Soal dan jawaban Persiapan Semester gasal Kelas XI Matematika Peminatan Bagian Keempat

$\begin{array}{ll}\\ 16.&\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}\\ 17.&\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}\\ 18.&\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}$

$\begin{array}{ll}\\ 19.&\textrm{Diketahui}\: \: \alpha -\beta =\displaystyle \frac{\pi }{3}\: \: \textrm{dan}\: \: \sin \alpha \sin \beta =\displaystyle \frac{1}{4}\\ &\textrm{dengan}\: \: \alpha \: \: \textrm{dan}\: \: \beta \: \: \textrm{adalah sudut} \: \: \textbf{lancip}\\ &\textrm{Nilai dari}\: \: \cos \left ( \alpha +\beta \right )\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&1&&&\textrm{d}.&\displaystyle \frac{1}{4}\\ \textrm{b}.&\displaystyle \frac{3}{4}&\textrm{c}.&\displaystyle \frac{1}{2}&\textrm{e}.&\color{red}\displaystyle 0 \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &\bullet \quad\alpha -\beta =\displaystyle \frac{\pi }{3}\: \: \textrm{dan}\: \: \sin \alpha \sin \beta =\displaystyle \frac{1}{4}\\ &\bullet \quad\textrm{dengan}\: \: \alpha \: \: \textrm{dan}\: \: \beta \: \: \textrm{sudut} \: \: \textbf{lancip}\\ &\qquad \textrm{akibatnya semua sudut dikuadran I}\\ &\qquad \textrm{sehingga}\color{purple}\begin{cases} \sin & =+ \\ \cos & =+ \\ \tan & =+ \end{cases}\\ &\textrm{ditanya}\: \: \cos \left ( \alpha +\beta \right ),\: \: \textrm{maka}\\ &\textrm{sebagai langkah awal kita adalah}:\\ &\cos \left ( \alpha -\beta \right )=\cos \left ( \displaystyle \frac{\pi }{3} \right )\\ &\Leftrightarrow \: \cos \alpha \cos \beta +\sin \alpha \sin \beta =\displaystyle \frac{1}{2}\\ &\Leftrightarrow \: \cos \alpha \cos \beta+\displaystyle \frac{1}{4} =\displaystyle \frac{1}{2}\\ &\Leftrightarrow \: \cos \alpha \cos \beta =\displaystyle \frac{1}{2}-\displaystyle \frac{1}{4}=\displaystyle \frac{1}{4}\\ &\textbf{Selanjutnya nilai dari}\\ &\cos \left ( \alpha +\beta \right )\\ &=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ &=\displaystyle \frac{1}{4}-\displaystyle \frac{1}{4}=0\\ \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 20.&\textrm{Nilai}\: \: \sin 75^{\circ}-\sin 165^{\circ}\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \frac{1}{4}\sqrt{2}&&&\textrm{d}.&\color{red}\displaystyle \frac{1}{2}\sqrt{2}\\\\ \textrm{b}.&\displaystyle \frac{1}{4}\sqrt{3}&\textrm{c}.&\displaystyle \frac{1}{4}\sqrt{6}&\textrm{e}.&\displaystyle \frac{1}{2}\sqrt{6} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\sin 75^{\circ}-\sin 165^{\circ}\\ &=2\cos \left ( \displaystyle \frac{75^{\circ}+165^{\circ}}{2} \right )\sin \left ( \displaystyle \frac{75^{\circ}-165^{\circ}}{2} \right )\\ &=2\cos \displaystyle \frac{240^{\circ}}{2}\sin \left (-\displaystyle \frac{90^{\circ}}{2} \right ) \\ &=2\cos 120^{\circ}\sin \left ( -45^{\circ} \right )\\ &=2\left (-\cos 60^{\circ} \right )\left (- \sin 45^{\circ} \right )\\ &=2\left ( -\displaystyle \frac{1}{2} \right )\left ( -\displaystyle \frac{1}{2}\sqrt{2} \right )\\ &=\color{red}\displaystyle \frac{1}{2}\sqrt{2} \end{aligned} \end{array}$.

Soal dan jawaban Persiapan Semester gasal Kelas XI Matematika Peminatan Bagian Ketiga

$\begin{array}{ll}\\ 11.&\textrm{Grafik fungsi trigonometri pada gambar}\\ &\textrm{berikut adalah}\: .... \end{array}$.

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

$\begin{array}{ll}\\ 12.&\textrm{Grafik fungsi trigonometri pada gambar}\\ &\textrm{berikut adalah}\: .... \end{array}$.

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

$\begin{array}{ll}\\ 13.&\textrm{Himpunan penyelesaian dari persamaan}\\ &\sin x=\sin \displaystyle \frac{2}{10}\pi \: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \displaystyle \left \{ \displaystyle \frac{22}{10}\pi ,\displaystyle \frac{8}{10}\pi \right \} \\\\ &\textrm{b}.\quad \displaystyle \left \{ \displaystyle \frac{2}{10}\pi ,\displaystyle \frac{28}{10}\pi \right \} \\\\ &\textrm{c}.\quad \displaystyle \color{red}\left \{ \displaystyle \frac{2}{10}\pi ,\displaystyle \frac{8}{10}\pi \right \} \\\\ &\textrm{d}.\quad \displaystyle \left \{ \displaystyle \frac{22}{10}\pi ,\displaystyle \frac{28}{10}\pi \right \} \\\\ &\textrm{e}.\quad \displaystyle \left \{ \displaystyle \frac{12}{10}\pi ,\displaystyle \frac{8}{10}\pi \right \} \\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\begin{aligned} & \sin x=\sin \displaystyle \frac{2}{10}\pi \\ &\Leftrightarrow \: \: x_{1}=\displaystyle \frac{2}{10}\pi+k.2\pi \: \: \: \: \color{blue}\textrm{atau}\\ &\Leftrightarrow \quad x_{2} =\left (\pi -\displaystyle \frac{2}{10}\pi \right )+k.2\pi=\displaystyle \frac{8}{10}\pi +k.2\pi \\ &k=0\Rightarrow x_{1}=\displaystyle \frac{2}{10}\pi\: \: (\color{blue}\textrm{mm})\: \: \color{black}\textrm{atau}\\ &\qquad\qquad x_{2}=\displaystyle \frac{8}{10}\pi\: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x_{1,2}=....+2\pi \quad (\color{red}\textrm{tidak memenuhi})\\ \end{aligned} \\ &\textbf{HP}=\color{red}\left \{\displaystyle \frac{2}{10}\pi,\: \displaystyle \frac{8}{10}\pi \right \} \end{array} \end{array}$.

$\begin{array}{ll}\\ 14.&\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}$.

$\begin{array}{ll}\\ 15.&\textrm{Himpunan penyelesaian dari persamaan}\\ &\cos 2x-2\cos x=-1\: \: \textrm{untuk}\: \: 0< x< 2\pi \\ &\textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \color{red}\displaystyle \left \{ 0,\displaystyle \frac{1}{2}\pi ,\displaystyle \frac{3}{2}\pi, 2\pi \right \} \\\\ &\textrm{b}.\quad \displaystyle \left \{ 0,\displaystyle \frac{1}{2}\pi ,\displaystyle \frac{2}{3}\pi, 2\pi \right \} \\\\ &\textrm{c}.\quad \displaystyle \left \{ 0,\displaystyle \frac{1}{2}\pi ,\pi ,\displaystyle \frac{3}{2}\pi \right \} \\\\ &\textrm{d}.\quad \displaystyle \left \{ 0,\displaystyle \frac{1}{2}\pi ,\displaystyle \frac{2}{3}\pi \right \} \\\\ &\textrm{e}.\quad \displaystyle \left \{ 0,\displaystyle \frac{1}{2}\pi ,\pi \right \} \\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\begin{aligned} & \cos 2x-2\cos x=-1\\ &\Leftrightarrow \cos 2x-2\cos x+1=0\\ &\Leftrightarrow \left ( 2\cos ^{2}x-1 \right )-2\cos x+1=0 \\ &\Leftrightarrow 2\cos x\left ( \cos x-1 \right )=0\\ &\Leftrightarrow \cos x=0\: \: \textrm{atau}\: \: \cos x=1\\ &\Leftrightarrow \cos x=\cos \displaystyle \frac{1}{2}\pi \: \: \textrm{atau}\: \: \cos x=\cos 0\\ &\Leftrightarrow x_{1,2}=\pm \displaystyle \frac{1}{2}\pi +k.2\pi \: \: \textrm{atau}\: \: x_{3}=k.2\pi\\ &\textrm{maka}\\ &k=0\Rightarrow x_{1}=-\displaystyle \frac{1}{2}\pi\: \: (\color{red}\textrm{tm})\: \: \color{black}\textrm{atau}\\ &\qquad\qquad x_{2}=\displaystyle \frac{1}{2}\pi\: \: (\color{blue}\textrm{mm})\\ &\qquad\qquad x_{3}=\displaystyle 0\: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x_{1}=\displaystyle \frac{3}{2}\pi \quad (\color{blue}\textrm{mm})\\ &\qquad\qquad x_{2}=\displaystyle \frac{5}{2}\pi\: \: (\color{red}\textrm{tm})\\ &\qquad\qquad x_{3}=\displaystyle 2\pi\: \: (\color{blue}\textrm{mm})\\ \end{aligned} \end{array} \end{array}$

Soal dan jawaban Persiapan Semester gasal Kelas XI Matematika Peminatan Bagian Kedua

$\begin{array}{ll}\\ 6.&\textrm{Nilai dari}\: \: \displaystyle \frac{\sin 49^{\circ}}{\cos 41^{\circ}}-\displaystyle \frac{\cos 17^{\circ}}{\sin 73^{\circ}}\\ &\textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&-1&&&\textrm{d}.&0,143\\ \textrm{b}.&\displaystyle -0,321&\textrm{c}.&\color{red}0&\textrm{e}.&\displaystyle 0,321 \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\sin 49^{\circ}}{\cos 41^{\circ}}-\displaystyle \frac{\cos 17^{\circ}}{\sin 73^{\circ}}\\ &=\displaystyle \frac{\sin 49^{\circ}}{\cos \left (90^{\circ}-49^{\circ} \right )}-\displaystyle \frac{\cos 17^{\circ}}{\sin \left (90^{\circ}-17^{\circ} \right )}\\ &=\displaystyle \frac{\sin 49^{\circ}}{\sin 49^{\circ}}-\displaystyle \frac{\cos 17^{\circ}}{\cos 17^{\circ}}\\ &=1-1\\ &=0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Nilai dari}\\ &p=r\sin \alpha \cos \beta \\ &q=r\sin \alpha \sin \beta \\ &s=r\cos \alpha \\ &\textrm{maka pernyataan berikut yang}\\ &\textrm{tepat adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\color{red}p^{2}+t^{2}+s^{2}=r^{2}&&&\\ \textrm{b}.&p^{2}-t^{2}+s^{2}=r^{2} \\ \textrm{c}.&p^{2}+t^{2}-s^{2}=r^{2}&\\ \textrm{d}.&-p^{2}+t^{2}+s^{2}=r^{2}&\\ \textrm{e}.&-p^{2}-t^{2}+s^{2}=r^{2} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Saat}\\ &p^{2}+q^{2}\: \: \textrm{maka hasilnya adalah}\\ &\color{purple}\begin{array}{lll}\\ p^{2}&=r^{2}\sin^{2} \alpha \cos^{2} \beta&\\ q^{2}&=r^{2}\sin^{2} \alpha \sin^{2} \beta&+\\\hline &=r^{2}\sin ^{2}\alpha \left ( \cos ^{2}\beta +\sin ^{2}\beta \right )\\ &=r^{2}\sin ^{2}\alpha (1)\\ &=r^{2}\sin ^{2}\alpha \end{array}\\ &\textrm{Dan saat}\\ &p^{2}+q^{2}+s^{2}\: \: \textrm{akan diperoleh hasil}\\ &\color{purple}\begin{array}{lll}\\ p^{2}+q^{2}&=r^{2}\sin ^{2}\alpha &\\ \qquad s^{2}&=r^{2}\cos ^{2}\alpha &+\\\hline &=r^{2}\sin ^{2}\alpha+r^{2}\cos ^{2}\alpha\\ &=r^{2}\left ( \sin ^{2}\alpha+\cos ^{2}\alpha \right )\\ &=r^{2}(1)\\ &=r^{2} \end{array}\\ \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Nilai dari}\\ & \displaystyle \frac{\cos \left ( 90^{\circ}+\theta \right )\sec \left ( 2\pi -\theta \right )\tan \left ( \pi -\theta \right )}{\sec \left ( \theta -2\pi \right )\sin \left ( 540^{\circ}+\theta \right )\cot \left ( \theta -90^{\circ} \right )}\\ &\textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&-1&&&\textrm{d}.&-\tan \theta \\ \textrm{b}.&\displaystyle 0&\textrm{c}.&\color{red}1&\textrm{e}.&\displaystyle \tan \theta \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\color{purple}\textrm{Ingat kembali sudut-sudut}\\ &\color{purple}\textrm{yang berelasi dari kudran selain I}\\ &\color{purple}\textrm{ke kuadran I beserta tandanya}\\ &\displaystyle \frac{\cos \left ( 90^{\circ}+\theta \right )\sec \left ( 2\pi -\theta \right )\tan \left ( \pi -\theta \right )}{\sec \left ( \theta -2\pi \right )\sin \left ( 540^{\circ}+\theta \right )\cot \left ( \theta -90^{\circ} \right )}\\ &=\displaystyle \frac{\left (-\sin \theta \right ) .\sec \theta .\left (-\tan \theta \right )}{\sec \theta .\left (-\sin \theta \right ). \left (-\tan \theta \right )}\\ &=1 \end{aligned} \end{array}$.

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

$\begin{array}{ll}\\ 10.&\textrm{Jika diketahui}\: \: \: \displaystyle \frac{3}{2}\pi <x<2\pi \\ &\textrm{dan}\: \: \: \tan x=m,\\ &\textrm{maka nilai dari}\: \: \sin x \cos x \: \: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle -\frac{1}{m^{2}+1}&&&\textrm{d}.&\displaystyle -\frac{m}{m^{2}-1} \\\\ \textrm{b}.& \color{red}\displaystyle -\frac{m}{m^{2}+1}&\textrm{c}.&\displaystyle \frac{m}{m^{2}+1}&\textrm{e}.&\displaystyle \frac{m}{m^{2}-1} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\: \: \displaystyle \frac{3}{2}\pi <x<2\pi\\ &\textrm{ini daerah Kwadran IV, akibatnya adalah nilai}\\ &\begin{cases} \sin x & = -\\ \cos x & =+ \\ \tan x & =- \end{cases}\\ &\textbf{Selanjutnya ada pernyataan}\: \: \tan x=m\\ &\textrm{ini artinya}\: \: \tan x=\displaystyle \frac{m}{1}\\ &\textbf{Perhatikanlah ilustrasi gambar berikut} \end{aligned} \end{array}$.

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

Soal dan jawaban Persiapan Semester gasal Kelas XI Matematika Peminatan Bagian Pertama

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

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

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

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

$\begin{array}{ll}\\ 4.&\textrm{Nilai dari}\: \: \left ( \cos ^{2}17^{\circ}-\sin ^{2}73^{\circ} \right ) \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\color{red}0&&&\textrm{d}.&1\\ \textrm{b}.&\displaystyle \frac{1}{3}&\textrm{c}.&\displaystyle \frac{2}{\sqrt{3}}&\textrm{e}.&\displaystyle \frac{1}{2}\sqrt{3} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\left ( \cos ^{2}17^{\circ}-\sin ^{2}73^{\circ} \right )\\ &=\left ( \cos ^{2}17^{\circ}-\left (\sin 73^{\circ} \right )^{2} \right )\\ &=\left ( \cos ^{2}17^{\circ}-\left (\sin \left ( 90^{\circ}-17^{\circ} \right ) \right )^{2} \right )\\ &=\left ( \cos ^{2}17^{\circ}-\cos ^{2}17^{\circ} \right )\\ &=0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Jika diketahui}\\ & \displaystyle \frac{x\csc ^{2}30^{\circ}\sec ^{2}45^{\circ}}{8\cos ^{2}45^{\circ}\sin ^{2}60^{\circ}}=\tan ^{2}60^{\circ}-\tan ^{2}30^{\circ},\\ & \textrm{maka nilai}\: \: x\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&-2&&&\textrm{d}.&\color{red}1\\ \textrm{b}.&\displaystyle -1&\textrm{c}.&0&\textrm{e}.&\displaystyle 2 \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{x\csc ^{2}30^{\circ}\sec ^{2}45^{\circ}}{8\cos ^{2}45^{\circ}\sin ^{2}60^{\circ}}=\tan ^{2}60^{\circ}-\tan ^{2}30^{\circ}\\ &\displaystyle \frac{x\left ( 4 \right )\left ( \displaystyle \frac{4}{2} \right )}{8\left ( \displaystyle \frac{2}{4} \right )\left ( \displaystyle \frac{3}{4} \right )}=3-\left ( \displaystyle \frac{1}{3} \right )\\ &\displaystyle \frac{8x}{3}=\displaystyle \frac{8}{3}\\ &x=1 \end{aligned} \end{array}$

Lanjutan 6 Materi Rumus Jumlah dan Selisih Sinus dan Cosinus

Untuk rumus jum lah dan selisih sinus dan cosinus untuk sudut berupa X dan Y adalah sebagai berikut:

$\begin{cases} \sin X+\sin Y & =2\sin \frac{1}{2}\left (X+Y \right )\cos \frac{1}{2}\left ( X-Y \right ) \\ \sin X-\sin Y & =2\cos \frac{1}{2}\left (X+Y \right )\sin \frac{1}{2}\left ( X-Y \right )\\ \cos X+\cos Y & =2\cos \frac{1}{2}\left (X+Y \right )\cos \frac{1}{2}\left ( X-Y \right ) \\ \cos X-\cos Y & =-2\sin \frac{1}{2}\left (X+Y \right )\sin \frac{1}{2}\left ( X-Y \right ) \end{cases}$.

Berikut contoh proses pembuktian rumus no.1, yaitu:

$\begin{array}{lll}\\ \sin \left (\alpha +\beta \right )=\sin \alpha \cos \beta +\cos \alpha \sin \beta &&\\ \sin \left (\alpha -\beta \right )=\sin \alpha \cos \beta -\cos \alpha \sin \beta &+&\\\hline \sin \left (\alpha +\beta \right )+\sin \left (\alpha -\beta \right )=2\sin \alpha \cos \beta &&\\\\ \color{red}\textrm{saat}\quad\color{blue}X=\alpha +\beta \qquad\qquad X=\alpha +\beta &&\\ \qquad\quad Y=\alpha -\beta\quad +\qquad Y=\alpha -\beta &-&\\\hline \qquad\quad X+Y=2\alpha\qquad\quad X-Y=2\beta &&\\ \qquad\quad 2\alpha =X+Y\qquad\quad 2\beta =X-Y&&\\ \qquad\quad \alpha =\displaystyle \frac{X+Y}{2}\qquad\quad \beta =\displaystyle \frac{X-Y}{2}\\ \textbf{Sehingga rumus akan menjadi}&&\\ \sin X+\sin Y=2\sin \displaystyle \frac{(X+Y)}{2}\cos \displaystyle \frac{(X-Y)}{2} \end{array}$.

Dan berikut proses pembuktian rumus no. 2, yaitu:

$\begin{array}{lll}\\ \sin \left (\alpha +\beta \right )=\sin \alpha \cos \beta +\cos \alpha \sin \beta &&\\ \sin \left (\alpha -\beta \right )=\sin \alpha \cos \beta -\cos \alpha \sin \beta &-&\\\hline \sin \left (\alpha +\beta \right )-\sin \left (\alpha -\beta \right )=2\cos \alpha \sin \beta &&\\\\ \color{red}\textrm{saat}\quad\color{blue}X=\alpha +\beta \qquad\qquad X=\alpha +\beta &&\\ \qquad\quad Y=\alpha -\beta\quad +\qquad Y=\alpha -\beta &-&\\\hline \qquad\quad X+Y=2\alpha\qquad\quad X-Y=2\beta &&\\ \qquad\quad 2\alpha =X+Y\qquad\quad 2\beta =X-Y&&\\ \qquad\quad \alpha =\displaystyle \frac{X+Y}{2}\qquad\quad \beta =\displaystyle \frac{X-Y}{2}\\ \textbf{Sehingga rumus akan menjadi}&&\\ \sin X-\sin Y=2\cos \displaystyle \frac{(X+Y)}{2}\sin \displaystyle \frac{(X-Y)}{2} \end{array}$

Berikut untuk proses pembktian rumus no. 3, yaitu:

$\begin{array}{lll}\\ \cos \left (\alpha +\beta \right )=\cos \alpha \cos \beta -\sin \alpha \sin \beta &&\\ \cos \left (\alpha -\beta \right )=\cos \alpha \cos \beta +\sin \alpha \sin \beta &+&\\\hline \cos \left (\alpha +\beta \right )+\cos \left (\alpha -\beta \right )=2\cos \alpha \cos \beta &&\\\\ \color{red}\textrm{saat}\quad\color{blue}X=\alpha +\beta \qquad\qquad X=\alpha +\beta &&\\ \qquad\quad Y=\alpha -\beta\quad +\qquad Y=\alpha -\beta &-&\\\hline \qquad\quad X+Y=2\alpha\qquad\quad X-Y=2\beta &&\\ \qquad\quad 2\alpha =X+Y\qquad\quad 2\beta =X-Y&&\\ \qquad\quad \alpha =\displaystyle \frac{X+Y}{2}\qquad\quad \beta =\displaystyle \frac{X-Y}{2}\\ \textbf{Sehingga rumus akan menjadi}&&\\ \cos X+\cos Y=2\cos \displaystyle \frac{(X+Y)}{2}\cos \displaystyle \frac{(X-Y)}{2} \end{array}$

Dan berikut tyang terakhir untuk bukti pada rumus terakhir no.4, yaitu:

$\begin{array}{lll}\\ \cos \left (\alpha +\beta \right )=\cos \alpha \cos \beta -\sin \alpha \sin \beta &&\\ \cos \left (\alpha -\beta \right )=\cos \alpha \cos \beta +\sin \alpha \sin \beta &-&\\\hline \cos \left (\alpha +\beta \right )-\cos \left (\alpha -\beta \right )=-2\cos \alpha \sin \beta &&\\\\ \color{red}\textrm{saat}\quad\color{blue}X=\alpha +\beta \qquad\qquad X=\alpha +\beta &&\\ \qquad\quad Y=\alpha -\beta\quad +\qquad Y=\alpha -\beta &-&\\\hline \qquad\quad X+Y=2\alpha\qquad\quad X-Y=2\beta &&\\ \qquad\quad 2\alpha =X+Y\qquad\quad 2\beta =X-Y&&\\ \qquad\quad \alpha =\displaystyle \frac{X+Y}{2}\qquad\quad \beta =\displaystyle \frac{X-Y}{2}\\ \textbf{Sehingga rumus akan menjadi}&&\\ \cos X-\cos Y=-2\sin \displaystyle \frac{(X+Y)}{2}\sin \displaystyle \frac{(X-Y)}{2} \end{array}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Sederhanakanlah bentuk berikut ini}\\ &\begin{array}{ll}\\ \textrm{a}.&\sin \left ( 2x+60^{\circ} \right )-\sin \left ( 2x-60^{\circ} \right )\\ \textrm{b}.&\sin \left ( x+\displaystyle \frac{1}{4}m \right )+\sin \left ( x-\displaystyle \frac{1}{4}m \right )\\ \textrm{c}.&\cos \left ( 2x+4y \right )-\cos \left ( 2x-4y \right )\\ \textrm{d}.&\cos \left ( \displaystyle \frac{5}{4}m+3x \right )+\cos \left ( \displaystyle \frac{5}{4}m-3x \right ) \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\\\ &\begin{aligned}\textrm{a}.\quad &\sin \left ( 2x+60^{\circ} \right )-\sin \left ( 2x-60^{\circ} \right )\\ &=2\cos \displaystyle \frac{\left (2x+60^{\circ} \right )+\left ( 2x-60^{\circ} \right )}{2}\sin \displaystyle \frac{\left (2x+60^{\circ} \right )-\left ( 2x-60^{\circ} \right )}{2}\\ &=2\cos \displaystyle \frac{4x}{2}\sin \displaystyle \frac{120^{\circ}}{2}\\ &=2\cos 2x\sin 60^{\circ}\\ &=2\cos 2x.\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\\ &=\sqrt{3}\cos 2x \end{aligned} \\ &\begin{aligned}\textrm{b}.\quad &\sin \left ( x+\displaystyle \frac{1}{4}m \right )+\sin \left ( x-\displaystyle \frac{1}{4}m \right )\\ &=2\sin \displaystyle \frac{\left (x+\displaystyle \frac{1}{4}m+x-\frac{1}{4}m \right )}{2}\cos \displaystyle \frac{\left (x+\displaystyle \frac{1}{4}m-\left (x-\frac{1}{4}m \right ) \right )}{2}\\ &=2\sin \displaystyle \frac{2x}{2}\cos \displaystyle \frac{\displaystyle \frac{2}{4}m}{2}\\ &=2\sin x\cos \displaystyle \frac{1}{4}m \end{aligned} \\ &\begin{aligned}\textrm{c}.\quad &\cos \left ( 2x+4y \right )-\cos \left ( 2x-4y \right )\\ &=-2\sin \displaystyle \frac{\left (2x+4y \right )+\left ( 2x-4y \right )}{2}\sin \displaystyle \frac{\left (2x+4y \right )-\left ( 2x-4y \right )}{2}\\ &=-2\sin \displaystyle \frac{4x}{2}\sin \displaystyle \frac{8y}{2}\\ &=-2\sin 2x\sin 4y\\ \end{aligned} \\ &\begin{aligned}\textrm{d}.\quad &\cos \left ( \displaystyle \frac{5}{4}m+3x \right )+\cos \left ( \displaystyle \frac{5}{4}m-3x \right )\\ &=2\cos \displaystyle \frac{\left (\displaystyle \frac{5}{4}m+3x+\displaystyle \frac{5}{4}m-3x \right )}{2}\cos \displaystyle \frac{\left (\displaystyle \frac{5}{4}m+3x-\left (\displaystyle \frac{5}{4}m-3x \right ) \right )}{2}\\ &=2\cos \displaystyle \frac{5m}{2}\cos \displaystyle \frac{6x}{2}\\ &=2\cos \displaystyle \frac{5x}{2}\cos 3x\\ \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Buktikanlah bahwa}\\ &\begin{array}{llllllll}\\ \displaystyle \frac{\sin 3\gamma +\sin \gamma }{\cos 3\gamma +\cos \gamma }=\tan 2\gamma \end{array}\\\\ &\textbf{Bukti}\\ &\begin{array}{|l|l|}\hline \begin{aligned}\displaystyle \frac{\sin 3\alpha +\sin \alpha }{\cos 3\alpha +\cos \alpha }&=\color{red}\displaystyle \frac{2\sin \displaystyle \frac{\left (3\alpha +\alpha \right )}{2}\cos \displaystyle \frac{\left ( 3\alpha -\alpha \right )}{2}}{2\cos \displaystyle \frac{\left ( 3\alpha +\alpha \right )}{2}\cos \displaystyle \frac{\left ( 3\alpha -\alpha \right )}{2}}\\ &=\displaystyle \frac{\sin 2\alpha }{\cos 2\alpha }\\ &=\tan 2\alpha \qquad \blacksquare\\ &\\ &\\ &\quad\qquad\qquad\textbf{atau}\Rightarrow\qquad \\ &\textrm{mungkin lumayan rumit}\\ &\\ & \end{aligned}&\begin{aligned}\displaystyle \frac{\sin 3\alpha +\sin \alpha }{\cos 3\alpha +\cos \alpha }&=\displaystyle \frac{3\sin \alpha -4\sin ^{3}\alpha +\sin \alpha }{4\cos ^{3}\alpha -3\cos \alpha +\cos \alpha }\\ &=\displaystyle \frac{4\sin \alpha -4\sin ^{3}\alpha }{4\cos ^{3}\alpha -2\cos \alpha }\\ &=\displaystyle \frac{4\sin \alpha \left ( 1-\sin ^{2}\alpha \right )}{2\cos \alpha \left ( 2\cos ^{2}\alpha -1 \right )}=\displaystyle \frac{4\sin \alpha }{2\cos \alpha }\times \displaystyle \frac{\cos ^{2}\alpha }{\cos 2\alpha }\\ &=\displaystyle \frac{2\tan \alpha \cos ^{2}\alpha }{\cos 2\alpha }\\ &=\displaystyle \frac{2\tan \alpha \cos ^{2}\alpha }{\cos^{2}\alpha -\sin ^{2}\alpha }\\ &=\displaystyle \frac{2\tan \alpha }{\displaystyle \frac{\cos ^{2}\alpha }{\cos ^{2}\alpha }-\displaystyle \frac{\sin ^{2}\alpha }{\cos ^{2}\alpha }}=\displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha }\\ &=\tan 2\alpha \qquad \blacksquare \end{aligned} \\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Buktikanlah bahwa}\\ &\begin{array}{lll}\\ \displaystyle \frac{\sin 5\theta -\sin 3\theta }{\cos 3\theta +\cos 5\theta }=\tan \theta \end{array}\\\\ &\textbf{Bukti}\\ &\begin{aligned}\displaystyle \frac{\sin 5\theta -\sin 3\theta }{\cos 3\theta +\cos 5\theta}&=\color{red}\displaystyle \frac{2\cos \displaystyle \frac{\left (5\theta +3\theta \right )}{2}\sin \displaystyle \frac{\left ( 5\theta -3\theta \right )}{2}}{2\cos \displaystyle \frac{\left ( 3\theta +5\theta \right )}{2}\cos \displaystyle \frac{\left ( 3\theta -5\theta \right )}{2}}\\ &=\displaystyle \frac{\sin \theta }{\cos (-\theta ) }=\displaystyle \frac{\sin \theta }{\cos \theta }\\ &=\tan \theta \qquad \blacksquare\\\\ \textbf{Catatan}:\quad&\textrm{ingat bahwa saat}\: \: \cos\: \: \textrm{sudut}\\ &-\theta=\theta,\: \: \textrm{sehingga}\: \: \cos (-\theta )=\cos \theta \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Buktikanlah bahwa}\\ &\begin{array}{lll}\\ \displaystyle \frac{\sin 3\beta -\sin \beta }{\cos \beta -\cos 3\beta }=\cot 2\beta \end{array}\\\\ &\textbf{Bukti}\\ &\begin{aligned}\displaystyle \frac{\sin 3\beta -\sin \beta }{\cos \beta -\cos 3\beta}&=\color{red}\displaystyle \frac{2\cos \displaystyle \frac{\left (3\beta +\beta \right )}{2}\sin \displaystyle \frac{\left ( 3\beta -\beta \right )}{2}}{-2\sin \displaystyle \frac{\left ( \beta +3\beta \right )}{2}\sin \displaystyle \frac{\left ( \beta -3\beta \right )}{2}}\\ &=\displaystyle \frac{\cos 2\beta \times \sin \beta }{\sin 2\beta \left ( -\sin (-\beta ) \right ) }=\displaystyle \frac{\cos 2\beta }{\sin 2\beta }\\ &=\cot 2\beta \qquad \blacksquare\\\\ \textbf{Catatan}:\quad&\textrm{ingat bahwa saat}\: \: \sin\: \: \textrm{sudut}\\ &-\beta =-\beta ,\: \: \textrm{sehingga}\: \: \sin (-\beta )=-\sin \beta \\ &\textrm{dan}\: \: -\sin (-\beta )=-\left ( -\sin \beta \right )=\sin \beta \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Tentukan nilai dari}\\ &\begin{array}{ll}\\ \textrm{a}.&\cos 105^{\circ}+\cos 15^{\circ}\\ \textrm{b}.&\sin 105^{\circ}-\sin 15^{\circ}\end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{a}.\quad&\cos 105^{\circ}+\cos 15^{\circ}\\ &=2\cos \left ( \displaystyle \frac{105^{\circ}+15^{\circ}}{2} \right )\cos \left ( \displaystyle \frac{105^{\circ}-15^{\circ}}{2} \right )\\ &=2\cos \displaystyle \frac{120^{\circ}}{2}\cos \displaystyle \frac{90^{\circ}}{2}\\ &=2\cos 60^{\circ}\cos 45^{\circ}\\ &=2\left ( \displaystyle \frac{1}{2} \right )\left ( \displaystyle \frac{1}{2}\sqrt{2} \right )\\ &=\color{red}\displaystyle \frac{1}{2}\sqrt{2} \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad&\sin 105^{\circ}-\sin 15^{\circ}\\ &=2\cos \left ( \displaystyle \frac{105^{\circ}+15^{\circ}}{2} \right )\sin \left ( \displaystyle \frac{105^{\circ}-15^{\circ}}{2} \right )\\ &=2\cos \displaystyle \frac{120^{\circ}}{2}\sin \displaystyle \frac{90^{\circ}}{2}\\ &=2\cos 60^{\circ}\sin 45^{\circ}\\ &=2\left ( \displaystyle \frac{1}{2} \right )\left ( \displaystyle \frac{1}{2}\sqrt{2} \right )\\ &=\color{red}\displaystyle \frac{1}{2}\sqrt{2} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Tentukan nilai dari}\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle \frac{\cos 75^{\circ}+\cos 15^{\circ}}{\sin 75^{\circ}-\sin 15^{\circ}}\\\\ \textrm{b}.&\displaystyle \frac{\cos 195^{\circ}-\cos 105^{\circ}}{\sin 105^{\circ}-\sin 15^{\circ}} \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{a}.\quad&\displaystyle \frac{\cos 75^{\circ}+\cos 15^{\circ}}{\sin 75^{\circ}-\sin 15^{\circ}}\\ &=\displaystyle \frac{2\cos \left ( \displaystyle \frac{75^{\circ}+15^{\circ}}{2} \right )\cos \left ( \displaystyle \frac{75^{\circ}-15^{\circ}}{2} \right )}{2\cos \left ( \displaystyle \frac{75^{\circ}+15^{\circ}}{2} \right )\sin \left ( \displaystyle \frac{75^{\circ}-15^{\circ}}{2} \right )}\\ &=\displaystyle \frac{\cos \displaystyle \frac{90^{\circ}}{2}\cos \displaystyle \frac{60^{\circ}}{2}}{\cos \displaystyle \frac{90^{\circ}}{2}\sin \displaystyle \frac{60^{\circ}}{2}}\\ &=\displaystyle \frac{\cos 30^{\circ}}{\sin 30^{\circ}}=\cot 30^{\circ}\\ &=\color{red}\sqrt{3} \end{aligned} \\ &\begin{aligned}\textrm{b}.\quad&\displaystyle \frac{\cos 195^{\circ}-\cos 105^{\circ}}{\sin 105^{\circ}-\sin 15^{\circ}}\\ &=\displaystyle \frac{-2\sin \left ( \displaystyle \frac{195^{\circ}+105^{\circ}}{2} \right )\sin \left ( \displaystyle \frac{195^{\circ}-105^{\circ}}{2} \right )}{2\cos \left ( \displaystyle \frac{105^{\circ}+15^{\circ}}{2} \right )\sin \left ( \displaystyle \frac{105^{\circ}-15^{\circ}}{2} \right )}\\ &=-\displaystyle \frac{\sin \displaystyle \frac{300^{\circ}}{2}\sin \displaystyle \frac{90^{\circ}}{2}}{\cos \displaystyle \frac{120^{\circ}}{2}\sin \displaystyle \frac{90^{\circ}}{2}}\\ &=-\displaystyle \frac{\sin 150^{\circ}}{\cos 60^{\circ}}=-\displaystyle \frac{\sin \left ( 180^{\circ}-30^{\circ} \right )}{\cos 60^{\circ}}\\ &=-\displaystyle \frac{\sin 30^{\circ}}{\cos 60^{\circ}}\\ &=-\displaystyle \frac{\displaystyle \frac{1}{2}}{\displaystyle \frac{1}{2}}\\ &=\color{red}-1 \end{aligned} \end{array}$.

DAFTAR PUSTAKA

1. Noormandiri, B.K. 2017. Matematika untuk Kelas SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA
2. Sukino. 2017. Matematika untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.

Lanjutan 5 Materi Rumus Perkalian Sinus dan Cosinus

C.   Rumus Perkalian Sinus dan Cosinus

Untuk sudut  $\alpha \: \: \textrm{dan}\: \: \beta$ berlaku rumus-rumus

$\begin{aligned}&\begin{cases} 2\sin \alpha \cos \gamma &=\sin \left ( \alpha +\gamma \right )+\sin \left ( \alpha -\gamma \right )\\ 2\cos \alpha \sin \gamma &=\sin \left ( \alpha +\gamma \right )-\sin \left ( \alpha -\gamma \right ) \end{cases}\\ &\begin{cases} 2\cos \alpha \cos \gamma &=\cos \left ( \alpha +\gamma \right )+\cos \left ( \alpha -\gamma \right )\\ - 2\sin \alpha \sin \gamma &=\cos \left ( \alpha +\gamma \right )-\cos \left ( \alpha -\gamma \right ) \end{cases} \end{aligned}$.

Sebagai buktinya akan ditunjukkan pada tulisan berikut

Bukti untuk nomor pertama

$\begin{aligned}\sin &\color{blue}\left ( \alpha +\beta \right )+\sin \left ( \alpha -\beta \right )\\ &=\sin \alpha \cos \beta +\cos \alpha \sin \beta \\ &\quad +\sin \alpha \cos \beta -\cos \alpha \sin \beta \\ &=\color{red}2\sin \alpha \cos \beta \end{aligned}$.

Selanjutnya untuk bukti baris kedua yaitu:

$\begin{aligned}\sin &\color{blue}\left ( \alpha +\beta \right )-\sin \left ( \alpha -\beta \right )\\ &=\sin \alpha \cos \beta +\cos \alpha \sin \beta \\ &\quad -\left (\sin \alpha \cos \beta -\cos \alpha \sin \beta \right ) \\ &=\color{red}2\cos \alpha \sin \beta \end{aligned}$.

Dan bukti untuk rumus ketiga

$\begin{aligned}\cos &\color{blue}\left ( \alpha +\beta \right )+\cos \left ( \alpha -\beta \right )\\ &=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ &\quad +\cos \alpha \cos \beta+\sin \alpha \sin \beta \\ &=\color{red}2\cos \alpha \cos \beta \end{aligned}$.

Adapun bukti untuk rumus yang tertakhir adalah:

$\begin{aligned}\cos &\color{blue}\left ( \alpha +\beta \right )-\cos \left ( \alpha -\beta \right )\\ &=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ &\quad -\left (\cos \alpha \cos \beta +\sin \alpha \sin \beta \right ) \\ &=\color{red}-2\sin \alpha \sin \beta \end{aligned}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Nyatakanlah ke dalam bentuk jumlah}\\ &\textrm{atau selisih dari bentuk berikut}\\ &\textrm{a}.\quad 2\cos 80^{\circ}\sin 50^{\circ}\\ &\textrm{b}.\quad 2\sin 80^{\circ}\cos 50^{\circ}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&2\cos 80^{\circ}\sin 50^{\circ}\\ &=\sin \left ( 80^{\circ}+50^{\circ} \right )-\sin \left ( 80^{\circ}-50^{\circ} \right )\\ &=\sin 130^{\circ}-\sin 30^{\circ}\\ &=\sin 130^{\circ}-\displaystyle \frac{1}{2}\\ \textrm{b}.\quad&2\sin 80^{\circ}\cos 50^{\circ}\\ &=\sin \left ( 80^{\circ}+50^{\circ} \right )+\sin \left ( 80^{\circ}-50^{\circ} \right )\\ &=\sin 130^{\circ}+\sin 30^{\circ}\\ &=\sin 130^{\circ}+\displaystyle \frac{1}{2} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Nyatakanlah ke dalam bentuk jumlah}\\ &\textrm{atau selisih dari bentuk berikut}\\ &\textrm{a}.\quad 2\cos 75^{\circ}\cos 15^{\circ}\\ &\textrm{b}.\quad \sin \left ( \displaystyle \frac{3\pi }{8} \right )\sin \left ( \displaystyle \frac{\pi }{8} \right )\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&2\cos 75^{\circ}\cos 15^{\circ}\\ &=\cos \left ( 75^{\circ}+15^{\circ} \right )+\cos \left ( 75^{\circ}-15^{\circ} \right )\\ &=\cos 90^{\circ}+\cos 60^{\circ}\\ &=0+\displaystyle \frac{1}{2}=\color{red}\displaystyle \frac{1}{2}\\ \textrm{b}.\quad&\sin \left ( \displaystyle \frac{3\pi }{8} \right )\sin \left ( \displaystyle \frac{\pi }{8} \right )\\ &=-\displaystyle \frac{1}{2}\left ( 2\sin \left ( \displaystyle \frac{3\pi }{8} \right )\sin \left ( \displaystyle \frac{\pi }{8} \right ) \right )\\ &=-\displaystyle \frac{1}{2}\left (\cos \left ( \displaystyle \frac{3\pi }{8}+\frac{\pi }{8} \right )-\cos \left ( \displaystyle \frac{3\pi }{8}-\frac{\pi }{8} \right ) \right )\\ &=-\displaystyle \frac{1}{2}\left ( \cos \displaystyle \frac{4\pi }{8}-\cos \displaystyle \frac{2\pi }{8} \right )\\ &=-\displaystyle \frac{1}{2}\left ( \cos \displaystyle \frac{1}{2}\pi -\cos \displaystyle \frac{1}{4}\pi \right )\\ &=-\displaystyle \frac{1}{2}\left ( 0-\displaystyle \frac{1}{2}\sqrt{2} \right )\\ &=-\displaystyle \frac{1}{2}\left ( -\displaystyle \frac{1}{2}\sqrt{2} \right )\\ &=\color{red}\displaystyle \frac{1}{4}\sqrt{2} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Tentukan nilai dari bentuk berikut}\\\\ &\textrm{a}.\quad 2\sin 37\displaystyle \frac{1}{2}^{\circ}\cos 7\displaystyle \frac{1}{2}^{\circ}\\\\ &\textrm{b}.\quad 2\cos 82\displaystyle \frac{1}{2}^{\circ}\sin 37\displaystyle \frac{1}{2}^{\circ}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&2\sin 37\displaystyle \frac{1}{2}^{\circ}\cos 7\displaystyle \frac{1}{2}^{\circ}\\ &=\sin \left (37\displaystyle \frac{1}{2}+7\displaystyle \frac{1}{2} \right )^{\circ}+\displaystyle \sin \left ( 37\displaystyle \frac{1}{2}-7\displaystyle \frac{1}{2} \right )^{\circ}\\ &=\sin 45^{\circ}+\sin 30^{\circ}\\ &=\displaystyle \frac{1}{2}\sqrt{2}+\displaystyle \frac{1}{2}\\ &=\displaystyle \frac{1}{2}\left ( \sqrt{2}+1 \right )\\ \textrm{b}.\quad&2\cos 82\displaystyle \frac{1}{2}^{\circ}\sin 37\displaystyle \frac{1}{2}^{\circ}\\ &=\sin \left (82\displaystyle \frac{1}{2}+37\displaystyle \frac{1}{2} \right )^{\circ}-\displaystyle \sin \left ( 82\displaystyle \frac{1}{2}-37\displaystyle \frac{1}{2} \right )^{\circ}\\ &=\sin 120^{\circ}-\sin 45^{\circ}\\ &=\sin \left ( 180^{\circ}-60^{\circ} \right )-\sin 45^{\circ}\\ &=\sin 60^{\circ}-\sin 45^{\circ}\\ &=\displaystyle \frac{1}{2}\sqrt{3}-\frac{1}{2}\sqrt{2}\\ &=\displaystyle \frac{1}{2}\left ( \sqrt{3}-\sqrt{2} \right ) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Tunjukkan bahwa}\\ &\textrm{a}.\quad 2\cos \left ( \displaystyle \frac{1}{4}\pi +\theta \right )\cos \left ( \displaystyle \frac{3}{4}\pi -\theta \right )=\sin 2\theta -1\\ &\textrm{b}.\quad 2\sin \left ( 315^{\circ}+B \right )\sin \left ( 45^{\circ}-B \right )=\sin 2B-1\\\\ &\color{blue}\textbf{Bukti}:\\ &\begin{aligned}\textrm{a}.\quad&2\cos \left ( \displaystyle \frac{1}{4}\pi +\theta \right )\cos \left ( \displaystyle \frac{3}{4}\pi -\theta \right )\\ &=\cos \left ( \displaystyle \frac{1}{4}\pi +\theta +\displaystyle \frac{3}{4}\pi -\theta \right )+\cos \left ( \displaystyle \frac{1}{4}\pi +\theta -\left (\displaystyle \frac{3}{4}\pi -\theta \right ) \right )\\ &=\cos \left ( \displaystyle \frac{1}{4}\pi +\theta +\displaystyle \frac{3}{4}\pi -\theta \right )+\cos \left ( \displaystyle \frac{1}{4}\pi +\theta -\displaystyle \frac{3}{4}\pi +\theta \right )\\ &=\cos \pi +\cos \left ( 2\theta -\displaystyle \frac{1}{2}\pi \right )\\ &=-1+\cos \left ( \displaystyle \frac{1}{2}\pi -2\theta \right )\\ &=-1+\sin 2\theta \qquad (\textbf{ingat sudut yang berelasi})\\ &=\color{red}\sin 2\theta -1\qquad\quad \color{black}\blacksquare\\ \textrm{b}.\quad&2\sin \left ( 315^{\circ}+B \right )\sin \left ( 45^{\circ}-B \right )\\ &=-\left (\cos \left ( 315^{\circ}+B+45^{\circ}-B \right )-\cos \left ( 315^{\circ}+B-\left ( 45^{\circ}-B \right ) \right ) \right )\\ &=-\left ( \cos 360^{\circ}-\cos \left ( 315^{\circ}-45^{\circ}+B+B \right ) \right )\\ &-=-\left ( 1-\cos \left (270^{\circ}+2B \right ) \right )\\ &=-\left (1-\sin 2B \right )\qquad (\textbf{ingat sudut yang berelasi})\\ &=\color{red}\sin 2B -1\qquad\quad \color{black}\blacksquare \end{aligned} \end{array}$

Lanjutan 4 Materi Rumus Sudut Paruh pada Trigonometri

Sebelumnya telah dituliskan bahwa

$\begin{array}{ll}\\ \bullet &\sin \alpha =2\sin \displaystyle \frac{1}{2}\alpha \cos \frac{1}{2}\alpha \\ \bullet &\cos \alpha =\cos ^{2}\displaystyle \frac{1}{2}\alpha -\sin ^{2}\displaystyle \frac{1}{2}\alpha \\ \bullet &\tan \alpha =\displaystyle \frac{2\tan \displaystyle \frac{1}{2}\alpha }{1-\tan ^{2}\displaystyle \frac{1}{2}\alpha } \end{array}$.

 B. 2. 1   Rumus Sudut Paruh untuk Sinus

Perhatikanlah Identitas trigonometri berikut

Perhatikanlah ilsutrasi segitiga ABC berikut

• $\sin ^{2}\alpha +\cos ^{2}\alpha =1$
• $\tan ^{2}\alpha -\sec ^{2}\alpha =-1$
• $\cot ^{2}\alpha -\csc ^{2}\alpha =-1$

Serta pada pembahasan sebelumnya untuk sudut rangkap bahwa

• $\sin 2\alpha =2\sin \alpha \cos \alpha$
• $\cos 2\alpha =\cos ^{2}\alpha -\sin ^{2} \alpha$, dan 
• $\cos 2\alpha =2\cos ^{2}\alpha -1$, serta
• $\cos 2\alpha =1 -2\sin ^{2} \alpha$

Maka rumus $\color{blue}\sin \displaystyle \frac{1}{2}\alpha$ adalah:

$\begin{aligned}\cos 2\alpha &=1 -2\sin ^{2} \alpha\\ 2\sin ^{2}\alpha &=1-\cos 2\alpha \\ \sin ^{2}\alpha &=\displaystyle \frac{1-\cos 2\alpha }{2}\\ \sin \alpha &=\sqrt{\displaystyle \frac{1-\cos 2\alpha }{2}}\\ \textrm{saat}\: \: &\textrm{sudut}\: \: \alpha \: \: \textrm{diganti}\: \: \displaystyle \frac{1}{2}\alpha ,\\ \textrm{maka}&\: \textrm{rumus akan menjadi}\\ \sin \left (\displaystyle \frac{1}{2}\alpha \right ) &=\sqrt{\displaystyle \frac{1-\cos 2\left (\displaystyle \frac{1}{2}\alpha \right ) }{2}}\\ \sin \displaystyle \frac{1}{2}\alpha &=\pm \sqrt{\color{red}\displaystyle \frac{1-\cos \alpha }{2}} \end{aligned}$.

B. 2. 2   Rumus Sudut Paruh untuk Cosinus

Sedangkan untuk rumus cosinusnya adalah:

$\begin{aligned}\cos 2\alpha &=2\cos ^{2} \alpha-1\\ 2\cos ^{2}\alpha &=1+\cos 2\alpha \\ \cos ^{2}\alpha &=\displaystyle \frac{1+\cos 2\alpha }{2}\\ \cos \alpha &=\sqrt{\displaystyle \frac{1+\cos 2\alpha }{2}}\\ \textrm{saat}\: \: &\textrm{sudut}\: \: \alpha \: \: \textrm{diganti}\: \: \displaystyle \frac{1}{2}\alpha ,\\ \textrm{maka}&\: \textrm{rumus akan menjadi}\\ \cos \left (\displaystyle \frac{1}{2}\alpha \right ) &=\sqrt{\displaystyle \frac{1+\cos 2\left (\displaystyle \frac{1}{2}\alpha \right ) }{2}}\\ \cos \displaystyle \frac{1}{2}\alpha &=\pm \sqrt{\color{red}\displaystyle \frac{1+\cos \alpha }{2}} \end{aligned}$

B. 2. 3   Rumus Sudut Paruh untuk Tangen

Adapun untuk rumus tangen adalah:

$\begin{aligned}\tan \displaystyle \frac{1}{2}\alpha &=\displaystyle \frac{\sin \displaystyle \frac{1}{2}\alpha }{\cos \displaystyle \frac{1}{2}\alpha }\\ &=\displaystyle \frac{\pm \sqrt{\displaystyle \frac{1-\cos \alpha }{2}}}{\pm \sqrt{\displaystyle \frac{1+\cos \alpha }{2}}}\\ &=\pm \sqrt{\color{red}\displaystyle \frac{1-\cos \alpha }{1+\cos \alpha }} \end{aligned}$

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tunjukkan bahwa nilai}\: \: \tan \displaystyle \frac{1}{2}\alpha =\color{red}\displaystyle \frac{1-\cos \alpha }{\sin \alpha }\\\\ &\color{blue}\textbf{Bukti}:\\ &\textrm{Sebelumnya diketahui bahwa}:\\ &\begin{aligned}\tan \displaystyle \frac{1}{2}\alpha &=\pm \sqrt{\color{red}\displaystyle \frac{1-\cos \alpha }{1+\cos \alpha }},\: \textrm{maka}\\ &=\pm \sqrt{\displaystyle \frac{1-\cos \alpha }{1+\cos \alpha }\times \displaystyle \frac{1-\cos \alpha }{1-\cos \alpha }}\\ &=\pm \sqrt{\displaystyle \frac{\left ( 1-\cos \alpha \right )^{2}}{1-\cos ^{2}\alpha }}\\ &=\pm \sqrt{\displaystyle \frac{\left ( 1-\cos \alpha \right )^{2}}{\sin ^{2}\alpha }}\\ &=\displaystyle \color{red}\displaystyle \frac{1-\cos \alpha }{\sin \alpha }\qquad \color{black}\blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tunjukkan bahwa nilai}\: \: \tan \displaystyle \frac{1}{2}\alpha =\color{red}\displaystyle \frac{\sin \alpha }{1+\cos \alpha }\\\\ &\color{blue}\textbf{Bukti}:\\ &\textrm{Sebelumnya diketahui bahwa}:\\ &\begin{aligned}\tan \displaystyle \frac{1}{2}\alpha &=\displaystyle \color{red}\displaystyle \frac{1-\cos \alpha }{\sin \alpha },\: \color{black}\textrm{maka}\\ &=\displaystyle \frac{1-\cos \alpha }{\sin \alpha }\times \frac{1+\cos \alpha }{1+\cos \alpha }\\ &=\displaystyle \frac{1-\cos ^{2}\alpha }{\sin \alpha \left ( 1+\cos \alpha \right )}\\ &=\displaystyle \frac{\sin ^{2}\alpha }{\sin \alpha \left ( 1+\cos \alpha \right )}\\ &=\displaystyle \color{red}\displaystyle \frac{\sin \alpha }{1+\cos \alpha }\qquad \color{black}\blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Tunjukkan bahwa nilai}\: \: \cos^{2} \displaystyle \frac{1}{2}\alpha =\color{red}\displaystyle \frac{1+\sec \alpha }{2\sec \alpha }\\\\ &\color{blue}\textbf{Bukti}:\\ &\textrm{Sebelumnya diketahui bahwa}:\\ &\begin{aligned}\cos^{2} \displaystyle \frac{1}{2}\alpha &=\displaystyle \color{red}\displaystyle \frac{1+\cos \alpha }{2 },\: \color{black}\textrm{maka}\\ &=\displaystyle \displaystyle \frac{1+\cos \alpha }{2}\times \displaystyle \frac{\displaystyle \frac{1}{\cos \alpha }}{\displaystyle \frac{1}{\cos \alpha }}\\ &=\displaystyle \frac{\displaystyle \frac{1}{\cos \alpha }+\displaystyle \frac{\cos \alpha }{\cos \alpha }}{\displaystyle \frac{2}{\cos \alpha }}\\ &=\displaystyle \frac{\sec \alpha +1}{2\sec \alpha }\\ &=\displaystyle \color{red}\displaystyle \frac{1+\sec \alpha }{2\sec \alpha }\qquad \color{black}\blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Tunjukkan bahwa nilai}\: \: \sin \displaystyle \frac{1}{2}\alpha =\color{red}\displaystyle \frac{\sqrt{1+\sin \alpha }-\sqrt{1-\sin \alpha }}{2 }\\\\ &\color{blue}\textbf{Bukti}:\\ &\textrm{Sebelumnya diketahui bahwa}:\\ &\begin{aligned}\sin^{2} \displaystyle \frac{1}{2}\alpha &=\displaystyle \color{red}\displaystyle \frac{1-\cos \alpha }{2 },\: \color{black}\textrm{maka}\\ &=\displaystyle \frac{1-\sqrt{\cos ^{2}\alpha }}{2}\\ &=\displaystyle \frac{1-\sqrt{1-\sin ^{2}\alpha }}{2}\\ &=\displaystyle \frac{1-\sqrt{\left (1+\sin \alpha \right ).\left (1-\sin \alpha \right )}}{2}\\ &=\displaystyle \frac{1-\sqrt{\left (1+\sin \alpha \right ).\left (1-\sin \alpha \right )}}{2}\\ &=\displaystyle \frac{2+\sin \alpha -\sin \alpha -2\sqrt{\left (1+\sin \alpha \right ).\left (1-\sin \alpha \right )}}{4}\\ &=\displaystyle \frac{1+\sin \alpha +1-\sin \alpha -2\sqrt{\left (1+\sin \alpha \right ).\left (1-\sin \alpha \right )}}{4}\\ &=\displaystyle \frac{\left ( \sqrt{1+\sin \alpha }-\sqrt{1-\sin \alpha }\right )^{2}}{4}\\ &=\left (\displaystyle \frac{\left ( \sqrt{1+\sin \alpha }-\sqrt{1-\sin \alpha }\right )}{2} \right )^{2}\\ \sin \displaystyle \frac{1}{2}\alpha &=\displaystyle \color{red}\displaystyle \frac{\sqrt{1+\sin \alpha }-\sqrt{1-\sin \alpha }}{2 }\qquad \color{black}\blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Dengan menggunakan rumus sudut}\\ &\textrm{paruh, tentukanlah nilai dari}\\ &\textrm{a}.\quad \cos \displaystyle \frac{\pi }{8},\qquad \textrm{b}.\quad \sin 15^{\circ}\qquad \textrm{c}.\quad \tan 15^{\circ}\\\\ &\color{blue}\textbf{Jawab}:\\ &\textrm{Ingat baik sudut}\: \: \displaystyle \frac{\pi }{8}\: \: \textrm{atau}\: \: \displaystyle \frac{\pi }{4}\: \: \textrm{di kuadran I}\\ &\textrm{maka}\: \: \cos \displaystyle \frac{\pi }{8}\: \: \textrm{bertanda positif, sehingga}\\ &\begin{aligned}\textrm{a}.\quad \cos \displaystyle \frac{\pi }{8}&=\sqrt{\displaystyle \frac{1+\cos \displaystyle \frac{\pi }{4}}{2}}=\sqrt{\displaystyle \frac{1+\displaystyle \frac{1}{2}\sqrt{2}}{2}}\\ &=\sqrt{\displaystyle \frac{2+\sqrt{2}}{4}}=\color{red}\displaystyle \frac{1}{2}\sqrt{2+\sqrt{2}} \end{aligned}\\ &\textrm{Adapun untuk sudut}\: \: 15^{\circ}\: \: \textrm{ataupun}\: \: 30^{\circ}\\ &\textrm{akan berada di kuadran I akibatnya }\\ &\textrm{tandanya positif, sehingga}\\ &\begin{aligned}\textrm{b}.\quad \sin 15^{\circ}&=\sqrt{\displaystyle \frac{1-\cos 30^{\circ}}{2}}\\ &=\sqrt{\displaystyle \frac{1-\displaystyle \frac{1}{2}\sqrt{3}}{2}}=\sqrt{\displaystyle \frac{2-\sqrt{3}}{4}}\\ &=\color{red}\displaystyle \frac{1}{2}\sqrt{2-\sqrt{3}} \end{aligned}\\ &\textrm{Dan untuk sudut}\: \: \tan \displaystyle \frac{1}{2}\alpha =\displaystyle \frac{1-\cos \alpha }{\sin \alpha }\\ &\begin{aligned}\textrm{c}.\quad \tan 15^{\circ}&=\displaystyle \frac{1-\cos 30^{\circ} }{\sin 30^{\circ} }\\ &=\displaystyle \frac{1-\displaystyle \frac{1}{2}\sqrt{3}}{\displaystyle \frac{1}{2}}\\ &=\displaystyle \frac{\displaystyle \frac{\left ( 2-\sqrt{3} \right )}{2}}{\displaystyle \frac{1}{2}}\\ &=\color{red}2-\sqrt{3} \end{aligned} \end{array}$.

DAFTAR PUSTAKA

1. Sukino. 2017. Matematika untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.

Lanjutan 3 Materi Rumus Sudut Ganda(Rangkap) dan Sudut Paruh Pada Trigonometri

B. 1   Rumus Sudut Rangkap

Pada materi sebelumnya sudah dibahas tentang hal ini yaitu,

$\begin{array}{ll}\\ \bullet &\sin 2\alpha =2\sin \alpha \cos \alpha \\ \bullet &\cos 2\alpha =\cos ^{2}\alpha -\sin ^{2}\alpha \\ \bullet &\tan 2\alpha =\displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha } \end{array}$.

B. 2   Rumus Sudut Rangkap Tiga

Berikut untuk rumus sudut rangkap tiga:

$\begin{aligned}\bullet \: \sin 3\alpha &=3\sin \alpha -4\sin ^{3}\alpha \\ \bullet \: \cos 3\alpha &=4\cos ^{3}\alpha -3\cos \alpha \\ \bullet \: \tan 3\alpha &=\displaystyle \frac{3\tan x-\tan ^{3}\alpha }{1-3\tan ^{2}\alpha } \end{aligned}$

B. 3   Rumus Sudut Paruh

Demikian juga untuk sudut paruhnya, juga sudah dibahas pada materi sebelumnya yang mana sebelumnya sudut paruh itu diperoleh dari aplikasi rumus jum lah dan selisih dua sudut ketika sudutnya berupa setengan sudut dan sama pula, yaitu:

$\begin{array}{ll}\\ \bullet &\sin \alpha =2\sin \displaystyle \frac{1}{2}\alpha \cos \frac{1}{2}\alpha \\ \bullet &\cos \alpha =\cos ^{2}\displaystyle \frac{1}{2}\alpha -\sin ^{2}\displaystyle \frac{1}{2}\alpha \\ \bullet &\tan \alpha =\displaystyle \frac{2\tan \displaystyle \frac{1}{2}\alpha }{1-\tan ^{2}\displaystyle \frac{1}{2}\alpha } \end{array}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Diketahui}\: \: \sin A=\displaystyle \frac{4}{5},\\ & \textrm{untuk}\: A\: \textrm{sudut tumpul. Tentukanlah}\\ &\textrm{a}.\quad \sin 2A\\ &\textrm{b}.\quad \cos 2A\\ &\textrm{c}.\quad \tan 2A\\\\ &\color{blue}\textbf{Jawab}:\\ &\textrm{Dengan menggunakan identitas}\\ &\sin ^{2}\alpha +\cos ^{2}\alpha =1,\: \: \textrm{dapat diperoleh}\\ &\textrm{nilai}\: \: \cos A,\: \: \textrm{yaitu}:\\ &\cos ^{2}\alpha =1-\sin ^{2}A=1-\left ( \displaystyle \frac{4}{5} \right )^{2}\\ &\qquad =1-\displaystyle \frac{16}{25}=\frac{9}{25}\Rightarrow \cos A=\pm \displaystyle \frac{3}{5}\\ &\textrm{ambil}\: \: \cos A=-\displaystyle \frac{3}{5},\: \: \textrm{karena di kuadran II}\\ &\textrm{ingat sudut tumpul}=\textrm{sudut di kuadran II}\\ &\begin{aligned}\textrm{a}.\: \: \sin 2A&=2\sin A\cos A\\ &=2\left ( \displaystyle \frac{4}{5} \right )\left ( -\displaystyle \frac{3}{5} \right )\\ &=-\displaystyle \frac{24}{25}\\ \textrm{b}.\: \: \cos 2A&=\cos ^{2}A-\sin ^{2}A\\ &=\left ( -\displaystyle \frac{3}{5} \right )^{2}-\left ( \displaystyle \frac{4}{5} \right )^{2}\\ &=\displaystyle \frac{9}{25}-\frac{16}{25}=-\frac{7}{25}\\ \textrm{c}.\: \: \tan 2A&=\displaystyle \frac{\sin 2A}{\cos 2A}\\ &=\displaystyle \frac{-\displaystyle \frac{24}{25}}{-\displaystyle \frac{7}{25}}=\displaystyle \frac{24}{7} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.& \textrm{Tentukanlah nilai dari}\\ &\textrm{a}.\quad \sin 15^{\circ}\\ &\textrm{b}.\quad \cos 15^{\circ}\\ &\textrm{c}.\quad \tan 15^{\circ}\\\\ &\color{blue}\textbf{Jawab}:\\ &\textrm{Banyak cara yang bisa digunakan}.\\ &\textrm{berikut salah satu cara itu, yaitu}\\ &\begin{aligned}\textnormal{a.}\quad\quad \cos \left (2\alpha \right )&=1-2\sin ^{2}\alpha, &\textnormal{pilih untuk}\quad \alpha =15^{0} \\ \cos \left (2\times 15^{0} \right )&=1-2\sin^{2} 15^{0}\\ \cos 30^{0}&=1-2\sin ^{2}15^{0}\\ \displaystyle \frac{1}{2}\sqrt{3}&=1-2\sin ^{2}15^{0}\\ \sqrt{3}&=2-4\sin ^{2}15^{0}\\ \sin ^{2}15^{0}&=\displaystyle \frac{2-\sqrt{3}}{4}\\ \sin 15^{0}&=\displaystyle \frac{1}{2}\sqrt{2-\sqrt{3}}\\ \textnormal{b.}\quad\quad \sin ^{2}\alpha +&\cos ^{2}\alpha =1,&\textnormal{rumus identitas}\\ \sin^{2}15^{0}+&\cos ^{2}15^{0}=1\\ \cos ^{2}15^{0}&=1-\sin ^{2}15^{0}\\ &=\displaystyle 1-\left ( \frac{2-\sqrt{3}}{4} \right ),&\textnormal{lihat jawaban poin a)}\\ &=\displaystyle \frac{2+\sqrt{3}}{4}\\ \cos 15^{0}&=\displaystyle \frac{1}{2}\sqrt{2+\sqrt{3}}\\ \textnormal{c.}\qquad\quad \tan \alpha &=\displaystyle \frac{\sin \alpha }{\cos \alpha }\\ \tan 15^{0}&=\displaystyle \frac{\sin 15^{0}}{\cos 15^{0}}\\ &=\displaystyle \frac{\displaystyle \frac{1}{2}\sqrt{2-\sqrt{3}}}{\displaystyle \frac{1}{2}\sqrt{2+\sqrt{3}}}\\ &=\sqrt{\displaystyle \frac{2-\sqrt{3}}{2+\sqrt{3}}}\times \sqrt{\frac{2-\sqrt{3}}{2-\sqrt{3}}},&\textnormal{sekawan dari penyebut}\\ &=\sqrt{\displaystyle \frac{4-2.2\sqrt{3}+3}{4-3}}\\ &=\sqrt{7-4\sqrt{3}} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Tunjukkan bahwa nilai}\\ & \displaystyle \frac{1-\cos 2\alpha }{\sin 2\alpha }=\color{red}\tan \alpha \\\\ &\color{blue}\textbf{Bukti}:\\ &\begin{aligned}\displaystyle \frac{1-\cos 2\alpha }{\sin 2\alpha }&=\displaystyle \frac{1-\left ( \cos ^{2}\alpha -\sin ^{2}\alpha \right ) }{2\sin \alpha \cos \alpha }\\ &=\displaystyle \frac{1-\left ( \left ( 1-\sin ^{2}\alpha \right )-\sin ^{2}\alpha \right )}{2\sin \alpha \cos \alpha }\\ &=\displaystyle \frac{1-1+\sin ^{2}\alpha +\sin ^{2}\alpha }{2\sin \alpha \cos \alpha }\\ &=\displaystyle \frac{2\sin ^{2}\alpha }{2\sin \alpha \cos \alpha }\\ &=\displaystyle \frac{\sin \alpha }{\cos \alpha }\\ &=\displaystyle \color{red}\tan \alpha \qquad \color{black}\blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Tunjukkan bahwa nilai}\\ &\textrm{a}.\quad \cos 3\alpha =4\cos ^{3}\alpha -3\cos \alpha \\ &\textrm{b}.\quad \tan 3\alpha =\displaystyle \frac{3\tan x-\tan ^{3}\alpha }{1-3\tan ^{2}\alpha } \\\\ &\color{blue}\textbf{Bukti}:\\ &\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad \cos 3\alpha &=\cos \left ( 2\alpha +\alpha \right )\\ &=\cos 2\alpha \cos \alpha -\sin 2\alpha \sin \alpha \\ &=\left ( 2\cos ^{2}\alpha -1 \right )\cos \alpha -\left ( 2\sin \alpha \cos \alpha \right )\sin \alpha \\ &=2\cos ^{3}\alpha -\cos \alpha -2\sin ^{2}\alpha \cos \alpha \\ &=2\cos ^{3}\alpha -\cos \alpha -2\left ( 1-\cos ^{2}\alpha \right ) \cos \alpha\\ &=2\cos ^{3}\alpha -\cos \alpha -2\cos \alpha +2\cos ^{2}\alpha\\ &=4\cos ^{3}\alpha -3\cos \alpha \qquad \blacksquare\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{b}.\quad \tan 3\alpha &=\tan \left ( 2\alpha +\alpha \right )\\ &=\displaystyle \frac{\tan 2\alpha +\tan \alpha }{1-\tan 2\alpha \tan \alpha }\\ &=\displaystyle \frac{\left ( \displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha } \right )+\tan \alpha }{1-\left ( \displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha } \right ).\tan \alpha }\\ &=\displaystyle \frac{\displaystyle \frac{2\tan \alpha +\tan \alpha -\tan ^{3}\alpha }{1-\tan ^{2}\alpha }}{\displaystyle \frac{\left ( 1-\tan ^{2}\alpha \right )-2\tan ^{2}\alpha }{1-\tan ^{2}\alpha }}\\ &=\displaystyle \frac{3\tan \alpha -\tan ^{3}\alpha }{1-3\tan ^{2}\alpha }\qquad \blacksquare \end{aligned} \\\hline \end{array} \end{array}$

Lanjutan 2 Materi Rumus-Rumus Trigonometri

 A. 3  Rumus  $\tan \left ( \alpha +\beta \right )$ dan $\tan \left ( \alpha -\beta \right )$.

Sebelumnya telah dibahas pada materi sebelumnya dan disertai pula dengan contoh soal rumus jumlah dan selisih dua sudut untuk sinus dan cosinus, yaitu:

Jika dua sudut yang dimaksud misalkan alfa dan beta, maka

$\begin{cases} \sin \left ( \alpha +\gamma \right ) & =\sin \alpha \cos \gamma +\cos \alpha \sin \gamma \\ \sin \left ( \alpha -\gamma \right ) & =\sin \alpha \cos \gamma -\cos \alpha \sin \gamma \\ \cos \left ( \alpha +\gamma \right ) & =\cos \alpha \cos \gamma -\sin \alpha \sin \gamma \\ \cos \left ( \alpha +\gamma \right ) & =\cos \alpha \cos \gamma -\sin \alpha \sin \gamma \end{cases}$.

Sengan menggunakan fakta yang ada-rumus yang telah diketahui-kita akan terbantu dalam menemukan rumus untuk tangen, yaitu:

$\begin{aligned} \tan \left ( \alpha +\beta \right )&=\displaystyle \frac{\sin \left ( \alpha +\beta \right )}{\cos \left ( \alpha +\beta \right )}\\ &=\displaystyle \frac{\left (\sin \alpha \cos \beta +\cos \alpha \sin \beta \right )\times \left ( \displaystyle \frac{1}{\cos \alpha \cos \beta } \right ) }{\left (\cos \alpha \cos \beta -\sin \alpha \sin \beta \right ) \times \left ( \displaystyle \frac{1}{\cos \alpha \cos \beta } \right )}\\ &=\displaystyle \frac{\displaystyle \frac{\sin \alpha }{\cos \alpha }+\displaystyle \frac{\sin \beta }{\cos \beta }}{1-\displaystyle \frac{\sin \alpha \sin \beta }{\cos \alpha \cos \beta }}\\ &=\displaystyle \color{red}\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\color{black}.\qquad \blacksquare \end{aligned}$.

Selanjutnya dengan untuk mendapatkan rumus  $\tan \left ( \alpha +\beta \right )$ adalah dengan mengganti  $\beta =-\beta$, maka

$\begin{aligned} \tan \left ( \alpha +\beta \right )&=\displaystyle \frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\\ &\qquad \textrm{dengan mengganti}\: \color{blue}\beta =-\beta \: \: \color{black}\textrm{maka,}\\ &=\displaystyle \frac{\tan \alpha +\tan \left ( -\beta \right )}{1-\tan \alpha \tan \left ( -\beta \right )}\\ \tan \left ( \alpha -\beta \right )&=\displaystyle \color{red}\frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }\color{black}.\qquad \blacksquare \end{aligned}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tunjukkan bahwa nilai}\: \: \tan 60^{\circ}=\color{red}\sqrt{3}\\\\ &\color{blue}\textbf{Bukti}:\\ &\begin{aligned}\tan 60^{\circ}&=\tan \left ( 30^{\circ}+30^{\circ} \right )\\ &=\displaystyle \frac{\tan 30^{\circ} +\tan 30^{\circ} }{1-\tan 30^{\circ} \tan 30^{\circ} }\\ &=\displaystyle \frac{2\tan 30^{\circ}}{1-\tan^{2} 30^{\circ}}\\ &=\displaystyle \frac{2\left ( \displaystyle \frac{1}{3}\sqrt{3} \right )}{1-\left ( \displaystyle \frac{1}{3}\sqrt{3} \right )^{2}}=\displaystyle \frac{\displaystyle \frac{2}{3}\sqrt{3}}{1-\displaystyle \frac{1}{9}\sqrt{9}}\\ &=\displaystyle \frac{\displaystyle \frac{2}{3}\sqrt{3}}{1-\displaystyle \frac{3}{9}}=\displaystyle \frac{\displaystyle \frac{2}{3}\sqrt{3}}{\displaystyle \frac{6}{9}}=\displaystyle \frac{\displaystyle \frac{2}{3}\sqrt{3}}{\displaystyle \frac{2}{3}}\\ &=\displaystyle \color{red}\sqrt{3}\qquad \color{black}\blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tunjukkan bahwa nilai}\\ & \tan 90^{\circ}=\color{red}\textrm{Tidak Terdefinisi}\\\\ &\color{blue}\textbf{Bukti}:\\ &\begin{aligned}\tan 90^{\circ}&=\tan \left ( 45^{\circ}+45^{\circ} \right )\\ &=\displaystyle \frac{\tan 45^{\circ} +\tan 45^{\circ} }{1-\tan 45^{\circ} \tan 45^{\circ} }\\ &=\displaystyle \frac{2\tan 45^{\circ}}{1-\tan^{2} 45^{\circ}}\\ &=\displaystyle \frac{2\left ( 1 \right )}{1-\left ( 1 \right )^{2}}=\displaystyle \frac{\displaystyle 2}{1-1}\\ &=\displaystyle \frac{2}{0}\\ &=\color{red}\textrm{Tidak Terdefinisi}\qquad \color{black}\blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Tentukan nilai dari}\: \: \tan 75^{\circ}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\tan 75^{\circ}&=\tan \left ( 30^{\circ}+45^{\circ} \right )\\ &=\displaystyle \frac{\tan 30^{\circ} +\tan 45^{\circ} }{1-\tan 30^{\circ} \tan 45^{\circ} }\\ &=\displaystyle \frac{\displaystyle \frac{1}{3}\sqrt{3}+1}{1-\left ( \displaystyle \frac{1}{3}\sqrt{3} \right )\left ( 1 \right )}\\ &=\displaystyle \frac{1+\displaystyle \frac{1}{3}\sqrt{3}}{1-\displaystyle \frac{1}{3}\sqrt{3}}=\displaystyle \frac{\displaystyle \frac{1}{3}\left (3+\sqrt{3} \right )}{\displaystyle \frac{1}{3}\left (3-\sqrt{3} \right )}\\ &=\displaystyle \frac{3+\sqrt{3} }{3-\sqrt{3} }\\ &=\displaystyle \frac{3+\sqrt{3} }{3-\sqrt{3} }\times \displaystyle \frac{3+\sqrt{3} }{3+\sqrt{3} }\\ &=\displaystyle \frac{3^{2}+3\sqrt{3}+3\sqrt{3}+\sqrt{9}}{3^{2}-\left ( \sqrt{3} \right )^{2}}\\ &=\displaystyle \frac{12+6\sqrt{3}}{9-3}=\displaystyle \frac{12+6\sqrt{3}}{6}\\ &=\displaystyle \frac{6}{6}\left ( 2+\sqrt{3} \right )\\ &=\color{red}2+\sqrt{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Tentukan nilai dari}\: \: \tan 105^{\circ}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\tan 105^{\circ}&=\tan \left ( 45^{\circ}+60^{\circ} \right )\\ &=\displaystyle \frac{\tan 45^{\circ} +\tan 60^{\circ} }{1-\tan 45^{\circ} \tan 60^{\circ} }\\ &=\displaystyle \frac{1+\sqrt{3}+1}{1-1.\left ( \sqrt{3} \right )}\\ &=\displaystyle \frac{1+\sqrt{3}}{1-\sqrt{3}}\\ &=\displaystyle \frac{1+\sqrt{3} }{1-\sqrt{3} }\\ &=\displaystyle \frac{1+\sqrt{3} }{1-\sqrt{3} }\times \displaystyle \frac{1+\sqrt{3} }{1+\sqrt{3} }\\ &=\displaystyle \frac{1^{2}+\sqrt{3}+\sqrt{3}+\left (\sqrt{3} \right )^{2}}{1^{2}-\left ( \sqrt{3} \right )^{2}}\\ &=\displaystyle \frac{4+2\sqrt{3}}{1-3}=\displaystyle \frac{4+2\sqrt{3}}{-2}\\ &=\color{red}-2-\sqrt{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Sederhanakan bentuk dari}\: \: \tan \left ( 270^{\circ}+A \right )\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\tan \left ( 270^{\circ}+A \right )&=\displaystyle \frac{\tan 270^{\circ}+\tan A}{1-\tan 270^{\circ}\tan A}\\ &=\displaystyle \frac{\textbf{TD}+\tan A}{1-\textbf{TD}\tan A}\\ &\quad \textrm{dengan}\: \: \textbf{TD}\: \textrm{adalah Tidak Terdefinisi}\\ &=\: \textrm{Bentuk yang harus dihindari}\\ \textrm{maka gunakan}&\: \textrm{bentuk berikut ini}\\ \tan \left ( 270^{\circ}+A \right )&=\displaystyle \frac{\sin \left ( 270^{\circ}+A \right )}{\cos \left ( 270^{\circ}+A \right )}\\ &=\displaystyle \frac{-\cos A}{\sin A}\\ &= \color{red}-\cot A\\\\ \color{purple}\textbf{Catatan}:&\: \textrm{lihatlah materi sebelumnya} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Sederhanakan bentuk dari}\: \: \tan \left ( 270^{\circ}-A \right )\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\tan \left ( 270^{\circ}-A \right )&=\displaystyle \frac{\sin \left ( 270^{\circ}-A \right )}{\cos \left ( 270^{\circ}-A \right )}\\ &=\displaystyle \frac{-\cos A}{-\sin A}\\ &= \color{red}\cot A \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Tunjukkan bahwa}\\ &\begin{array}{lll}\\ \textrm{a}.&\tan \left ( 2\alpha \right )=\displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha }\\ \textrm{b}.&\tan \alpha =\displaystyle \frac{2\tan \left ( \displaystyle \frac{1}{2}\alpha \right )}{1-\tan \left ( \displaystyle \frac{1}{2}\alpha \right )} \end{array}\\\\ &\color{blue}\textbf{Bukti}\\ &\begin{aligned}\textnormal{a.}\quad \tan \left ( \alpha +\beta \right )&=\displaystyle \frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta } ,&\textnormal{dan jika}\quad \beta =\alpha ,&&\textnormal{maka}\\ \tan \left ( \alpha +\alpha \right )&=\displaystyle \frac{\tan \alpha +\tan \alpha }{1-\tan \alpha \tan \alpha } \\ \tan 2\alpha &= \displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha } \alpha\qquad \blacksquare \\ \textnormal{b.}\qquad\quad \tan 2\alpha &=\displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha } ,&\textnormal{dan jika}\quad \alpha =\displaystyle \frac{1}{2}\alpha ,&&\textnormal{maka}\\ \tan 2\left ( \displaystyle \frac{1}{2}\alpha \right )&=\displaystyle \frac{2\tan \left ( \displaystyle \frac{1}{2}\alpha \right )}{1-\tan ^{2}\left ( \displaystyle \frac{1}{2}\alpha \right )}\\ \tan \alpha &=\displaystyle \frac{2\tan \left ( \displaystyle \frac{1}{2}\alpha \right )}{1-\tan ^{2}\left ( \displaystyle \frac{1}{2}\alpha \right )}\qquad \blacksquare \end{aligned}\\ \end{array}$.

Lanjutan 1 Materi Rumus-Rumus Trigonometri

A. 2  Rumus  $\cos \left ( \alpha +\beta \right )$ dan $\cos \left ( \alpha -\beta \right )$.

Dalam penentuan rumus $\cos \left ( \alpha -\beta \right )$, pada uraian berikut akan ditunjukkan penentuan rumus yang dimaksud dengan bantuan segitiga ABC

Perhatikanlah ilustrai berikut

Jika diurai gambar di atas adalah 

Mungkin gambarnya ada tang kurang jelas, mari kita perjelas lagi gambar di atas

Perhatikan bahwa

dan

Sehingga

$\begin{aligned}\left [ ABC \right ]&=\left [ ACD \right ]+\left [ BCD \right ]\\ \displaystyle \frac{1}{2}ab\sin \left ( 90^{0}-\alpha +\beta \right )&=\displaystyle \frac{1}{2}ab\cos \alpha \cos \beta +\displaystyle \frac{1}{2}ab\sin \alpha \sin \beta \\ \sin \left ( 90^{0}-\left ( \alpha -\beta \right ) \right )&=\color{red}\cos \alpha \cos \beta +\sin \alpha \sin \beta \\ \cos \left ( \alpha -\beta \right )&=\color{red}\cos \alpha \cos \beta +\sin \alpha \sin \beta.\qquad \color{black}\blacksquare \end{aligned}$.

Selanjutnya dengan untuk mendapatkan rumus  $\cos \left ( \alpha +\beta \right )$ adalah dengan mengganti  $\beta =-\beta$, maka

$\begin{aligned}&\cos \left ( \alpha -(-\beta ) \right )=\cos \left ( \alpha +\color{red}\beta \right )\\ &=\cos\alpha \cos \left ( -\beta \right )+\sin \alpha \sin \left ( -\beta \right )\\ &=\cos \alpha \cos \beta +\sin \alpha \left ( -\sin \beta \right )\\ &=\cos \alpha \cos \beta -\sin \alpha \sin \beta \qquad \blacksquare \end{aligned}$.

Catatan:

$\left [ ABC \right ]=\textbf{luas segitiga ABC}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tunjukkan bahwa nilai}\: \: \cos 60^{\circ}=\color{red}\displaystyle \frac{1}{2}\\\\ &\color{blue}\textbf{Bukti}:\\ &\begin{aligned}\cos 60^{\circ}&=\cos \left ( 30^{\circ}+30^{\circ} \right )\\ &=\cos 30^{\circ}\cos 30^{\circ}-\sin 30^{\circ}\sin 30^{\circ}\\ &=\cos^{2} 30^{\circ}-\sin^{2} 30^{\circ}\\ &=\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )^{2}-\left ( \displaystyle \frac{1}{2} \right )^{2}\\ &=\displaystyle \frac{1}{4}\sqrt{9}-\frac{1}{4}=\displaystyle \frac{3}{4}-\frac{1}{4}=\displaystyle \frac{2}{4}\\ &=\displaystyle \color{red}\frac{1}{2}\qquad \color{black}\blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tunjukkan bahwa nilai}\: \: \cos 90^{\circ}=\color{red}0\\\\ &\begin{aligned}\color{blue}\textbf{Bukti}&\: \: \textbf{pertama}\\ \cos 90^{\circ}&=\cos \left ( 60^{\circ}+30^{\circ} \right )\\ &=\cos 60^{\circ}\cos 30^{\circ}-\sin 60^{\circ}\sin 30^{\circ}\\ &=\left ( \displaystyle \frac{1}{2} \right )\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )-\left (\displaystyle \frac{1}{2}\sqrt{3} \right )\left (\displaystyle \frac{1}{2} \right )\\ &=\displaystyle \frac{1}{4}\sqrt{3}-\displaystyle \frac{1}{4}\sqrt{3}\\ &=\displaystyle \color{red}0\qquad \color{black}\blacksquare\\ \color{blue}\textbf{Bukti}&\: \: \textbf{kedua}\\ \cos 90^{\circ}&=\cos \left ( 30^{\circ}+60^{\circ} \right )\\ &=....+....\\ &=....\\ &=....\\ \color{blue}\textbf{Bukti}&\: \: \textbf{ketiga}\\ \cos 90^{\circ}&=\cos \left ( 45^{\circ}+45^{\circ} \right )\\ &=....+....\\ &=....\\ &=....\\ \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Tentukan nilai dari}\: \: \cos 75^{\circ}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\cos 75^{\circ}&=\cos \left ( 30^{\circ}+45^{\circ} \right )\\ &=\cos 30^{\circ}\cos 45^{\circ}-\sin 30^{\circ}\sin 45^{\circ}\\ &=\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\left ( \displaystyle \frac{1}{2}\sqrt{2} \right )-\left ( \displaystyle \frac{1}{2} \right )\left ( \displaystyle \frac{1}{2}\sqrt{2} \right )\\ &=\displaystyle \frac{1}{4}\sqrt{6}-\displaystyle \frac{1}{4}\sqrt{2}\\ &=\displaystyle \color{red}\frac{1}{4}\left ( \sqrt{6}-\sqrt{2} \right ) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Tentukan nilai dari}\: \: \cos 105^{\circ}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\cos 105^{\circ}&=\cos \left ( 45^{\circ}+60^{\circ} \right )\\ &=\cos 45^{\circ}\cos 60^{\circ}-\sin 45^{\circ}\sin 60^{\circ}\\ &=\left ( \displaystyle \frac{1}{2}\sqrt{2} \right )\left ( \displaystyle \frac{1}{2} \right )-\left ( \displaystyle \frac{1}{2}\sqrt{2} \right )\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\\ &=\displaystyle \frac{1}{4}\sqrt{2}-\displaystyle \frac{1}{4}\sqrt{6}\\ &=\displaystyle \color{red}\frac{1}{4}\left ( \sqrt{2}-\sqrt{6} \right ) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Sederhanakan bentuk dari}\: \: \cos \left ( 270^{\circ}+A \right )\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\cos \left ( 270^{\circ}+A \right )&=\cos 270^{\circ}\cos A-\sin 270^{\circ}\sin A\\ &=\left ( 0 \right )\cos A-\left (-1 \right )\sin A\\ &=0+\sin A\\ &= \color{red}\sin A \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Sederhanakan bentuk dari}\: \: \cos \left ( 270^{\circ}-A \right )\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\cos \left ( 270^{\circ}-A \right )&=\cos 270^{\circ}\cos A+\sin 270^{\circ}\sin A\\ &=\left ( 0 \right )\cos A+\left ( -1 \right )\sin A\\ &=0-\sin A\\ &= \color{red}-\sin A \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Tunjukkan bahwa}\\ &\begin{array}{lll}\\ \textrm{a}.&\cos \left ( 2\alpha \right )= \cos^{2} \alpha-\sin^{2} \alpha\\ \textrm{b}.&\cos \alpha =\cos^{2} \left ( \displaystyle \frac{1}{2}\alpha \right )-\sin^{2} \left ( \displaystyle \frac{1}{2}\alpha \right ) \end{array}\\\\ &\color{blue}\textbf{Bukti}\\ &\begin{aligned}\textnormal{a.}\quad \cos \left ( \alpha +\beta \right )&=\cos \alpha \cos \beta - \sin \alpha \sin \beta ,&\textnormal{dan jika}\quad \beta =\alpha ,&&\textnormal{maka}\\ \cos \left ( \alpha +\alpha \right )&=\cos \alpha \cos \alpha -\sin \alpha \sin \alpha \\ \cos 2\alpha &= \cos^{2} \alpha-\sin^{2} \alpha\qquad \blacksquare \\ \textnormal{b.}\qquad\quad \cos 2\alpha &=\cos^{2} \alpha-\sin^{2} \alpha ,&\textnormal{dan jika}\quad \alpha =\displaystyle \frac{1}{2}\alpha ,&&\textnormal{maka}\\ \cos 2\left ( \displaystyle \frac{1}{2}\alpha \right )&=\cos^{2} \left ( \displaystyle \frac{1}{2}\alpha \right )-\sin^{2} \left ( \displaystyle \frac{1}{2}\alpha \right )\\ \cos \alpha &=\cos^{2} \left ( \displaystyle \frac{1}{2}\alpha \right )-\sin^{2} \left ( \displaystyle \frac{1}{2}\alpha \right )\qquad \blacksquare \end{aligned}\\ \end{array}$

DAFTAR PUSTAKA

1. Kanginan, M. 2007. Matematika untuk Kelas X Semester 2 Sekolah Menengah Atas. Bandung: GRAFINDO MEDIA PRATAMA.
2. Noormandiri, B.K. 2017. Matematika Jilid 2 untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam Berdasarkan Kurikulum 2013 Edisi Revisi 2016. Jakarta: ERLANGGA.

Rumus-Rumus Trigonometri

Sebelumnya kita buka arsip lama di blog ini, yaitu:

 Materi Lawas

A. Rumus Jumlah Dan selisih Sudut

A. 1  Rumus  $\sin \left ( \alpha +\beta \right )$ dan $\sin \left ( \alpha -\beta \right )$.

Dalam penentuan rumus $\sin \left ( \alpha +\beta \right )$, pada uraian berikut akan ditunjukkan penentuan rumus yang dimaksud dengan bantuan segitiga ABC

Perhatikanlah ilustrai berikut

Bukti:

$\begin{aligned}\displaystyle \frac{AC}{\sin 90^{0}}&=\displaystyle \frac{CA'}{\sin \alpha }=\displaystyle \frac{AA'}{\sin \angle C}\\ AA'&=AC.\sin \angle C\\ &=AC.\sin \left ( 90^{0}-\alpha \right )\\ &=AC.\cos \alpha\\ &\textnormal{dengan cara yang kurang lebih }\\ &\textrm{sama akan diperoleh juga}\\ AA'&=AB.\cos \beta\\ &\textnormal{selanjutnya kita tentukan luasnya, yaitu}\\ \left [ ABC \right ]&=\left [ AA'C \right ]+\left [ AA'B \right ]\\ \displaystyle \frac{1}{2}.AB.AC.\sin \left ( \alpha +\beta \right )&=\displaystyle \frac{1}{2}.AC.AA'.\sin \alpha +\displaystyle \frac{1}{2}.AB.AA'.\sin \beta \\ \sin \left ( \alpha +\beta \right )&=\displaystyle \frac{AC.AA'.\sin \alpha }{AB.AC}+\displaystyle \frac{AB.AA'.\sin \beta }{AB.AC}\\ &=\displaystyle \frac{AA'}{AB}.\sin \alpha +\displaystyle \frac{AA'}{AC}.\sin \beta \\ &=\displaystyle \frac{\left ( AB.\cos \beta \right )}{AB}.\sin \alpha +\displaystyle \frac{\left ( AC.\cos \alpha \right )}{AC}.\sin \beta \\ \sin \left ( \alpha +\beta \right )&=\sin \alpha .\cos \beta +\cos \alpha .\sin \beta \quad \blacksquare \end{aligned}$.

Selanjutnya dengan untuk mendapatkan rumus  $\sin \left ( \alpha -\beta \right )$ adalah dengan mengganti  $\beta =-\beta$, maka

$\begin{aligned}&\sin \left ( \alpha +(-\beta ) \right )=\sin \left ( \alpha -\color{red}\beta \right )\\ &=\sin\alpha \cos \left ( -\beta \right )+\cos \alpha \sin \left ( -\beta \right )\\ &=\sin \alpha \cos \beta +\cos \alpha \left ( -\sin \beta \right )\\ &=\sin \alpha \cos \beta -\cos \alpha \sin \beta \qquad \blacksquare \end{aligned}$.

Catatan:

$\left [ ABC \right ]=\textbf{luas segitiga ABC}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tunjukkan bahwa nilai}\: \: \sin 60^{\circ}=\color{red}\displaystyle \frac{1}{2}\sqrt{3}\\\\ &\color{blue}\textbf{Bukti}:\\ &\begin{aligned}\sin 60^{\circ}&=\sin \left ( 30^{\circ}+30^{\circ} \right )\\ &=\sin 30^{\circ}\cos 30^{\circ}+\cos 30^{\circ}\sin 30^{\circ}\\ &=\sin 30^{\circ}\cos 30^{\circ}+\sin 30^{\circ}\cos 30^{\circ}\\ &=2\sin 30^{\circ}\cos 30^{\circ}\\ &=2\left ( \displaystyle \frac{1}{2} \right )\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\\ &=\displaystyle \color{red}\frac{1}{2}\sqrt{3}\qquad \color{black}\blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tunjukkan bahwa nilai}\: \: \sin 90^{\circ}=\color{red}1\\\\ &\begin{aligned}\color{blue}\textbf{Bukti}&\: \: \textbf{pertama}\\ \sin 90^{\circ}&=\sin \left ( 60^{\circ}+30^{\circ} \right )\\ &=\sin 60^{\circ}\cos 30^{\circ}+\cos 60^{\circ}\sin 30^{\circ}\\ &=\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )+\left (\displaystyle \frac{1}{2} \right )\left (\displaystyle \frac{1}{2} \right )\\ &=\displaystyle \frac{1}{4}\sqrt{9}+\displaystyle \frac{1}{4}\\ &=\displaystyle \frac{3}{4}+\displaystyle \frac{1}{4}\\ &=\displaystyle \frac{4}{4}\\ &=\displaystyle \color{red}1\qquad \color{black}\blacksquare\\ \color{blue}\textbf{Bukti}&\: \: \textbf{kedua}\\ \sin 90^{\circ}&=\sin \left ( 30^{\circ}+60^{\circ} \right )\\ &=....+....\\ &=....\\ &=....\\ \color{blue}\textbf{Bukti}&\: \: \textbf{ketiga}\\ \sin 90^{\circ}&=\sin \left ( 45^{\circ}+45^{\circ} \right )\\ &=....+....\\ &=....\\ &=....\\ \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Tentukan nilai dari}\: \: \sin 75^{\circ}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\sin 75^{\circ}&=\sin \left ( 30^{\circ}+45^{\circ} \right )\\ &=\sin 30^{\circ}\cos 45^{\circ}+\cos 30^{\circ}\sin 45^{\circ}\\ &=\left ( \displaystyle \frac{1}{2} \right )\left ( \displaystyle \frac{1}{2}\sqrt{2} \right )+\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\left ( \displaystyle \frac{1}{2}\sqrt{2} \right )\\ &=\displaystyle \frac{1}{4}\sqrt{2}+\displaystyle \frac{1}{4}\sqrt{6}\\ &=\displaystyle \color{red}\frac{1}{4}\left ( \sqrt{2}+\sqrt{6} \right ) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Tentukan nilai dari}\: \: \sin 105^{\circ}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\sin 105^{\circ}&=\sin \left ( 45^{\circ}+60^{\circ} \right )\\ &=\sin 45^{\circ}\cos 60^{\circ}+\cos 45^{\circ}\sin 60^{\circ}\\ &=\left ( \displaystyle \frac{1}{2}\sqrt{2} \right )\left ( \displaystyle \frac{1}{2} \right )+\left ( \displaystyle \frac{1}{2}\sqrt{2} \right )\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\\ &=\displaystyle \frac{1}{4}\sqrt{2}+\displaystyle \frac{1}{4}\sqrt{6}\\ &=\displaystyle \color{red}\frac{1}{4}\left ( \sqrt{2}+\sqrt{6} \right ) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Sederhanakan bentuk dari}\: \: \sin \left ( 270^{\circ}+A \right )\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\sin \left ( 270^{\circ}+A \right )&=\sin 270^{\circ}\cos A+\cos 270^{\circ}\sin A\\ &=\left ( -1 \right )\cos A+\left ( 0 \right )\sin A\\ &=-\cos A+0\\ &= \color{red}-\cos A \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Sederhanakan bentuk dari}\: \: \sin \left ( 270^{\circ}-A \right )\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\sin \left ( 270^{\circ}-A \right )&=\sin 270^{\circ}\cos A-\cos 270^{\circ}\sin A\\ &=\left ( -1 \right )\cos A-\left ( 0 \right )\sin A\\ &=-\cos A-0\\ &= \color{red}-\cos A \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Tunjukkan bahwa}\\ &\begin{array}{lll}\\ \textrm{a}.&\sin \left ( 2\alpha \right )=2\sin \alpha \cos \alpha\\ \textrm{b}.&\sin \alpha =2\sin \left ( \displaystyle \frac{1}{2}\alpha \right ).\cos \left ( \displaystyle \frac{1}{2}\alpha \right ) \end{array}\\\\ &\color{blue}\textbf{Bukti}\\ &\begin{aligned}\textnormal{a.}\quad \sin \left ( \alpha +\beta \right )&=\sin \alpha \cos \beta + \cos \alpha \sin \beta ,&\textnormal{dan jika}\quad \beta =\alpha ,&&\textnormal{maka}\\ \sin \left ( \alpha +\alpha \right )&=\sin \alpha \cos \alpha +\cos \alpha \sin \alpha \\ \sin 2\alpha &=2\sin \alpha \cos \alpha\qquad \blacksquare \\ \textnormal{b.}\qquad\quad \sin 2\alpha &=2\sin \alpha \cos \alpha ,&\textnormal{dan jika}\quad \alpha =\displaystyle \frac{1}{2}\alpha ,&&\textnormal{maka}\\ \sin 2\left ( \displaystyle \frac{1}{2}\alpha \right )&=2\sin \left ( \displaystyle \frac{1}{2}\alpha \right )\cos \left ( \displaystyle \frac{1}{2}\alpha \right )\\ \sin \alpha &=2\sin \left ( \displaystyle \frac{1}{2}\alpha \right )\cos \left ( \displaystyle \frac{1}{2}\alpha \right )\qquad \blacksquare \end{aligned}\\ \end{array}$.

DAFTAR PUSTAKA

1. Kanginan, M. 2007. Matematika untuk Kelas X Semester 2 Sekolah Menengah Atas. Bandung: GRAFINDO MEDIA PRATAMA.
2. Noormandiri, B.K. 2017. Matematika Jilid 2 untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam Berdasarkan Kurikulum 2013 Edisi Revisi 2016. Jakarta: ERLANGGA.