{sini akan diperoleh}\\ &\textbf{penyelesaian umumnya}\\ \textrm{yaitu}\quad &:\color{red}2n\pi +\displaystyle \frac{\pi }{6} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 30.&\textrm{Jika diketahui}\\ & f\left ( \displaystyle \frac{6}{\sqrt{\sin ^{2}x+4}} \right )=\tan x,\: \: \pi \leq x\leq 2\pi ,\\ &\textrm{maka nilai}\: \: f(3)=....\\ &\begin{array}{llllll}\\ \textrm{a}.&\color{red}0\\ \textrm{b}.&1\\ \textrm{c}.&\displaystyle \frac{\pi }{2}\\ \textrm{d}.&\pi \\ \textrm{e}.&2\pi \\ \end{array}\\ &\\ &\textbf{Jawab}:\quad\color{red}\textbf{a}\\ &\begin{aligned}\textrm{Diketahui bahwa}\: &\\ f\left ( \displaystyle \frac{6}{\sqrt{\sin ^{2}x+4}} \right )&=\tan x,\: \: \pi \leq x\leq 2\pi\\ f(3)&=....\\ \textrm{maka selanjutnya}\, &\\ \displaystyle \frac{6}{\sqrt{\sin ^{2}x+4}}&=3\\ \sqrt{\sin ^{2}x+4}&=2\\ \sin ^{2}x+4&=4\\ \sin ^{2}x&=0\\ \sin x&=\begin{cases} x_{1}=0^{\circ} &+k.2\pi \\ x_{2}=180^{\circ} & +k.2\pi \end{cases}\\ \tan \pi =\tan 2\pi &=0\\ \therefore \quad f(3)&=\color{red}0 \end{aligned} \end{array}$

Latihan Soal 2 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

$\begin{array}{ll}\\ 11.&\textrm{Periode dari fungsi}\: \: f(x)=-2\cos 3x\: \: \textrm{adalah}\: ....\\\\ &\textrm{a}.\quad 90^{\circ} \\\\ &\textrm{b}.\quad 100^{\circ} \\\\ &\textrm{c}.\quad \color{red}120^{\circ} \\\\ &\textrm{d}.\quad 150^{\circ} \\\\ &\textrm{e}.\quad 180^{\circ} \\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Periode dari}\: :\: f(x)=-2\cos 3x\\ &\textrm{adalah}\\ &=\displaystyle \frac{360^{\circ}}{3}\\ &=\color{red}120^{\circ}\\ &\\ &\textrm{Ingat bahwa}\\ &f(x)=a\cos bx,\: \: \textrm{maka periodenya}\\ &=\displaystyle \frac{360^{\circ}}{b} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 12.&\textrm{Perhatikanlah grafik berikut} \end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{Gambar di atas adalah grafik fungsi dari}\\ &\textrm{a}.\quad f(x)=\cos 2x \\\\ &\textrm{b}.\quad f(x)=\cos 3x \\\\ &\textrm{c}.\quad \color{red}f(x)=3\cos x \\\\ &\textrm{d}.\quad f(x)=3\cos 3x \\\\ &\textrm{e}.\quad f(x)=\displaystyle \frac{1}{3}\cos x \\\\ &\textbf{Jawab}:\\ &\textrm{Gambar cukup jelas}\\ &\textrm{dengan periode}\: \: 360^{\circ}\\ &\textrm{gambar dari grafik}\: \: f(x)=\color{red}3\cos x \end{array}$.

$\begin{array}{ll}\\ 13.&\textrm{Nilai dari}\: \: \displaystyle \frac{\sin \displaystyle \frac{3}{4}\pi +\tan \pi +\cos \pi }{\sin \displaystyle \frac{1}{2}\pi +\cos 2\pi -3\cos \displaystyle \frac{1}{3}\pi }=\: ....\\\\ &\textrm{a}.\quad 4 \\\\ &\textrm{b}.\quad 2-\sqrt{2} \\\\ &\textrm{c}.\quad \sqrt{2}-2 \\\\ &\textrm{d}.\quad -4 \\\\ &\textrm{e}.\quad \color{red}-1 \\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\sin \displaystyle \frac{3}{4}\pi +\tan \pi +\cos \pi }{\sin \displaystyle \frac{1}{2}\pi +\cos 2\pi -3\cos \displaystyle \frac{1}{3}\pi }\\ &=\displaystyle \frac{\sin 150^{\circ} +\tan 180^{\circ} +\cos 180^{\circ} }{\sin \displaystyle 90^{\circ} +\cos 360^{\circ} -3\cos \displaystyle 60^{\circ} }\\ &=\displaystyle \frac{\displaystyle \frac{1}{2}+0+(-1)}{1+1-3\left (\displaystyle \frac{1}{2} \right ) }\\ &=\displaystyle \frac{-\displaystyle \frac{1}{2}}{2-\displaystyle \frac{3}{2}}\\ &=-\displaystyle \frac{\displaystyle \frac{1}{2}}{\displaystyle \frac{1}{2}}\\ &=\color{red}-1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 14.&\textrm{Grafik fungsi trigonometri pada gambar}\\ &\textrm{berikut adalah}\: .... \end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{a}.\quad \displaystyle y=2\sin \left ( x-\displaystyle \frac{1}{2}\pi \right ) \\\\ &\textrm{b}.\quad \displaystyle y=2\sin \left ( \displaystyle \frac{1}{2}\pi -x \right ) \\\\ &\textrm{c}.\quad \displaystyle \color{red}y=2\sin \left ( 2x+\displaystyle \frac{1}{6}\pi \right ) \\\\ &\textrm{d}.\quad \displaystyle y=-2\sin \left ( \displaystyle \frac{1}{2}\pi +x \right ) \\\\ &\textrm{e}.\quad \displaystyle y=-2\sin \left ( \displaystyle \frac{1}{2}\pi -2x \right ) \\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Dari grafik tampak jelas bahwa}\\ &\textrm{gambar di atas adalah garfik}\\ &\textrm{fungsi sinus di geser ke}\: \: \textbf{kiri}\\ &\textrm{dengan}\: \: \textbf{amplitudo}\: \: 2\\ &\textrm{dan}\: \: \textbf{periodenya}\: \: 2\pi,\: \textrm{maka}\\ &\textrm{bentuk persamaan}\: \: \textbf{grafik fungsinya}\\ &y=\color{red}2\sin \left ( 2x+k \right )\\ &\textrm{dengan}\: \: +k\: \: \textrm{adalah}\\ &\textbf{besar geseran ke kirinya} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 15.&\textrm{Grafik fungsi trigonometri pada gambar}\\ &\textrm{berikut adalah}\: .... \end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{a}.\quad \displaystyle y=-2\sin (3x+45)^{\circ} \\\\ &\textrm{b}.\quad \displaystyle y=-2\sin (3x-15)^{\circ} \\\\ &\textrm{c}.\quad \displaystyle y=-2\sin (3x-45)^{\circ} \\\\ &\textrm{d}.\quad \displaystyle y=2\sin (3x+15)^{\circ} \\\\ &\textrm{e}.\quad \displaystyle \color{red}y=2\sin (3x-45)^{\circ} \\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Dari grafik tampak jelas bahwa}\\ &\textrm{gambar di atas adalah garfik}\\ &\textrm{fungsi sinus di geser ke}\: \: \textbf{kanan}\\ &\textrm{dengan}\: \: \textbf{amplitudo}\: \: 2\\ &\textrm{dan}\: \: \textbf{periodenya}\: \: \displaystyle \frac{360^{\circ}}{3}=120^{\circ},\: \textrm{maka}\\ &\textrm{bentuk persamaan}\: \: \textbf{grafik fungsinya}\\ &y=\color{red}2\sin 3\left ( x-k \right )\\ &\textrm{dengan}\: \: -k\: \: \textrm{adalah}\\ &\textbf{besar geseran ke kanan}\: \: 15^{\circ}\\ &\textrm{Jadi},\: \: y=\color{red}2\sin 3(x-15)^{\circ}=2\sin (3x-45)^{\circ} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 16.&\textrm{Grafik fungsi trigonometri pada gambar}\\ &\textrm{berikut adalah}\: .... \end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{a}.\quad \displaystyle y=-\cos (2x-30)^{\circ} \\\\ &\textrm{b}.\quad \displaystyle y=\sin (2x-60)^{\circ} \\\\ &\textrm{c}.\quad \displaystyle \color{red}y=\cos (2x+30)^{\circ} \\\\ &\textrm{d}.\quad \displaystyle y=\sin (2x-80)^{\circ} \\\\ &\textrm{e}.\quad \displaystyle y=\sin (2x+60)^{\circ} \\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Dari grafik tampak jelas bahwa}\\ &\textrm{gambar di atas adalah garfik}\\ &\textrm{fungsi cosinus di geser ke}\: \: \textbf{kiri}\\ &\textrm{dengan}\: \: \textbf{amplitudo}\: \: 1\\ &\textrm{dan}\: \: \textbf{periodenya}\: \: \displaystyle \frac{360^{\circ}}{2}=180^{\circ},\: \textrm{maka}\\ &\textrm{bentuk persamaan}\: \: \textbf{grafik fungsinya}\\ &y=\color{red}\cos 2\left ( x+k \right )\\ &\textrm{dengan}\: \: +k\: \: \textrm{adalah}\\ &\textbf{besar geseran ke kiri}\: \: 15^{\circ}\\ &\textrm{Jadi},\: \: y=\color{red}\cos 2(x+15)^{\circ}=\sin (2x+30)^{\circ} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 17.&\textrm{Himpunan penyelesaian dari persamaan}\\ &\sin x=\sin \displaystyle \frac{2}{10}\pi \: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \displaystyle \left \{ \displaystyle \frac{22}{10}\pi ,\displaystyle \frac{8}{10}\pi \right \} \\\\ &\textrm{b}.\quad \displaystyle \left \{ \displaystyle \frac{2}{10}\pi ,\displaystyle \frac{28}{10}\pi \right \} \\\\ &\textrm{c}.\quad \displaystyle \color{red}\left \{ \displaystyle \frac{2}{10}\pi ,\displaystyle \frac{8}{10}\pi \right \} \\\\ &\textrm{d}.\quad \displaystyle \left \{ \displaystyle \frac{22}{10}\pi ,\displaystyle \frac{28}{10}\pi \right \} \\\\ &\textrm{e}.\quad \displaystyle \left \{ \displaystyle \frac{12}{10}\pi ,\displaystyle \frac{8}{10}\pi \right \} \\\\ &\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\begin{aligned} & \sin x=\sin \displaystyle \frac{2}{10}\pi \\ &\Leftrightarrow \: \: x_{1}=\displaystyle \frac{2}{10}\pi+k.2\pi \: \: \: \: \color{blue}\textrm{atau}\\ &\Leftrightarrow \quad x_{2} =\left (\pi -\displaystyle \frac{2}{10}\pi \right )+k.2\pi=\displaystyle \frac{8}{10}\pi +k.2\pi \\ &k=0\Rightarrow x_{1}=\displaystyle \frac{2}{10}\pi\: \: (\color{blue}\textrm{mm})\: \: \color{black}\textrm{atau}\\ &\qquad\qquad x_{2}=\displaystyle \frac{8}{10}\pi\: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x_{1,2}=....+2\pi \quad (\color{red}\textrm{tidak memenuhi})\\ \end{aligned} \\ &\textbf{HP}=\color{red}\left \{\displaystyle \frac{2}{10}\pi,\: \displaystyle \frac{8}{10}\pi \right \} \end{array} \end{array}$.

$\begin{array}{ll}\\ 18.&\textrm{Himpunan penyelesaian dari persamaan}\\ &\tan \left ( 2x-\displaystyle \frac{1}{4}\pi \right )=\tan \displaystyle \frac{1}{4}\pi \: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \displaystyle \left \{ \displaystyle \frac{1}{3}\pi ,\pi ,\displaystyle \frac{5}{3}\pi ,\displaystyle \frac{7}{3}\pi \right \} \\\\ &\textrm{b}.\quad \displaystyle \left \{ \displaystyle \frac{1}{4}\pi ,\displaystyle \frac{3}{5}\pi ,\frac{5}{4}\pi ,\frac{8}{5}\pi \right \} \\\\ &\textrm{c}.\quad \displaystyle \left \{ \displaystyle \frac{1}{4}\pi ,\displaystyle \frac{3}{4}\pi ,\frac{6}{4}\pi ,\displaystyle \frac{7}{4}\pi \right \} \\\\ &\textrm{d}.\quad \displaystyle \left \{ \displaystyle \frac{2}{4}\pi ,\displaystyle \frac{3}{4}\pi ,\pi ,\displaystyle \frac{7}{4}\pi \right \} \\\\ &\textrm{e}.\quad \displaystyle \color{red}\left \{ \displaystyle \frac{1}{4}\pi ,\frac{3}{4}\pi ,\frac{5}{4}\pi ,\displaystyle \frac{7}{4}\pi \right \} \\\\ &\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\begin{aligned} & \tan \left ( 2x-\displaystyle \frac{1}{4}\pi \right )=\tan \displaystyle \frac{1}{4}\pi \\ &\Leftrightarrow \: \: 2x-\displaystyle \frac{1}{4}\pi=\displaystyle \frac{1}{4}\pi+k.\pi\\ &\Leftrightarrow \quad 2x =\displaystyle \frac{2}{4}\pi +k.\pi \\ &\Leftrightarrow \quad x =\displaystyle \frac{1}{4}\pi +k.\frac{\pi}{2} \\ &k=0\Rightarrow x=\displaystyle \frac{1}{4}\pi\: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x=\displaystyle \frac{1}{4}\pi+\frac{\pi}{2}=\displaystyle \frac{3}{4}\pi \: \: (\color{blue}\textrm{mm}) \\ &k=2\Rightarrow x=\displaystyle \frac{1}{4}\pi+\pi =\displaystyle \frac{5}{4}\pi \: \: (\color{blue}\textrm{mm}) \\ &k=3\Rightarrow x=\displaystyle \frac{1}{4}\pi+\frac{3\pi}{2}=\displaystyle \frac{7}{4}\pi \: \: (\color{blue}\textrm{mm}) \\ &k=4\Rightarrow x=\displaystyle \frac{1}{4}\pi+2\pi =\displaystyle \frac{9}{4}\pi \: \: (\color{red}\textrm{tidak memenuhi}) \\ \end{aligned} \\ &\textbf{HP}=\color{red}\color{red}\left \{\displaystyle \frac{1}{4}\pi,\: \displaystyle \frac{3}{4}\pi ,\frac{5}{4}\pi ,\frac{7}{4}\pi \right \} \end{array} \end{array}$.

$\begin{array}{ll}\\ 19.&\textrm{Himpunan penyelesaian dari persamaan}\\ &\cos (x-30)^{\circ}= \displaystyle \frac{1}{2}\sqrt{2} \: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \displaystyle \left \{ 75^{\circ},285^{\circ} \right \}\\\\ &\textrm{b}.\quad \displaystyle \left \{ 75^{\circ},343^{\circ} \right \}\\\\ &\textrm{c}.\quad \displaystyle \left \{ 75^{\circ},344^{\circ} \right \}\\\\ &\textrm{d}.\quad \displaystyle \color{red}\left \{ 75^{\circ},345^{\circ} \right \}\\\\ &\textrm{e}.\quad \displaystyle \left \{ 75^{\circ},346^{\circ} \right \}\\\\ &\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\begin{aligned} & \cos (x-30)^{\circ}= \displaystyle \frac{1}{2}\sqrt{2} \\ &\cos (x-30)^{\circ}= \cos 45^{\circ}\\ &\Leftrightarrow \: \: x-30^{\circ}=\pm 45^{\circ}+k.360^{\circ}\\ &\Leftrightarrow \quad x =30^{\circ}\pm 45^{\circ}+k.360^{\circ} \\ &k=0\Rightarrow x_{1}=75^{\circ}\: \: (\color{blue}\textrm{mm})\\ &\qquad\textrm{atau}\: \: x_{2}=-15^{\circ}\: \: \color{red}(tm)\\ &k=1\Rightarrow x_{1}=75^{\circ}+360^{\circ} \: \: (\color{red}\textrm{tm})\\ &\qquad \textrm{atau}\: \: \: x_{2}=-15^{\circ}+360^{\circ}=345^{\circ}\: \: \color{blue}(mm) \end{aligned} \\ &\textbf{HP}=\color{red}\color{red}\left \{\displaystyle 75^{\circ},345^{\circ} \right \} \end{array} \end{array}$.

$\begin{array}{ll}\\ 20.&\textrm{Himpunan penyelesaian dari persamaan}\\ &\cos (x-30)^{\circ}= \displaystyle \frac{1}{2}\sqrt{3} \: \: \textrm{untuk}\: \: 0^{\circ}< x< 360^{\circ}\\ &\textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \displaystyle \left \{ 100^{\circ},330^{\circ} \right \}\\\\ &\textrm{b}.\quad \displaystyle \color{red}\left \{ 30^{\circ},330^{\circ} \right \}\\\\ &\textrm{c}.\quad \displaystyle \left \{ 120^{\circ},300^{\circ} \right \}\\\\ &\textrm{d}.\quad \displaystyle \left \{ 60^{\circ},120^{\circ} \right \}\\\\ &\textrm{e}.\quad \displaystyle \left \{ 50^{\circ},300^{\circ} \right \}\\\\ &\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\begin{aligned} & \cos x^{\circ}= \displaystyle \frac{1}{2}\sqrt{3} \\ &\cos x^{\circ}= \cos 30^{\circ}\\ &\Leftrightarrow \quad x =\pm 30^{\circ}+k.360^{\circ} \\ &k=0\Rightarrow x_{1}=30^{\circ}\: \: (\color{blue}\textrm{mm})\\ &\qquad\textrm{atau}\: \: x_{2}=-30^{\circ}\: \: \color{red}(tm)\\ &k=1\Rightarrow x_{1}=30^{\circ}+360^{\circ}=390^{\circ} \: \: (\color{red}\textrm{tm})\\ &\qquad \textrm{atau}\: \: \: x_{2}=-30^{\circ}+360^{\circ}=330^{\circ}\: \: \color{blue}(mm) \end{aligned} \\ &\textbf{HP}=\color{red}\color{red}\left \{\displaystyle 30^{\circ},330^{\circ} \right \} \end{array} \end{array}$

Latihan Soal 1 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

 $\begin{array}{ll}\\ 1.&\textrm{Nilai}\: \: 75^{\circ}\: \: \textrm{jika dinyatakan ke radian}\\ &\textrm{adalah}\: \: ....\: \: \textrm{radian}\\\\ &\textrm{a}.\quad \displaystyle \frac{1}{3}\pi \\\\ &\textrm{b}.\quad \displaystyle \frac{5}{6}\pi \\\\ &\textrm{c}.\quad \displaystyle \color{red}\frac{5}{12}\pi \\\\ &\textrm{d}.\quad \displaystyle \frac{7}{12}\pi \\\\ &\textrm{e}.\quad \displaystyle \frac{9}{12}\pi \\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{Diketah}&\textrm{ui bahwa}\\ 180^{\circ}&=\pi \: \: \: radian\\ 1^{\circ}&=\displaystyle \frac{\pi }{180}\: \: \: radian\\ 75\times 1^{\circ}&=75\times \displaystyle \frac{\pi }{180}\: \: \: radian\\ 75^{\circ}&=\displaystyle \frac{5}{12}\pi \: \: \: radian \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Jika}\: \: \tan \theta =\displaystyle \frac{5}{12}\: \: \textrm{untuk}\: \: 0^{\circ}\leq \theta \leq 90^{\circ}\\ &\textrm{maka}\: \: \cos \theta \: \: \textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \displaystyle \frac{5}{13} \\\\ &\textrm{b}.\quad \displaystyle \color{red}\frac{12}{13} \\\\ &\textrm{c}.\quad \displaystyle \frac{13}{5} \\\\ &\textrm{d}.\quad \displaystyle \frac{13}{12} \\\\ &\textrm{e}.\quad \displaystyle \frac{12}{5} \\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikanlah gambar segitiga berikut} \end{array}$.

$. \qquad\begin{aligned}\textrm{Diketah}&\textrm{ui bahwa}\\ \tan \theta &=\displaystyle \frac{5}{12}\: ,\: \textrm{untuk}\: 0^{\circ}\leq \theta \leq 90^{\circ}\\ &\color{red}\textrm{lihat gambar di atas}\\ &\textrm{dengan dalil Pythagoras akan}\\ &\textrm{didapatkan sisimiringnya}=13\\ \textrm{jadi}&,\: \textrm{nilai dari}\\ \cos \theta &=\displaystyle \frac{12}{13} \end{aligned}$.

$\begin{array}{ll}\\ 3.&\textrm{Perhatikanlah gambar berikut}\\ \end{array}$.

$.\qquad\begin{array}{ll}\\ &\textrm{Panjang BC adalah}\: ....\\ &\textrm{a}.\quad \displaystyle 20\sin 36^{\circ} \\ &\textrm{b}.\quad \displaystyle 20\cos 36^{\circ} \\ &\textrm{c}.\quad \color{red}\displaystyle 20\tan 36^{\circ} \\ &\textrm{d}.\quad \displaystyle 15\\ &\textrm{e}.\quad \displaystyle 16 \\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{Diketah}&\textrm{ui bahwa}\\ \tan 36^{\circ} &=\displaystyle \frac{BC}{20}\\ \Leftrightarrow &\: \color{red}BC\color{black}=20\tan 36^{\circ} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Nilai}\: \: \tan 300^{\circ} \: \: \: \textrm{adalah}\: \: ....\\\\ &\textrm{a}.\quad \displaystyle \sqrt{3} \\\\ &\textrm{b}.\quad \displaystyle \frac{1}{3}\sqrt{3} \\\\ &\textrm{c}.\quad \displaystyle -\frac{1}{3}\sqrt{3} \\\\ &\textrm{d}.\quad \displaystyle \frac{1}{2}\sqrt{3} \\\\ &\textrm{e}.\quad \displaystyle \color{red}-\sqrt{3} \\\\ &\textbf{Jawab}:\\ &\begin{aligned}\tan 300^{\circ}&=\tan \left ( 360^{\circ}-60^{\circ} \right )\\ &=-\tan 60^{\circ}\\ &=\color{red}-\sqrt{3}\\ \textbf{catatan}&: \textrm{ingat sudut berelasi} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Nilai}\: \: \tan 60^{\circ}-\sin 120^{\circ}-\tan 210^{\circ} \: \: \: \textrm{adalah}\: \: ....\\\\ &\textrm{a}.\quad \displaystyle \frac{1}{6}\sqrt{6} \\\\ &\textrm{b}.\quad \displaystyle \frac{1}{3}\sqrt{6} \\\\ &\textrm{c}.\quad \displaystyle -\frac{1}{2}\sqrt{6} \\\\ &\textrm{d}.\quad \displaystyle \frac{1}{3}\sqrt{3} \\\\ &\textrm{e}.\quad \displaystyle \color{red}\frac{1}{6}\sqrt{3} \\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\tan 60^{\circ}-\sin 120^{\circ}-\tan 210^{\circ}\\ &=\tan 60^{\circ}-\sin \left ( 180^{\circ}-60^{\circ} \right )-\tan \left ( 180^{\circ}+30^{\circ} \right )\\ &=\tan 60^{\circ}-\sin 60^{\circ}-\tan 30^{\circ}\\ &=\sqrt{3}-\displaystyle \frac{1}{2}\sqrt{3}-\displaystyle \frac{1}{3}\sqrt{3}\\ &=\left (1-\displaystyle \frac{1}{2}-\frac{1}{3} \right )\sqrt{3}\\ &=\displaystyle \color{red}\frac{1}{6}\sqrt{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Nilai}\: \: x \: \: \textrm{positif terkecil yang memenuhi}\\ &\sin x=-\displaystyle \frac{1}{2}\sqrt{3}\: \: \textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \displaystyle 30^{\circ} \\\\ &\textrm{b}.\quad \displaystyle 60^{\circ} \\\\ &\textrm{c}.\quad \displaystyle 120^{\circ} \\\\ &\textrm{d}.\quad \color{red}\displaystyle 240^{\circ} \\\\ &\textrm{e}.\quad \displaystyle 300^{\circ} \\\\ &\textbf{Jawab}:\\ &\begin{aligned}\sin x&=-\frac{1}{2}\sqrt{3}\\ \textrm{Gun}&\textrm{akan rumus persamaan}\\ &\textrm{sederhana, yaitu}:\\ \sin x&=-\sin 60^{\circ}\\ &=\sin \left ( 180^{\circ}+60^{\circ} \right )\\ &=\sin 240^{\circ}\\ x&=\color{red}240^{\circ} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Jika}\: \: \cos x=\displaystyle \frac{2\sqrt{5}}{5} \: \: \textrm{maka nilai}\\ &\cot x\left ( \displaystyle \frac{\pi }{2}-x \right )\: \: \textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \color{red}\displaystyle \frac{1}{2} \\\\ &\textrm{b}.\quad \displaystyle \frac{1}{3} \\\\ &\textrm{c}.\quad \displaystyle \frac{1}{6} \\\\ &\textrm{d}.\quad \displaystyle \frac{1}{7} \\\\ &\textrm{e}.\quad \displaystyle \frac{1}{8} \\\\ &\textbf{Jawab}:\\ &\begin{aligned}\cos x&=\frac{2\sqrt{5}}{5},\: \: \textrm{maka}\\ \sin^{2} x&+\cos ^{2}x=1\\ \sin x&=1-\cos ^{2}x\\ &=\sqrt{1-\cos ^{2}x}=\sqrt{1-\left (\displaystyle \frac{2\sqrt{5}}{5} \right )^{2}}\\ &=\sqrt{1-\displaystyle \frac{20}{25}}=\sqrt{\displaystyle \frac{5}{25}}=\displaystyle \frac{\sqrt{5}}{5}\\ \cot &\left ( \displaystyle \frac{\pi }{2}-x \right )=\tan x,\: \: \textrm{maka}\\ \tan x&=\displaystyle \frac{\sin x}{\cos x}\\ &=\displaystyle \frac{\sqrt{5}}{2\sqrt{5}}\\ &=\color{red}\displaystyle \frac{1}{2} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Pada setiap}\: \: \alpha \: \: \textrm{berlaku}\\ &\tan \alpha +\cos+\tan (-\alpha ) +\cos (-\alpha )=....\\ &\begin{array}{llllll}\\ \textrm{a}.&0\\ \textrm{b}.&2\tan \alpha \\ \textrm{c}.&\color{red}2\cos \alpha \\ \textrm{d}.&2\left ( \tan \alpha +\cos \alpha \right )\\ \textrm{e}.&2\\\\ &&\color{blue}\textbf{SAT Subjeck Test} \end{array}\\ &\\ &\textbf{Jawab}:\quad\color{red}\textbf{c}\\ &\begin{aligned}\tan \alpha &+\cos+\tan (-\alpha ) +\cos (-\alpha )\\ &=\tan \alpha +\cos-\tan \alpha +\cos \alpha \\ &=\color{red}2\cos \alpha \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Jika}\: \: \sin x=-\sin 35^{\circ}\: \: \textrm{untuk}\: \: 90^{\circ}\leq x\leq 270^{\circ}\\ &\textrm{maka nilai}\: \: x\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\color{red}215^{\circ}\\ \textrm{b}.&235^{\circ}\\ \textrm{c}.&240^{\circ}\\ \textrm{d}.&255^{\circ}\\ \textrm{e}.&270^{\circ} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{Diketahui}\: \textrm{bahwa}\: \: \sin x=-\sin 35^{\circ}\\ & \textrm{untuk}\: \: 90^{\circ}\leq x\leq 270^{\circ},\: \textrm{dengan}\\ &\color{blue}\sin x=-\sin 35^{\circ}\\ &\qquad \textrm{(hanya terjadi dikuadran III dan IV)}\\ &\textrm{karena ba}\textrm{tasnya}\: \textrm{hanya untuk kuadran III saja, }\\ &\textrm{maka}\\ &=\sin \left ( 180^{\circ}+35^{\circ} \right )\\ &=\color{red}\sin 215^{\circ} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Jika}\: \: \tan y=\tan 83^{\circ}\: \: \textrm{untuk}\: \: 90^{\circ}< y< 270^{\circ}\\ &\textrm{maka nilai}\: \: x\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&173^{\circ}\\ \textrm{b}.&187^{\circ}\\ \textrm{c}.&\color{red}263^{\circ}\\ \textrm{d}.&268^{\circ}\\ \textrm{e}.&293^{\circ} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\textrm{Diketahui}&\: \textrm{bahwa nilai}\\ \tan y&=\tan 83^{\circ}\: \: \textrm{untuk}\: \: 90^{\circ}< y< 270^{\circ}\\ \tan y&=\tan \left ( 180^{\circ}+83^{\circ} \right ),\\ & \textrm{karena berada dikuadran III}\\ &=\color{red}\tan 263^{\circ} \end{aligned} \end{array}$.