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

 $\begin{array}{l}\\ 71.& \text{Grafik fungsi trigonometri pada gambar}\\ & \text{berikut adalah}....\end{array}$.

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

$\begin{array}{l}\\ 72.& \text{Grafik fungsi trigonometri pada gambar}\\ & \text{berikut adalah}....\end{array}$.

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

$\begin{array}{l}\\ 73.& \text{Himpunan penyelesaian dari persamaan}\\ & \sin x=\sin {\displaystyle \frac{2}{10}\pi \text{untuk}{0}^{\circ }\le x\le {360}^{\circ }}\\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \left\{{\displaystyle \frac{22}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}}\\ \\ & \text{b}.{\displaystyle \left\{{\displaystyle \frac{2}{10}\pi ,{\displaystyle \frac{28}{10}\pi }}\right\}}\\ \\ & \text{c}.{\displaystyle \left\{{\displaystyle \frac{2}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}}\\ \\ & \text{d}.{\displaystyle \left\{{\displaystyle \frac{22}{10}\pi ,{\displaystyle \frac{28}{10}\pi }}\right\}}\\ \\ & \text{e}.{\displaystyle \left\{{\displaystyle \frac{12}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \sin x=\sin {\displaystyle \frac{2}{10}\pi }\\ & \iff x_1={\displaystyle \frac{2}{10}\pi +k.2\pi \text{atau}}\\ & \iff x_2=(\pi -{\displaystyle \frac{2}{10}\pi })+k.2\pi ={\displaystyle \frac{8}{10}\pi +k.2\pi }\\ & k=0\Rightarrow x_1={\displaystyle \frac{2}{10}\pi (\text{mm})\text{atau}}\\ & x_2={\displaystyle \frac{8}{10}\pi (\text{mm})}\\ & k=1\Rightarrow x_{1,2}=....+2\pi (\text{tidak memenuhi})\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle \frac{2}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}\end{array}\end{array}$.

$\begin{array}{l}\\ 74.& \text{Himpunan penyelesaian dari persamaan}\\ & \tan (2x-{\displaystyle \frac{1}{4}\pi })=\tan {\displaystyle \frac{1}{4}\pi \text{untuk}{0}^{\circ }\le x\le {360}^{\circ }}\\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \left\{{\displaystyle \frac{1}{3}\pi ,\pi ,{\displaystyle \frac{5}{3}\pi ,{\displaystyle \frac{7}{3}\pi }}}\right\}}\\ \\ & \text{b}.{\displaystyle \left\{{\displaystyle \frac{1}{4}\pi ,{\displaystyle \frac{3}{5}\pi ,\frac{5}{4}\pi ,\frac{8}{5}\pi }}\right\}}\\ \\ & \text{c}.{\displaystyle \left\{{\displaystyle \frac{1}{4}\pi ,{\displaystyle \frac{3}{4}\pi ,\frac{6}{4}\pi ,{\displaystyle \frac{7}{4}\pi }}}\right\}}\\ \\ & \text{d}.{\displaystyle \left\{{\displaystyle \frac{2}{4}\pi ,{\displaystyle \frac{3}{4}\pi ,\pi ,{\displaystyle \frac{7}{4}\pi }}}\right\}}\\ \\ & \text{e}.{\displaystyle \left\{{\displaystyle \frac{1}{4}\pi ,\frac{3}{4}\pi ,\frac{5}{4}\pi ,{\displaystyle \frac{7}{4}\pi }}\right\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \tan (2x-{\displaystyle \frac{1}{4}\pi })=\tan {\displaystyle \frac{1}{4}\pi }\\ & \iff 2x-{\displaystyle \frac{1}{4}\pi ={\displaystyle \frac{1}{4}\pi +k.\pi }}\\ & \iff 2x={\displaystyle \frac{2}{4}\pi +k.\pi }\\ & \iff x={\displaystyle \frac{1}{4}\pi +k.\frac{\pi }{2}}\\ & k=0\Rightarrow x={\displaystyle \frac{1}{4}\pi (\text{mm})}\\ & k=1\Rightarrow x={\displaystyle \frac{1}{4}\pi +\frac{\pi }{2}={\displaystyle \frac{3}{4}\pi (\text{mm})}}\\ & k=2\Rightarrow x={\displaystyle \frac{1}{4}\pi +\pi ={\displaystyle \frac{5}{4}\pi (\text{mm})}}\\ & k=3\Rightarrow x={\displaystyle \frac{1}{4}\pi +\frac{3\pi }{2}={\displaystyle \frac{7}{4}\pi (\text{mm})}}\\ & k=4\Rightarrow x={\displaystyle \frac{1}{4}\pi +2\pi ={\displaystyle \frac{9}{4}\pi (\text{tidak memenuhi})}}\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle \frac{1}{4}\pi ,{\displaystyle \frac{3}{4}\pi ,\frac{5}{4}\pi ,\frac{7}{4}\pi }}\right\}\end{array}\end{array}$.

$\begin{array}{l}\\ 75.& \text{Himpunan penyelesaian dari persamaan}\\ & \cos 2x-2\cos x=-1\text{untuk}0<x<2\pi \\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \{0,{\displaystyle \frac{1}{2}\pi ,{\displaystyle \frac{3}{2}\pi ,2\pi }}\}}\\ \\ & \text{b}.{\displaystyle \{0,{\displaystyle \frac{1}{2}\pi ,{\displaystyle \frac{2}{3}\pi ,2\pi }}\}}\\ \\ & \text{c}.{\displaystyle \{0,{\displaystyle \frac{1}{2}\pi ,\pi ,{\displaystyle \frac{3}{2}\pi }}\}}\\ \\ & \text{d}.{\displaystyle \{0,{\displaystyle \frac{1}{2}\pi ,{\displaystyle \frac{2}{3}\pi }}\}}\\ \\ & \text{e}.{\displaystyle \{0,{\displaystyle \frac{1}{2}\pi ,\pi }\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \cos 2x-2\cos x=-1\\ & \iff \cos 2x-2\cos x+1=0\\ & \iff (2{\cos }^{2}x-1)-2\cos x+1=0\\ & \iff 2\cos x(\cos x-1)=0\\ & \iff \cos x=0\text{atau}\cos x=1\\ & \iff \cos x=\cos {\displaystyle \frac{1}{2}\pi \text{atau}\cos x=\cos 0}\\ & \iff x_{1,2}=\pm {\displaystyle \frac{1}{2}\pi +k.2\pi \text{atau}{x}_3=k.2\pi }\\ & \text{maka}\\ & k=0\Rightarrow x_1=-{\displaystyle \frac{1}{2}\pi (\text{tm})\text{atau}}\\ & x_2={\displaystyle \frac{1}{2}\pi (\text{mm})}\\ & x_3={\displaystyle 0(\text{mm})}\\ & k=1\Rightarrow x_1={\displaystyle \frac{3}{2}\pi (\text{mm})}\\ & x_2={\displaystyle \frac{5}{2}\pi (\text{tm})}\\ & x_3={\displaystyle 2\pi (\text{mm})}\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 76.& \text{Himpunan penyelesaian dari persamaan}\\ & 3\tan (2x-{\displaystyle \frac{1}{3}\pi })=-\sqrt{3}\\ & \text{untuk}0\le x\le \pi \text{adalah}....\\ \\ & \text{a}.{\displaystyle \left\{{\displaystyle \frac{1}{12}\pi ,{\displaystyle \frac{7}{12}\pi }}\right\}}\\ \\ & \text{b}.{\displaystyle \left\{{\displaystyle \frac{2}{12}\pi ,{\displaystyle \frac{9}{12}\pi }}\right\}}\\ \\ & \text{c}.{\displaystyle \left\{{\displaystyle \frac{3}{12}\pi ,{\displaystyle \frac{7}{12}\pi }}\right\}}\\ \\ & \text{d}.{\displaystyle \left\{{\displaystyle \frac{3}{12}\pi ,{\displaystyle \frac{9}{12}\pi }}\right\}}\\ \\ & \text{e}.{\displaystyle \left\{{\displaystyle \frac{5}{12}\pi ,{\displaystyle \frac{7}{12}\pi }}\right\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& 3\tan (2x-{\displaystyle \frac{1}{3}\pi })=-\sqrt{3}\\ & \tan (2x-{\displaystyle \frac{1}{3}\pi })=-\frac{\sqrt{3}}{3}\\ & (\text{kuadran IV, karena Y negatif, X positif})\\ & \tan (2x-{\displaystyle \frac{1}{3}\pi })=-\tan {\displaystyle \frac{1}{6}\pi ,\mathbf{\text{menjadi}}}\\ & \tan (2x-{\displaystyle \frac{1}{3}\pi })=\tan (2\pi -{\displaystyle \frac{1}{6}\pi })=\tan {\displaystyle \frac{11}{6}\pi }\\ & (2x-{\displaystyle \frac{1}{3}\pi })={\displaystyle \frac{11}{6}\pi }\\ & \iff 2x={\displaystyle \frac{1}{3}\pi +{\displaystyle \frac{11}{6}\pi +k.\pi ={\displaystyle \frac{13}{6}\pi +k.\pi }}}\\ & \iff x={\displaystyle \frac{13}{12}\pi +{\displaystyle \frac{k.\pi }{2}}}\\ & k=0\Rightarrow x={\displaystyle \frac{13}{12}\pi ={\displaystyle \frac{1}{12}\pi (\text{mm})}}\\ & k=1\Rightarrow x={\displaystyle \frac{13}{12}\pi +{\displaystyle \frac{1}{2}\pi ={\displaystyle \frac{19}{12}\pi ={\displaystyle \frac{7}{12}\pi (\text{mm})}}}}\\ & k=2\Rightarrow x={\displaystyle \frac{13}{12}\pi +\pi \text{tidak memenuhi}}\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle \frac{1}{12}\pi ,{\displaystyle \frac{7}{12}\pi }}\right\}\end{array}\end{array}$.

$\begin{array}{l}\\ 77.& \text{Salah satu nilai}x\text{yang memenuhi}\\ & \text{persamaan}\cos x+\sin x={\displaystyle \frac{1}{2}\sqrt{6}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{24}\pi }& & & \text{d}.& {\displaystyle \frac{1}{8}\pi }\\ \\ \text{b}.& {\displaystyle \frac{1}{15}\pi }& \text{c}.& {\displaystyle \frac{1}{12}\pi }& \text{e}.& {\displaystyle \frac{1}{6}\pi }\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \text{Diketahui bahwa}\\ & \sin x+\cos x={\displaystyle \frac{1}{2}\sqrt{6}(\mathbf{\text{ingat}}:a=1,b=1)}\\ & \sin x+\cos x=k\cos (x-\theta )={\displaystyle \frac{1}{2}\sqrt{6}}\\ & \{\begin{array}{ll}k& =\sqrt{1^2+1^2}=\sqrt{2}\\ \tan & \theta ={\displaystyle \frac{a}{b}={\displaystyle \frac{1}{1}=1\Rightarrow \theta ={45}^{\circ }={\displaystyle \frac{1}{4}\pi }}}\end{array}\\ & \text{sudut}\theta \text{di kuadran I, karena}a,b>0\\ & \text{selanjutnya}\\ & \begin{array}{rl}& \sin x+\cos x=k\cos (x-\theta )={\displaystyle \frac{1}{2}\sqrt{6}}\\ & \iff \sqrt{2}\cos (x-{\displaystyle \frac{1}{4}\pi })={\displaystyle \frac{1}{2}\sqrt{6}}\\ & \iff \cos (x-{\displaystyle \frac{1}{4}\pi })={\displaystyle \frac{{\displaystyle \frac{1}{2}\sqrt{6}}}{\sqrt{2}}={\displaystyle \frac{1}{2}\sqrt{3}}}\\ & \iff \cos (x-{\displaystyle \frac{1}{4}\pi })=\cos {\displaystyle \frac{1}{6}\pi }\\ & \iff x-{\displaystyle \frac{1}{4}\pi =\pm {\displaystyle \frac{1}{6}\pi +k.2\pi }}\\ & \iff x={\displaystyle \frac{1}{4}\pi \pm {\displaystyle \frac{1}{6}\pi +k.2\pi }}\\ & \iff x_1={\displaystyle \frac{5}{12}\pi +k.2\pi \mathbf{\text{atau}}}\\ & x_2={\displaystyle \frac{1}{12}\pi +k.2\pi }\\ & k=0\Rightarrow x_1={\displaystyle \frac{5}{12}\pi (\text{memenuhi})}\\ & \Rightarrow x_2={\displaystyle \frac{1}{12}\pi (\text{memenuhi})}\\ & \text{Langkah berikutnya tidak diperlukan}\\ & \text{karena jawaban sudah kita dapatkan}\\ & \text{yaitu}:{\displaystyle \frac{1}{12}\pi }\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 78.& \text{Himpunan penyelesaian persamaan}\\ & \cos x^{\circ }-\sqrt{3}\sin x^{\circ }=1\\ & \text{untuk}0\le x<360\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \{0,240\}}& & & \text{d}.& {\displaystyle \{180,240\}}\\ \\ \text{b}.& {\displaystyle \{150,270\}}& \text{c}.& {\displaystyle \{180,300\}}& \text{e}.& {\displaystyle \{210,270\}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \text{Diketahui dari soal bahwa}\\ & \cos x^{\circ }-\sqrt{3}\sin x^{\circ }=1,\\ & \text{lalu kita ubah posisinya menjadi}\\ \\ & -\sqrt{3}\sin x+\cos x=1(\mathbf{\text{ingat}}:a=-\sqrt{3},b=1)\\ & -\sqrt{3}\sin x+\cos x=k\cos (x-\theta )=1\\ & \{\begin{array}{ll}k& =\sqrt{{(-\sqrt{3})}^2+{\left(1\right)}^2}=\sqrt{4}=2\\ \tan & \theta ={\displaystyle \frac{a}{b}={\displaystyle \frac{-\sqrt{3}}{1}=-\sqrt{3}\Rightarrow \theta ={300}^{\circ }}}\end{array}\\ & \text{sudut}\theta \text{di kuadran IV, karena}a<0,b>0\\ & \text{selanjutnya}\\ & \begin{array}{rl}& -\sqrt{3}\sin x+\cos x=k\cos (x-\theta )=1\\ & \iff 2\cos (x-{300}^{\circ })=1\\ & \iff \cos (x-{300}^{\circ })={\displaystyle \frac{1}{2}}\\ & \iff \cos (x-{300}^{\circ })=\cos {60}^{\circ }\\ & \iff x-{300}^{\circ }=\pm {60}^{\circ }+k{.360}^{\circ }\\ & \iff x={300}^{\circ }\pm {60}^{\circ }+k{.360}^{\circ }\\ & k=0\Rightarrow x_1={300}^{\circ }+{60}^{\circ }={360}^{\circ }=0^{\circ }(\text{mm})\\ & \text{atau}\\ & x_2={300}^{\circ }-{60}^{\circ }={240}^{\circ }(\text{mm})\\ & k=1\Rightarrow x={300}^{\circ }\pm {60}^{\circ }+{360}^{\circ }(\text{tm})\end{array}\\ & \mathbf{\text{HP}}=\{0^{\circ },{240}^{\circ }\}\end{array}\end{array}$

$\begin{array}{l}\\ 79.& \text{Diketahui}\alpha -\beta ={\displaystyle \frac{\pi }{3}\text{dan}\sin \alpha \sin \beta ={\displaystyle \frac{1}{4}}}\\ & \text{dengan}\alpha \text{dan}\beta \text{adalah sudut}\mathbf{\text{lancip}}\\ & \text{Nilai dari}\cos (\alpha +\beta )\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 1& & & \text{d}.& {\displaystyle \frac{1}{4}}\\ \text{b}.& {\displaystyle \frac{3}{4}}& \text{c}.& {\displaystyle \frac{1}{2}}& \text{e}.& {\displaystyle 0}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & \bullet \alpha -\beta ={\displaystyle \frac{\pi }{3}\text{dan}\sin \alpha \sin \beta ={\displaystyle \frac{1}{4}}}\\ & \bullet \text{dengan}\alpha \text{dan}\beta \text{sudut}\mathbf{\text{lancip}}\\ & \text{akibatnya semua sudut dikuadran I}\\ & \text{sehingga}\{\begin{array}{ll}\sin & =+\\ \cos & =+\\ \tan & =+\end{array}\\ & \text{ditanya}\cos (\alpha +\beta ),\text{maka}\\ & \text{sebagai langkah awal kita adalah}:\\ & \cos (\alpha -\beta )=\cos \left({\displaystyle \frac{\pi }{3}}\right)\\ & \iff \cos \alpha \cos \beta +\sin \alpha \sin \beta ={\displaystyle \frac{1}{2}}\\ & \iff \cos \alpha \cos \beta +{\displaystyle \frac{1}{4}={\displaystyle \frac{1}{2}}}\\ & \iff \cos \alpha \cos \beta ={\displaystyle \frac{1}{2}-{\displaystyle \frac{1}{4}={\displaystyle \frac{1}{4}}}}\\ & \mathbf{\text{Selanjutnya nilai dari}}\\ & \cos (\alpha +\beta )\\ & =\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ & ={\displaystyle \frac{1}{4}-{\displaystyle \frac{1}{4}=0}}\end{array}\end{array}$.

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

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

 $\begin{array}{l}\\ 61.& \text{Nilai}{105}^{\circ }\text{jika dinyatakan ke radian}\\ & \text{adalah}....\text{radian}\\ \\ & \text{a}.{\displaystyle \frac{1}{3}\pi }\\ \\ & \text{b}.{\displaystyle \frac{5}{6}\pi }\\ \\ & \text{c}.{\displaystyle \frac{5}{12}\pi }\\ \\ & \text{d}.{\displaystyle \frac{7}{12}\pi }\\ \\ & \text{e}.{\displaystyle \frac{9}{12}\pi }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Diketah}& \text{ui bahwa}\\ {180}^{\circ }& =\pi radian\\ 1^{\circ }& ={\displaystyle \frac{\pi }{180}radian}\\ 105\times 1^{\circ }& =105\times {\displaystyle \frac{\pi }{180}radian}\\ {105}^{\circ }& ={\displaystyle \frac{7}{12}\pi radian}\end{array}\end{array}$.

$\begin{array}{l}\\ 62.& \text{Nilai}\tan {240}^{\circ }\text{adalah}....\\ \\ & \text{a}.{\displaystyle \sqrt{3}}\\ \\ & \text{b}.{\displaystyle \frac{1}{3}\sqrt{3}}\\ \\ & \text{c}.{\displaystyle -\frac{1}{3}\sqrt{3}}\\ \\ & \text{d}.{\displaystyle \frac{1}{2}\sqrt{3}}\\ \\ & \text{e}.{\displaystyle -\sqrt{3}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\tan {240}^{\circ }& =\tan ({180}^{\circ }+{60}^{\circ })\\ & =\tan {60}^{\circ }\\ & =\sqrt{3}\\ \mathbf{\text{catatan}}& :\text{ingat sudut berelasi}\end{array}\end{array}$.

$\begin{array}{l}\\ 63.& \text{Perhatikanlah gambar berikut}\end{array}$.

$.\begin{array}{l}\\ & \text{Pada gambar di atas perbandingan}\\ & \sin \theta \text{adalah}....\\ & \text{a}.\sqrt{{\displaystyle \frac{a^2-d^2}{f^2+g^2}}}\\ & \text{b}.\sqrt{{\displaystyle \frac{a^2+b^2}{f^2+g^2}}}\\ & \text{c}.\sqrt{{\displaystyle \frac{a^2-b^2}{f^2-g^2}}}\\ & \text{d}.\sqrt{{\displaystyle \frac{a^2+b^2}{f^2-g^2}}}\\ & \text{e}.\sqrt{{\displaystyle \frac{a^2-b^2}{f^2+g^2}}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Dari so}& \text{al diketahui bahwa}\\ \sin \theta & ={\displaystyle \frac{c}{e}={\displaystyle \frac{\sqrt{a^2-b^2}}{\sqrt{f^2+g^2}}}}\\ & =\sqrt{{\displaystyle \frac{a^2-b^2}{f^2+g^2}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 64.& \text{Nilai dari}({\cos }^2{17}^{\circ }-{\sin }^2{73}^{\circ })\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 0& & & \text{d}.& 1\\ \text{b}.& {\displaystyle \frac{1}{3}}& \text{c}.& {\displaystyle \frac{2}{\sqrt{3}}}& \text{e}.& {\displaystyle \frac{1}{2}\sqrt{3}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& ({\cos }^2{17}^{\circ }-{\sin }^2{73}^{\circ })\\ & =({\cos }^2{17}^{\circ }-{(\sin {73}^{\circ })}^2)\\ & =({\cos }^2{17}^{\circ }-{(\sin ({90}^{\circ }-{17}^{\circ }))}^2)\\ & =({\cos }^2{17}^{\circ }-{\cos }^2{17}^{\circ })\\ & =0\end{array}\end{array}$.

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

$\begin{array}{l}\\ 66.& \text{Nilai dari}{\displaystyle \frac{\sin {49}^{\circ }}{\cos {41}^{\circ }}-{\displaystyle \frac{\cos {17}^{\circ }}{\sin {73}^{\circ }}}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -1& & & \text{d}.& 0,143\\ \text{b}.& {\displaystyle -0,321}& \text{c}.& 0& \text{e}.& {\displaystyle 0,321}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& {\displaystyle \frac{\sin {49}^{\circ }}{\cos {41}^{\circ }}-{\displaystyle \frac{\cos {17}^{\circ }}{\sin {73}^{\circ }}}}\\ & ={\displaystyle \frac{\sin {49}^{\circ }}{\cos ({90}^{\circ }-{49}^{\circ })}-{\displaystyle \frac{\cos {17}^{\circ }}{\sin ({90}^{\circ }-{17}^{\circ })}}}\\ & ={\displaystyle \frac{\sin {49}^{\circ }}{\sin {49}^{\circ }}-{\displaystyle \frac{\cos {17}^{\circ }}{\cos {17}^{\circ }}}}\\ & =1-1\\ & =0\end{array}\end{array}$.

$\begin{array}{l}\\ 67.& \text{Nilai dari}\\ & p=r\sin \alpha \cos \beta \\ & q=r\sin \alpha \sin \beta \\ & s=r\cos \alpha \\ & \text{maka pernyataan berikut yang}\\ & \text{tepat adalah}....\\ & \begin{array}{l}\\ \text{a}.& p^2+t^2+s^2=r^2& & & \\ \text{b}.& p^2-t^2+s^2=r^2\\ \text{c}.& p^2+t^2-s^2=r^2& \\ \text{d}.& -p^2+t^2+s^2=r^2& \\ \text{e}.& -p^2-t^2+s^2=r^2\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Saat}\\ & p^2+q^2\text{maka hasilnya adalah}\\ & \begin{array}{l}\\ p^2& =r^2{\sin }^2\alpha {\cos }^2\beta & \\ q^2& =r^2{\sin }^2\alpha {\sin }^2\beta & +\\ & =r^2{\sin }^2\alpha ({\cos }^2\beta +{\sin }^2\beta )\\ & =r^2{\sin }^2\alpha (1)\\ & =r^2{\sin }^2\alpha \end{array}\\ & \text{Dan saat}\\ & p^2+q^2+s^2\text{akan diperoleh hasil}\\ & \begin{array}{l}\\ p^2+q^2& =r^2{\sin }^2\alpha & \\ s^2& =r^2{\cos }^2\alpha & +\\ & =r^2{\sin }^2\alpha +r^2{\cos }^2\alpha \\ & =r^2({\sin }^2\alpha +{\cos }^2\alpha )\\ & =r^2(1)\\ & =r^2\end{array}\end{array}\end{array}$.

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

$\begin{array}{l}\\ 69.& \text{Diketahui bahwa}\\ & \sin \theta +\cos \theta ={\displaystyle \frac{1}{2}}\\ & \text{maka nilai dari}\\ & {\sin }^3\theta +{\cos }^3\theta \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{2}}& & & \text{d}.& {\displaystyle \frac{5}{8}}\\ \\ \text{b}.& {\displaystyle \frac{3}{4}}& \text{c}.& {\displaystyle \frac{9}{15}}& \text{e}.& {\displaystyle \frac{11}{16}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & \sin \theta +\cos \theta ={\displaystyle \frac{1}{2}}\\ & \iff {(\sin \theta +\cos \theta )}^2={\displaystyle \frac{1}{4}}\\ & \iff {\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta ={\displaystyle \frac{1}{4}}\\ & \iff 1+2\sin \theta \cos \theta ={\displaystyle \frac{1}{4}}\\ & \iff 2\sin \theta \cos \theta =-{\displaystyle \frac{3}{4}}\\ & \iff \sin \theta \cos \theta =-{\displaystyle \frac{3}{8}}\\ & \mathbf{\text{Selanjutnya}}\\ & {\sin }^3\theta +{\cos }^3\theta \\ & =(\sin \theta +\cos \theta )({\sin }^2\theta -\sin \theta \cos \theta +{\cos }^2\theta )\\ & =\left({\displaystyle \frac{1}{2}}\right)(1-(-{\displaystyle \frac{3}{8}}))\\ & ={\displaystyle \frac{1}{2}\times {\displaystyle \frac{11}{8}}}\\ & ={\displaystyle \frac{11}{16}}\end{array}\end{array}$.

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

$.\begin{array}{rl}& \text{maka nilai dari}\\ & \sin x\cos x\left(\text{ingat yang diminta di Kwadran IV}\right)\\ & =(-{\displaystyle \frac{m}{\sqrt{m^2+1}}})\times (+{\displaystyle \frac{1}{\sqrt{m^2+1}}})\\ & =-{\displaystyle \frac{m}{m^2+1}}\end{array}$

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

 $\begin{array}{l}\\ 51.& \text{Nilai dari}{\displaystyle \frac{1}{{\sec }^{2}A}+\frac{1}{{\csc }^{2}A}=....}\\ & \begin{array}{l}\\ \text{a}.& -\mathrm{\infty }\\ \text{b}.& {\displaystyle -1}\\ \text{c}.& {\displaystyle 0}\\ \text{d}.& {\displaystyle 1}\\ e.& {\displaystyle \mathrm{\infty }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& {\displaystyle \frac{1}{{\sec }^{2}A}+\frac{1}{{\csc }^{2}A}}\\ & ={\cos }^{2}A+{\sin }^2=1\end{array}\end{array}$

$\begin{array}{l}\\ 52.& \text{Nilai dari}{\displaystyle \frac{\tan B+\tan C}{\cot B+\cot C}=....}\\ & \begin{array}{l}\\ \text{a}.& \cot B\times \cot C\\ \text{b}.& {\displaystyle \tan B\times \tan C}\\ \text{c}.& {\displaystyle \sec B\times \csc C}\\ \text{d}.& {\displaystyle \tan B\times \cot C}\\ e.& {\displaystyle \tan B\times \csc C}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& {\displaystyle \frac{\tan B+\tan C}{\cot B+\cot C}}\\ & ={\displaystyle \frac{\tan B+\tan C}{{\displaystyle \frac{1}{\tan B}+\frac{1}{\tan C}}}}\\ & ={\displaystyle \frac{\tan B+\tan C}{\left({\displaystyle \frac{\tan B+\tan C}{\tan B\times \tan C}}\right)}}\\ & =\tan B\times \tan C\end{array}\end{array}$

$\begin{array}{l}\\ 53.& \text{Nilai dari}\\ & {\displaystyle \frac{\tan A}{\sec A-1}+\frac{\tan A}{\sec A+1}=....}\\ & \begin{array}{l}\\ \text{a}.& 2\tan A\\ \text{b}.& 2\cot A\\ \text{c}.& {\displaystyle 2\sec A}\\ \text{d}.& {\displaystyle 2\csc A}\\ e.& {\displaystyle 2\tan A.\sec A}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& {\displaystyle \frac{\tan A}{\sec A-1}+\frac{\tan A}{\sec A+1}}\\ & =\tan A\left({\displaystyle \frac{1}{{\displaystyle \frac{1}{\cos A}-1}}+\frac{1}{{\displaystyle \frac{1}{\cos A}+1}}}\right)\\ & ={\displaystyle \frac{\sin A}{\cos A}\left({\displaystyle \frac{\cos A}{1-\cos A}+\frac{\cos A}{1+\cos A}}\right)}\\ & ={\displaystyle \frac{\sin A}{1-\cos A}+\frac{\sin A}{1+\cos A}}\\ & ={\displaystyle \frac{\sin A(1+\cos A)+\sin A(1-\cos A)}{(1-\cos A)(1+\cos A)}}\\ & ={\displaystyle \frac{2\sin A}{1-{\cos }^2}}\\ & ={\displaystyle \frac{2\sin A}{{\sin }^{2}A}}\\ & ={\displaystyle \frac{2}{\sin A}}\\ & =2\csc A\end{array}\\ \\ & \text{Sebagai catatanya}\\ & \text{Anda bisa gunakan cara yang lain}\end{array}$

$\begin{array}{l}\\ 54.& \text{Nilai}x\text{yang memenuhi persamaan}\\ & \sin (2x-{20}^{\circ })=-\cos (3x+{50}^{\circ })\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -{30}^{\circ }\\ \text{b}.& -{25}^{\circ }\\ \text{c}.& {20}^{\circ }\\ \text{d}.& {25}^{\circ }\\ e.& {30}^{\circ }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\sin (2x-{20}^{\circ })& =-\cos (3x+{50}^{\circ })\\ \sin ({20}^{\circ }-2x)& =\cos (3x+{50}^{\circ })\\ \sin A& =\cos B,\text{artinya}\\ A+B& ={90}^{\circ },\text{maka}\\ ({20}^{\circ }-2x)+(3x+{50}^{\circ })& ={90}^{\circ }\\ x+{70}^{\circ }& ={90}^{\circ }\\ x& ={90}^{\circ }-{70}^{\circ }\\ & ={20}^{\circ }\end{array}\end{array}$

$\begin{array}{l}\\ 55.& \text{Nilai}x\text{yang memenuhi persamaan}\\ & \tan (2x+{60}^{\circ })=\cot ({90}^{\circ }-3x)\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {20}^{\circ }\\ \text{b}.& {30}^{\circ }\\ \text{c}.& {40}^{\circ }\\ \text{d}.& {50}^{\circ }\\ \text{e}.& {60}^{\circ }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}\tan (2x+{60}^{\circ })& =\cot ({90}^{\circ }-3x)\\ \tan (2x+{60}^{\circ })& =\tan 3x\\ 2x+{60}^{\circ }& =3x\\ 2x-3x& =-{60}^{\circ }\\ -x& =-{60}^{\circ }\\ x& ={60}^{\circ }\end{array}\end{array}$.

$\begin{array}{l}\\ 56.& \text{Jika nilai}\cot A+\cos A=x\\ & \text{dan}\cot A-\cos A=y,\\ & \text{maka nilai}(x^2-y^2)=....\\ & \begin{array}{l}\\ \text{a}.& 4\sqrt{xy}\\ \text{b}.& 2\sqrt{xy}\\ \text{c}.& xy\\ \text{d}.& 2\\ \text{e}.& 4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}xy& =(\cot A+\cos A)(\cot A-\cos A)\\ & ={\cot }^{2}A-{\cos }^{2}A\\ & ={\displaystyle \frac{{\cos }^{2}A}{{\sin }^{2}A}-{\cos }^{2}A}\\ & ={\displaystyle \frac{{\cos }^{2}A}{{\sin }^{2}A}-{\cos }^{2}A\times \frac{{\sin }^{2}A}{{\sin }^{2}A}}\\ & ={\displaystyle \frac{{\cos }^{2}A}{{\sin }^{2}A}(1-{\sin }^{2}A)}\\ & ={\displaystyle \frac{{\cos }^{4}A}{{\sin }^{2}A}}\\ \sqrt{xy}& ={\displaystyle \frac{\cos A}{\sin A}\times \cos A}\\ \text{Se}& \text{lanjutnya}\\ x^2-y^2& ={(\cot A+\cos A)}^2-{(\cot A-\cos A)}^2\\ (x+y)& (x-y)=(\cot A+\cos A+\cot A-\cos A)\\ & \times (\cot A+\cos A-(\cot A-\cos A))\\ x^2-y^2& =2\cot A\times 2\cos A\\ & =4\times {\displaystyle \frac{\cos A}{\sin A}\times \cos A}\\ & =4\sqrt{xy}\end{array}\end{array}$

$\begin{array}{l}\\ 57.& \text{Jika nilai}\cos A+\sin A=\sqrt{2}\cos A\\ & \text{maka nilai}(\cos A-\sin A)=....\\ & \begin{array}{l}\\ \text{a}.& -\sqrt{2}\cos A\\ \text{b}.& -\sqrt{2}\sin A\\ \text{c}.& \sqrt{2}\sin A\\ \text{d}.& {\displaystyle \frac{1}{\sqrt{2}}\sec A}\\ \text{e}.& {\displaystyle \frac{1}{\sqrt{2}}\csc A}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\cos A+\sin A& =\sqrt{2}\cos A\\ {(\cos A+\sin A)}^2& ={(\sqrt{2}\cos A)}^2\\ 1+2\sin A\cos A& =2{\cos }^{2}A\\ 2\sin A\cos A& =2{\cos }^{2}A-1\\ \text{maka}& \\ {(\cos A-\sin A)}^2& =1-2\sin A\cos A\\ & =1-(2{\cos }^{2}A-1)\\ & =2-2{\cos }^{2}A\\ & =2(1-{\cos }^{2}A)\\ & =2{\sin }^{2}A\\ \cos A-\sin A& =\sqrt{2{\sin }^{2}A}\\ & =\sin A\sqrt{2}\\ & =\sqrt{2}.\sin A\end{array}\end{array}$.

$\begin{array}{l}\\ 58.& \text{Nilai}\cos \gamma (\csc \gamma +\tan \gamma )\\ & \text{adalah ekivalen dengan}....\\ & \begin{array}{l}\\ \text{a}.& \cot \gamma +\sin \gamma \\ \text{b}.& \tan \gamma +\cos \gamma \\ \text{c}.& \cot \gamma -\sin \gamma \\ \text{d}.& \tan \gamma -\cos \gamma \\ \text{e}.& \cot \gamma -\tan \gamma \end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\cos \gamma (\csc \gamma +\tan \gamma )& =\cos \gamma \left({\displaystyle \frac{1}{\sin \gamma }+{\displaystyle \frac{\sin \gamma }{\cos \gamma }}}\right)\\ & =\cos \gamma \left({\displaystyle \frac{\cos \gamma +{\sin }^2\gamma }{\sin \gamma \cos \gamma }}\right)\\ & ={\displaystyle \frac{\cos \gamma +{\sin }^2\gamma }{\sin \gamma }}\\ & ={\displaystyle \frac{\cos \gamma }{\sin \gamma }+{\displaystyle \frac{{\sin }^2\gamma }{\sin \gamma }}}\\ & =\cot \gamma +\sin \gamma \end{array}\end{array}$.

$\begin{array}{l}\\ 59.& \text{Jika diketahui}\sin \theta \cos \theta ={\displaystyle \frac{3}{8},}\\ & \text{maka nilai}{\displaystyle \frac{1}{\sin \theta }-{\displaystyle \frac{1}{\cos \theta }\text{adalah}=....}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{4}}\\ \text{b}.& {\displaystyle \frac{1}{2}}\\ \text{c}.& {\displaystyle \frac{3}{4}}\\ \text{d}.& {\displaystyle \frac{4}{3}}\\ \text{e}.& {\displaystyle 4}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}& {\displaystyle \frac{1}{\sin \theta }-{\displaystyle \frac{1}{\cos \theta }}}\\ & =\sqrt{{\left({\displaystyle \frac{1}{\sin \theta }-{\displaystyle \frac{1}{\cos \theta }}}\right)}^2}\\ & =\sqrt{{\displaystyle \frac{1}{{\sin }^2\theta }-{\displaystyle \frac{2}{\sin \theta \cos \theta }+{\displaystyle \frac{1}{{\cos }^2\theta }}}}}\\ & =\sqrt{{\displaystyle \frac{{\cos }^2\theta +{\sin }^2\theta }{{\sin }^2\theta {\cos }^2\theta }-{\displaystyle \frac{2}{\sin \theta \cos \theta }}}}\\ & =\sqrt{{\displaystyle \frac{1}{{\left({\displaystyle \frac{3}{8}}\right)}^2}-{\displaystyle \frac{2}{\left({\displaystyle \frac{3}{8}}\right)}}}}=\sqrt{{\displaystyle \frac{1}{{\displaystyle \frac{9}{64}}}-{\displaystyle \frac{2}{{\displaystyle \frac{3}{8}}}}}}=\sqrt{{\displaystyle \frac{64}{9}-{\displaystyle \frac{16}{3}}}}\\ & =\sqrt{{\displaystyle \frac{64}{9}-{\displaystyle \frac{48}{9}}}}=\sqrt{{\displaystyle \frac{16}{9}}}={\displaystyle \frac{4}{3}}\end{array}\end{array}$.

$\begin{array}{l}\\ 60.& \text{Bentuk}{\displaystyle \frac{(\sin \alpha +\sin \beta )(\sin \alpha -\sin \beta )}{{(\cos \alpha \cos \beta )}^2}}\\ & \text{ekivalen dengan}....\\ & \begin{array}{l}\\ \text{a}.& \tan \alpha -\tan \beta \\ \text{b}.& \tan \alpha +\tan \beta \\ \text{c}.& {\tan }^2\alpha -{\tan }^2\beta \\ \text{d}.& {\tan }^2\alpha +{\tan }^2\beta \\ \text{e}.& {\tan }^3\alpha -{\tan }^3\beta \end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}& {\displaystyle \frac{(\sin \alpha +\sin \beta )(\sin \alpha -\sin \beta )}{{(\cos \alpha \cos \beta )}^2}}\\ & ={\displaystyle \frac{{\sin }^2\alpha -{\sin }^2\beta }{{(\cos \alpha \cos \beta )}^2}}\\ & ={\displaystyle \frac{{\sin }^2\alpha -{\sin }^2\alpha {\sin }^2\beta -{\sin }^2\beta +{\sin }^2\alpha {\sin }^2\beta }{{(\cos \alpha \cos \beta )}^2}}\\ & ={\displaystyle \frac{{\sin }^2\alpha (1-{\sin }^2\beta )-{\sin }^2\beta (1-{\sin }^2\alpha )}{{(\cos \alpha \cos \beta )}^2}}\\ & ={\displaystyle \frac{{\sin }^2\alpha {\cos }^2\beta -{\sin }^2\beta {\cos }^2\alpha }{{(\cos \alpha \cos \beta )}^2}}\\ & ={\displaystyle \frac{{\sin }^2\alpha {\cos }^2\beta }{{(\cos \alpha \cos \beta )}^2}-{\displaystyle \frac{{\sin }^2\beta {\cos }^2\alpha }{{(\cos \alpha \cos \beta )}^2}}}\\ & ={\displaystyle \frac{{\sin }^2\alpha }{{\cos }^2\alpha }-{\displaystyle \frac{{\sin }^2\beta }{{\cos }^2\beta }}}\\ & ={\tan }^2\alpha -{\tan }^2\beta \end{array}\end{array}$.

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)

 $\begin{array}{l}\\ 41.& \text{Jika}{\tan }^{2}x+\sec x=5\text{untuk}\\ & 0\le x\le {\displaystyle \frac{\pi }{2}\text{maka nilai}\cos x=....}\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& {\displaystyle \frac{1}{2}}\\ \text{c}.& {\displaystyle \frac{1}{3}}\\ \text{d}.& {\displaystyle \frac{1}{\sqrt{2}}}\\ \text{e}.& {\displaystyle \frac{1}{2}\sqrt{3}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Ingat bahwa}& 0\le x\le {\displaystyle \frac{\pi }{2}}\\ \text{berarti sudu}& \text{t}x\text{berada di kuadran I}\\ \text{sehingga ak}& \text{an menyebabkan nilai}\\ & \cos x=+\\ \text{Selanjutnya}& \\ {\tan }^{2}x+\sec x& =5\\ {\sec }^{2}x-1+\sec x& =5\\ {\sec }^{2}x+\sec x-6& =0\\ (\sec x+3)(\sec x-2)& =0\\ \sec x=-3\text{atau}& \sec x=2\\ \text{untuk}\sec x& =-3(\text{tidak memenuhi})\\ \text{untuk}\sec x& =2(\text{memenuhi})\\ \text{Selanjutnya}& \text{lagi}\\ \sec x& =2\\ {\displaystyle \frac{1}{\cos x}}& =2\\ \cos x& ={\displaystyle \frac{1}{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 42.& \text{Himpunan penyelesaian persamaan}\\ & 3\cos 2x+5\sin x+1=0\text{untuk}{0}^{\circ }\le x\le 2\pi \\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \left\{{\displaystyle \frac{7}{6}\pi ,\frac{11}{6}\pi }\right\}\\ \text{b}.& \left\{{\displaystyle \frac{5}{6}\pi ,\frac{11}{6}\pi }\right\}\\ \text{c}.& \left\{{\displaystyle \frac{1}{6}\pi ,\frac{7}{6}\pi }\right\}\\ \text{d}.& \left\{{\displaystyle \frac{1}{5}\pi ,\frac{5}{6}\pi }\right\}\\ \text{e}.& \left\{{\displaystyle \frac{5}{6}\pi ,\frac{7}{6}\pi }\right\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}3\cos 2x+5\sin x+1& =0\\ 3(1-2{\sin }^{2}x)+5\sin x+1& =0\\ -6{\sin }^2+5\sin x+4& =0\\ 6{\sin }^{2}x-5\sin x-4& =0\\ (3\sin x-4)(2\sin x+1)& =0\\ \sin x={\displaystyle \frac{4}{3}\text{atau}\sin x}& =-\frac{1}{2}\\ \sin x& =\sin {150}^{\circ }=\frac{5}{6}\pi \\ x& =\{\begin{array}{ll}{\displaystyle \frac{7}{6}\pi }& +k.2\pi \\ \pi -{\displaystyle \frac{7}{6}\pi }& +k.2\pi \end{array}\\ \text{saat}k& =0\\ x_1& ={\displaystyle \frac{7}{6}\pi }\\ x_2& =-{\displaystyle \frac{1}{6}\pi }\\ \text{saat}k& =1\\ x_1& ={\displaystyle \frac{7}{6}\pi +2\pi }\\ x_2& =-{\displaystyle \frac{1}{6}\pi +2\pi =\frac{11}{6}\pi }\end{array}\end{array}$

$\begin{array}{l}\\ 43.& \text{Himpunan penyelesaian dari}\\ & \sqrt{3}\sin 2x+2{\cos }^{2}x=-1\text{untuk}\\ & 0^{\circ }\le x\le {360}^{\circ }\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{{240}^{\circ },{300}^{\circ }\}\\ \text{b}.& \{{30}^{\circ },{60}^{\circ }\}\\ \text{c}.& \{{150}^{\circ },{315}^{\circ }\}\\ \text{d}.& \{{120}^{\circ },{300}^{\circ }\}\\ \text{e}.& \{{60}^{\circ },{150}^{\circ }\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\sqrt{3}\sin 2x+2{\cos }^{2}x& =-1\\ \sqrt{3}\sin 2x+1+\cos 2x& =-1\\ \sqrt{3}\sin 2x+\cos 2x& =-2\\ \sqrt{{\sqrt{3}}^2+1^2}\cos (2x-\alpha )& =-2\\ a=x=1,b& =y=\sqrt{3}\\ \alpha & =\arctan {\displaystyle \frac{b}{a}=\arctan \frac{\sqrt{3}}{1}}\\ \alpha & ={60}^{\circ }\\ \text{maka persamaan akan}& \text{menjadi}\\ 2\cos (2x-{60}^{\circ })& =-2\\ \cos (2x-{60}^{\circ })& =-1\\ \cos (2x-{60}^{\circ })& =\cos {180}^{\circ }\\ (2x-{60}^{\circ })& =\pm {180}^{\circ }+k{.360}^{\circ }\\ 2x& ={60}^{\circ }\pm {180}^{\circ }+k{.360}^{\circ }\\ x& ={30}^{\circ }\pm {90}^{\circ }+k{.180}^{\circ }\\ \text{saat}k& =0\\ x_1& ={120}^{\circ }\\ x_2& =-{60}^{\circ }\\ \text{saat}k& =1\\ x_3& ={120}^{\circ }+{360}^{\circ }=....\\ x_2& =-{60}^{\circ }+{360}^{\circ }={300}^{\circ }\end{array}\end{array}$

$\begin{array}{l}\\ 44.& \text{Jika}3\sin \theta +4\cos \theta =5\text{maka}\\ & \text{nilai dari}\sin \theta \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 0,3\\ \text{b}.& 0,60\\ \text{c}.& 0,75\\ \text{d}.& 0,80\\ \text{e}.& 1,20\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}3\sin \theta +4\cos \theta & =5\\ k\cos (\theta -\alpha )& =5\\ a=x=4,& b=y=3\\ \theta & =\arctan {\displaystyle \frac{b}{a}}\\ \theta & =\arctan {\displaystyle \frac{3}{4}}\\ \text{atau}& \\ \tan \theta & ={\displaystyle \frac{3}{4}}\\ \text{maka}& \\ \sin \theta & ={\displaystyle \frac{3}{\sqrt{3^2+4^2}}}\\ & ={\displaystyle \frac{3}{5}}\\ & =0,6\end{array}\end{array}$

$\begin{array}{l}\\ 45.& \text{Jika}\tan \theta +\sec \theta =x\text{maka}\\ & \text{nilai dari}\tan \theta \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{2x}{x^2-1}}\\ \text{b}.& {\displaystyle \frac{2x}{x^2+1}}\\ \text{c}.& {\displaystyle \frac{x^2+1}{2x}}\\ \text{d}.& {\displaystyle \frac{x^2-1}{2x}}\\ \text{e}.& {\displaystyle \frac{x^2-1}{x^2+1}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\text{Langkah}& 1\\ \tan \theta +\sec \theta & =x\\ {\displaystyle \frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta }}& =x\\ {\displaystyle \frac{\sin \theta +1}{\cos \theta }}& =x\\ \sin \theta +1& =x\cos \theta ...........1\\ \text{Langkah}& 2\\ {(\tan \theta +\sec \theta )}^2& =x^2\\ {\tan }^2\theta +2\tan \theta \sec \theta +{\sec }^2\theta & =x^2\\ {\sec }^2\theta -1+2\tan \theta \sec \theta & +{\sec }^2\theta =x^2\\ 2{\sec }^2\theta +2\tan \theta \sec \theta & =x^2+1\\ {\displaystyle \frac{2}{{\cos }^2\theta }+\frac{2\sin \theta }{{\cos }^2\theta }}& =x^2+1\\ 1+\sin \theta & =\left({\displaystyle \frac{x^2+1}{2}}\right){\cos }^2\theta .....2\\ \text{Langkah}& 3\\ \left({\displaystyle \frac{x^2+1}{2}}\right){\cos }^2& =x\cos \theta \\ \cos \theta & ={\displaystyle \frac{2x}{x^2+1}}\\ \text{maka}& (\mathbf{\text{dengan sisi segitiga}})\\ \tan \theta ={\displaystyle \frac{\sqrt{{(x^2+1)}^2-(2x{)}^2}}{2x}}& ={\displaystyle \frac{\sqrt{x^4+2x^2+1-4x^2}}{2x}}\\ \tan \theta & ={\displaystyle \frac{\sqrt{x^4-2x^2+1}}{2x}}\\ & ={\displaystyle \frac{\sqrt{{(x^2-1)}^2}}{2x}}\\ & ={\displaystyle \frac{x^2-1}{2x}}\end{array}\end{array}$

$\begin{array}{rl}.\mathbf{\text{Sehingga}}& \mathbf{\text{dari kasus di atas didapatkan}}\\ \sin \theta & ={\displaystyle \frac{x^2-1}{x^2+1}}\\ \cos \theta & ={\displaystyle \frac{2x}{x^2+1}}\\ \tan \theta & ={\displaystyle \frac{x^2-1}{2x}}\end{array}$

$\begin{array}{l}\\ 46.& \text{Jika}\sec x+\tan x={\displaystyle \frac{3}{2}\text{untuk}}\\ & 0\le x\le {\displaystyle \frac{\pi }{2}\text{maka nilai}\sin x=....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{5}{13}}\\ \text{b}.& {\displaystyle \frac{12}{13}}\\ \text{c}.& {\displaystyle 1}\\ \text{d}.& {\displaystyle \frac{2}{13}}\\ \text{e}.& {\displaystyle \frac{5}{12}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\sec x+\tan x& ={\displaystyle \frac{3}{2}}\\ \text{ingat saat}& \text{mengerjakan soal no}.14\\ \sec \theta +\tan \theta & =x\\ \sin \theta & ={\displaystyle \frac{x^2-1}{x^2+1},\text{maka}}\\ \sin x& ={\displaystyle \frac{{\left({\displaystyle \frac{3}{2}}\right)}^2-1}{{\left({\displaystyle \frac{3}{2}}\right)}^2+1}}\\ & ={\displaystyle \frac{{\displaystyle \frac{9}{4}-1}}{{\displaystyle \frac{9}{4}+1}}}\\ & =\frac{{\displaystyle \frac{5}{4}}}{{\displaystyle \frac{13}{4}}}\\ & ={\displaystyle \frac{5}{13}}\end{array}\end{array}$.

$\begin{array}{l}\\ 47.& \text{Nilai}\\ & {(\sin A+\cos A)}^2+{(\sin A-\cos A)}^2=....\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& {\displaystyle 2}\\ \text{c}.& {\displaystyle 3}\\ \text{d}.& {\displaystyle 3\cos A}\\ e.& {\displaystyle 4\sin A}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& {(\sin A+\cos A)}^2+{(\sin A-\cos A)}^2\\ & ={\sin }^{2}A+2\sin A\cos A+{\cos }^{2}A\\ & +{\sin }^{2}A-2\sin A\cos A+{\cos }^{2}A\\ & =1+1\\ & =2\end{array}\end{array}$

$\begin{array}{l}\\ 48.& \text{Nilai}\sqrt{{\displaystyle \frac{1+\sin A}{1-\sin A}}}=....\\ & \begin{array}{l}\\ \text{a}.& \sec A+\tan A\\ \text{b}.& {\sec }^{2}A+{\tan }^{2}A\\ \text{c}.& {\sec }^{2}A-{\tan }^{2}A\\ \text{d}.& {\tan }^{2}A-{\sec }^{2}A\\ e.& \sec A\times \tan A\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \sqrt{{\displaystyle \frac{1+\sin A}{1-\sin A}}}\\ & =\sqrt{{\displaystyle \frac{1+\sin A}{1-\sin A}\times \frac{1+\sin A}{1+\sin A}}}\\ & =\sqrt{{\displaystyle \frac{{(1+\sin A)}^2}{1-{\sin }^{2}A}}}\\ & =\sqrt{{\displaystyle \frac{{(1+\sin A)}^2}{{\cos }^{2}A}}}\\ & ={\displaystyle \frac{1+\sin A}{\cos A}}\\ & ={\displaystyle \frac{1}{\cos A}+\frac{\sin A}{\cos A}}\\ & =\sec A+\tan A\end{array}\end{array}$

$\begin{array}{l}\\ 49.& \text{Jika}{0}^{\circ }\le \theta ⩽{90}^{\circ },\text{maka nilai}\\ & \left({\displaystyle \frac{5\cos \theta -4}{3-5\sin \theta }-\frac{3+5\sin \theta }{4+5\cos \theta }}\right)=....\\ & \begin{array}{l}\\ \text{a}.& -1\\ \text{b}.& {\displaystyle 0}\\ \text{c}.& {\displaystyle \frac{1}{4}}\\ \text{d}.& {\displaystyle \frac{1}{2}}\\ e.& {\displaystyle 1}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \left({\displaystyle \frac{5\cos \theta -4}{3-5\sin \theta }-\frac{3+5\sin \theta }{4+5\cos \theta }}\right)\\ & =\left({\displaystyle \frac{5\cos \theta -4}{3-5\sin \theta }\times \frac{4+5\cos \theta }{4+5\cos \theta }}\right)\\ & -\left({\displaystyle \frac{3+5\sin \theta }{4+5\cos \theta }\times \frac{3-5\sin \theta }{3-5\sin \theta }}\right)\\ & ={\displaystyle \frac{25{\cos }^2-16-(9-25{\sin }^2\theta )}{(3-5\sin \theta )(4+5\cos \theta )}}\\ & ={\displaystyle \frac{25({\sin }^2\theta +{\cos }^2\theta )-25}{(3-5\sin \theta )(4+5\cos \theta )}}\\ & ={\displaystyle \frac{0}{(3-5\sin \theta )(4+5\cos \theta )}}\\ & =0\end{array}\end{array}$

$\begin{array}{l}\\ 50.& \text{Nilai}\\ & (1+\cot \theta -\csc \theta )(1+\tan \theta +\sec \theta )=....\\ & \begin{array}{l}\\ \text{a}.& -2\\ \text{b}.& {\displaystyle -1}\\ \text{c}.& {\displaystyle 0}\\ \text{d}.& {\displaystyle 1}\\ \text{e}.& {\displaystyle 2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& (1+\cot \theta -\csc \theta )(1+\tan \theta +\sec \theta )\\ & =(1+{\displaystyle \frac{\cos \theta }{\sin \theta }-\frac{1}{\sin \theta }})(1+\frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta })\\ & =\left({\displaystyle \frac{\sin \theta +\cos \theta -1}{\sin \theta }}\right)\left({\displaystyle \frac{\cos \theta +\sin \theta +1}{\cos \theta }}\right)\\ & ={\displaystyle \frac{{(\sin \theta +\cos \theta )}^2-1}{\sin \theta \cos \theta }}\\ & ={\displaystyle \frac{1+2\sin \theta \cos \theta -1}{\sin \theta \cos \theta }}\\ & =2\end{array}\end{array}$

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

 $\begin{array}{l}\\ 31.& \text{Nilai}x\text{yang memenuhi persamaan}\\ & 2{\cos }^{2}x+\cos x-1=0\text{untuk}0\le x\le \pi \\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{3}\pi \text{dan}\pi }\\ \text{b}.& {\displaystyle \frac{1}{3}\pi \text{dan}\frac{2}{3}\pi }\\ \text{c}.& {\displaystyle \frac{1}{3}\pi \text{dan}\frac{3}{4}\pi }\\ \text{d}.& {\displaystyle \frac{1}{4}\pi \text{dan}\frac{3}{4}\pi }\\ \text{e}.& {\displaystyle \frac{1}{4}\pi \text{dan}\frac{2}{3}\pi }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}2{\cos }^{2}x+\cos x-1& =0\\ (2\cos x-1)(\cos x+1)& =0\\ \cos x={\displaystyle \frac{1}{2}\text{atau}}& \cos x=-1\\ \cos x=\cos {60}^{\circ }=\cos \frac{1}{3}\pi & \\ \text{atau}\cos x=\cos {180}^{\circ }& =\cos \pi \end{array}\end{array}$

$\begin{array}{l}\\ 32.& \text{Untuk}x\text{yang memenuhi persamaan}\\ & {\tan }^{2}x-\tan x-6=0\text{pada}0\le x\le \pi ,\\ & \text{maka himpunan nilai}\sin x\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \left\{{\displaystyle \frac{3\sqrt{10}}{10},\frac{2\sqrt{5}}{5}}\right\}\\ \text{b}.& \left\{{\displaystyle \frac{3\sqrt{10}}{10},-\frac{2\sqrt{5}}{5}}\right\}\\ \text{c}.& \{-{\displaystyle \frac{3\sqrt{10}}{10},\frac{2\sqrt{5}}{5}}\}\\ \text{d}.& \left\{{\displaystyle \frac{\sqrt{10}}{10},\frac{\sqrt{5}}{5}}\right\}\\ \text{e}.& \left\{{\displaystyle \frac{\sqrt{10}}{10},\frac{2\sqrt{5}}{5}}\right\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{\tan }^{2}x-\tan x-6& =0\\ (\tan x-3)(\tan x+2)& =0\\ \tan x=3\text{atau}& \tan x=-2\\ \tan x={\displaystyle \frac{3}{1}\text{atau}}& \tan x=\frac{-2}{1}\\ \sin x={\displaystyle \frac{3}{\sqrt{1^2+3^2}}\text{atau}}& \sin x=\frac{2}{\sqrt{1^2+2^2}}\\ \sin x={\displaystyle \frac{3}{\sqrt{10}}\text{atau}}& \sin x=\frac{2}{\sqrt{5}}\\ \sin x={\displaystyle \frac{3}{10}\sqrt{10}\text{atau}}& \sin x=\frac{2}{5}\sqrt{5}\end{array}\end{array}$.

$\begin{array}{l}\\ 33.& \text{Jika}2{\sin }^{2}x+3\cos x=0\\ & \text{dan}{0}^0\le x\le {180}^0,\\ & \text{maka nilai textit{x} adalah}....\\ & \begin{array}{l}\\ \text{a}.& {30}^0\\ \text{b}.& {60}^0\\ \text{c}.& {120}^0\\ \text{d}.& {150}^0\\ \text{e}.& {170}^0\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}2& {\sin }^{2}x+3\cos x=0,\\ & \text{ingat identitas}{\sin }^{2}x+{\cos }^{2}x=1\\ 2& (1-{\cos }^{2}x)+3\cos x=0\\ 2& -2{\cos }^{2}x+3\cos x=0,\\ & \text{(dikali dengan -1)}\\ 2& {\cos }^{2}x-\cos x-2=0,\\ & \text{menjadi persamaan kuadrat dalam}\cos x\\ & (\cos x-2)(2\cos x+1)=0,(\text{difaktorkan})\\ & \cos x-2=0\text{atau}2\cos x+1=0\\ & \underset{\text{tidak memenuhi}}{\underbrace{\cos x=2}}\text{atau}\underset{\text{memenuhi}}{\underbrace{\cos x=-{\displaystyle \frac{1}{2}}}},\\ & \text{ingat rentang nilai cosinus adalah}:|\cos x|\le 1\\ & \text{Selanjutnya pilih yang memenuhi, yaitu}\\ & \cos x=-{\displaystyle \frac{1}{2}}\\ & \cos x=\cos ({180}^0-{60}^0)\\ & \cos x=\cos {120}^0\\ & \therefore x={120}^0\end{array}\end{array}$.

$\begin{array}{l}\\ 34.& \text{Akar-akar dari persamaan}\\ & 4{\sin }^{2}x+4\cos x=1\text{pada selang}\\ & -\pi \le x\le \pi \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{3}{2}\text{atau}-\frac{1}{2}}\\ \text{b}.& -{\displaystyle \frac{3}{2}\text{atau}\frac{1}{2}}\\ \text{c}.& {\displaystyle \frac{3}{2}\pi \text{atau}-\frac{1}{2}\pi }\\ \text{d}.& {\displaystyle \frac{1}{3}\pi \text{atau}-\frac{1}{3}\pi }\\ \text{e}.& {\displaystyle -\frac{2}{3}\pi \text{atau}\frac{2}{3}\pi }\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}4& {\sin }^{2}x+4\cos x=1\\ 4& (1-{\cos }^{2}x)+4\cos x-1=0\\ 4& -4{\cos }^{2}x+4\cos x-1=0\\ 4& {\cos }^{2}x-4\cos x-3=0\\ & (2\cos x-3)(2\cos x+1)=0\\ & (2\cos x-3)=0\text{atau}(2\cos x+1)=0\\ & \underset{\text{tidak memenuhi}}{\underbrace{\cos x={\displaystyle \frac{3}{2}}}}\text{atau}\underset{\text{memenuhi}}{\underbrace{\cos x=-{\displaystyle \frac{1}{2}}}}\\ & \text{pilih yang memenuhi persamaan, yaitu}\\ & \cos x=-{\displaystyle \frac{1}{2}}\\ & \cos x=\cos ({180}^0-{60}^0)\\ & \cos x=\cos {120}^0\\ & \therefore x={120}^0={\displaystyle \frac{2}{3}\pi }\end{array}\end{array}$.

$\begin{array}{l}\\ 35.& \text{Himpunan penyelesaian persamaan}\\ & {\sin }^{2}2x+2\sin x\cos x-2=0\\ & \text{pada selang}0\le x\le {360}^0\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{{45}^0,{135}^0\}\\ \text{b}.& \{{45}^0,{225}^0\}\\ \text{c}.& \{{135}^0,{180}^0\}\\ \text{d}.& \{{135}^0,{225}^0\}\\ \text{e}.& \{{135}^0,{315}^0\}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}& {\sin }^{2}2x+2\sin x\cos x-2=0\\ & {\sin }^{2}2x+\sin 2x-2=0,\\ & \text{(ingat bahwa)}:\sin 2x=2\sin x\cos x\\ & (\sin 2x+2)(\sin 2x-1)=0\\ & \sin 2x+2=0\text{atau}\sin 2x-1=0\\ & \underset{\text{tidak memenuhi}}{\underbrace{\sin 2x=-2}}\text{atau}\underset{\text{memenuhi}}{\underbrace{\sin 2x=1}}\\ & \text{pilih yang memenuhi persamaan, yaitu}\\ & \sin 2x=1\\ & \sin 2x=\sin {90}^0\\ & 2x={90}^0+k{.360}^0\text{atau}\\ & 2x=({180}^0-{90}^0)+k{.360}^0\\ & \text{(cukup ambil 1 persamaan saja, karena sama)}\\ & x={45}^0+k{.180}^0\\ & k=0\Rightarrow x={45}^0+{0.180}^0={45}^0\\ & k=1\Rightarrow x={45}^0+{1.180}^0={45}^0+{180}^0={225}^0\\ & k=2\Rightarrow x={45}^0+{2.180}^0={45}^0+{360}^0={405}^0\\ & \text{(tidak memenuhi rentang)}\\ & \therefore \mathbf{\text{HP}}=\{{45}^0,{225}^0\}\end{array}\end{array}$.

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

$\begin{array}{l}\\ 37.& \text{Nilai-nilai}x\text{yang terletak pada}0\le x\le 2\pi ,\\ & \text{yang memenuhi persamaan}\sqrt{3}\cos x+\sin x=\sqrt{2}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {75}^{\circ }\text{atau}{285}^{\circ }\\ \text{b}.& {75}^{\circ }\text{atau}{345}^{\circ }\\ \text{c}.& {15}^{\circ }\text{atau}{285}^{\circ }\\ \text{d}.& {15}^{\circ }\text{atau}{345}^{\circ }\\ \text{e}.& {15}^{\circ }\text{atau}{75}^{\circ }\end{array}\\ \\ & \text{Jawab}:\text{b}\\ & \begin{array}{rl}\sqrt{3}\cos x+\sin x& =\sqrt{2}\\ \sqrt{{\sqrt{3}}^2+1^2}& (\cos (\alpha -\arctan {\displaystyle \frac{1}{\sqrt{3}}}))=\sqrt{2}\\ 2\cos (x-{30}^{\circ })& =\sqrt{2}\\ \cos (x-{30}^{\circ })& ={\displaystyle \frac{\sqrt{2}}{2}=\frac{1}{2}\sqrt{2}}\\ \cos (x-{30}^{\circ })& =\cos {45}^{\circ }\\ (x-{30}^{\circ })& =\pm {45}^{\circ }+k{.360}^{\circ }\\ x& ={30}^{\circ }\pm {45}^{\circ }+k{.360}^{\circ }\\ \text{saat}k& =0\\ x_1& ={75}^{\circ }\\ x_2& =-{15}^{\circ }\\ \text{saat}k& =1\\ x_3& ={75}^{\circ }+{360}^{\circ }={435}^{\circ }\\ x_4& =-{15}^{\circ }+{360}^{\circ }={345}^{\circ }\end{array}\end{array}$

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

$\begin{array}{l}\\ 39.& \text{Himpunan penyelesaian dari}\sin x-\sqrt{3}\cos x=-1\\ & \text{untuk}{0}^{\circ }\le x\le {360}^{\circ }\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{0^{\circ },{120}^{\circ }\}\\ \text{b}.& \{{90}^{\circ },{330}^{\circ }\}\\ \text{c}.& \{{60}^{\circ },{180}^{\circ }\}\\ \text{d}.& \{{90}^{\circ },{120}^{\circ }\}\\ \text{e}.& \{{30}^{\circ },{270}^{\circ }\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}\sin x-\sqrt{3}\cos x& =-1\\ \sqrt{1^2+{(-\sqrt{3})}^2}\cos (x-\alpha )& =-1\\ a& =x=-\sqrt{3},b=y=1\\ & \mathbf{\text{kuadran II}}\\ \alpha & =\arctan \left({\displaystyle \frac{1}{-\sqrt{3}}}\right)=-{30}^{\circ }\\ & ={180}^{\circ }-{30}^{\circ }={150}^{\circ }\\ \text{maka persamaan akan}& \text{menjadi}\\ 2\cos (x-{150}^{\circ })& =-1\\ \cos (x-{150}^{\circ })& ={\displaystyle \frac{-1}{2}}\\ \cos (x-{150}^{\circ })& =\cos {120}^{\circ }\\ (x-{150}^{\circ })& =\pm {120}^{\circ }+k{.360}^{\circ }\\ x& ={150}^{\circ }\pm {120}^{\circ }+k{.360}^{\circ }\\ \text{saat}& k=0\\ x_1& ={270}^{\circ }\\ x_2& ={30}^{\circ }\\ \text{saat}& k=1\\ x_3& ={270}^{\circ }+{360}^{\circ }=....\\ x_4& ={30}^{\circ }+{360}^{\circ }=....\end{array}\end{array}$

$\begin{array}{l}\\ 40.& \text{Nilai}\tan x\text{yang memenuhi persamaan}\\ & \cos 2x+7\cos x-3=0\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \sqrt{3}\\ \text{b}.& {\displaystyle \frac{1}{2}\sqrt{3}}\\ \text{c}.& {\displaystyle \frac{1}{3}\sqrt{3}}\\ \text{d}.& {\displaystyle \frac{1}{2}}\\ \text{e}.& {\displaystyle \frac{1}{5}\sqrt{5}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\cos 2x+7\cos x-3& =0\\ 2{\cos }^{2}x-1+7\cos x-3& =0\\ 2{\cos }^{2}x+7\cos x-4& =0\\ (\cos x+4)(2\cos -1)& =0\\ \cos x=-4\text{atau}\cos x& ={\displaystyle \frac{1}{2}=\cos {60}^{\circ }}\\ x& ={60}^{\circ },\\ \text{maka}& \\ \tan {60}^{\circ }& =\sqrt{3}\\ \\ \text{ dan ingat bahwa}& \cos x=-4\text{tidak memenuhi}\end{array}\end{array}$

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

$\begin{array}{l}\\ 21.& \text{Jika diketahui}\tan (3x+{60}^0)=\sqrt{3},\\ & \text{maka nilai}x\text{yang memenuhi persamaan}\\ & \text{tersebut adalah}....\\ & \begin{array}{l}\\ \text{a}.& {90}^0\\ \text{b}.& {110}^0\\ \text{c}.& {120}^0\\ \text{d}.& {130}^0\\ \text{e}.& {230}^0\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\tan (3x+{60}^0)& =\sqrt{3}\\ \tan (3x+{60}^0)& =\tan {60}^0\\ (3x+{60}^0)& ={60}^0+k{.180}^0\\ 3x& =k{.180}^0\\ x& =k{.60}^0\\ k=0\to x& =0^0\\ k=1\to x& ={60}^0\\ k=2\to x& ={120}^0\\ k=3\to x& ={180}^0\\ k=4\to x& ={240}^0\\ k=5\to x& ={300}^0\\ k=\cdots \to x& =\cdots \\ \cdots \cdots \to & =....\mathbf{\text{dst}}\end{array}\end{array}$.

$\begin{array}{l}\\ 22.& \text{Himpunan penyelesaian dari }\\ & \sqrt{3}\tan x=-1,0^0\le x\le {360}^0\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{{30}^0,{150}^0\}\\ \text{b}.& \{{150}^0,{210}^0\}\\ \text{c}.& \{{150}^0,{330}^0\}\\ \text{d}.& \{{120}^0,{300}^0\}\\ \text{e}.& \{{60}^0,{240}^0\}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\sqrt{3}\tan x& =-1\\ \tan x& =-{\displaystyle \frac{1}{\sqrt{3}}=-\frac{1}{3}\sqrt{3}}\\ & \text{nilai di kuadran II}\\ \tan x& =\tan {150}^0\\ x& ={150}^0+k{.180}^0\\ k=0,\Rightarrow x& ={150}^0\\ k=1,\Rightarrow x& ={330}^0\\ k=2,\Rightarrow x& ={510}^0,\text{tidak}\text{memenuhi}\text{(tm)}\\ k=3,\mathbf{\text{dst}}& =\text{(tm)}\\ \therefore \mathbf{\text{HP}}& =\{{150}^0,{330}^0\}\end{array}\end{array}$.

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

$\begin{array}{l}\\ 24.& \text{Himpunan penyelesaian dari }\\ & \cos (x-{30}^0)=-\sin {50}^0,0^0\le x\le {360}^0\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \left\{{90}^0\right\}\\ \text{b}.& \{{140}^0,{150}^0\}\\ \text{c}.& \{{170}^0,{250}^0\}\\ \text{d}.& \{{80}^0,{280}^0\}\\ \text{e}.& \{{20}^0,{340}^0\}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\cos (x-{30}^0)& =-\sin {50}^0\\ \cos (x-{30}^0)& =\cos ({90}^0+{50}^0)\\ \cos (x-{30}^0)& =\cos {140}^0\\ x-{30}^0& =\pm {140}^0+k{.360}^0\\ x& ={30}^0\pm {140}^0+k{.360}^0\\ k=0,\Rightarrow x_1& ={30}^0+{140}^0={170}^0\\ \Rightarrow x_2& ={30}^0-{140}^0=-{110}^0(\text{tm})\\ k=1,\Rightarrow x_1& ={30}^0+{140}^0+{360}^0={530}^0(\text{tm})\\ \Rightarrow x_2& ={30}^0-{140}^0+{360}^0={250}^0\\ \therefore \mathbf{\text{HP}}& =\{{170}^0,{250}^0\}\end{array}\end{array}$.

$\begin{array}{l}\\ 25.& \text{Himpunan penyelesaian dari}\\ & \sin 2x={\displaystyle \frac{1}{2}\sqrt{3}\text{untuk}{0}^{\circ }\le x\le {360}^{\circ }}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{{30}^{\circ },{210}^{\circ }\}\\ \text{b}.& \{{60}^{\circ },{240}^{\circ }\}\\ \text{c}.& \{{30}^{\circ },{60}^{\circ },{210}^{\circ }\}\\ \text{d}.& \{{30}^{\circ },{60}^{\circ },{210}^{\circ },{240}^{\circ }\}\\ \text{e}.& \{{30}^{\circ },{60}^{\circ },{210}^{\circ },{240}^{\circ },{270}^{\circ }\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\sin 2x& ={\displaystyle \frac{1}{2}\sqrt{3}}\\ \sin 2x& =\sin {60}^{\circ }\\ 2x& =\{\begin{array}{ll}{60}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{60}^{\circ })& +k{.360}^{\circ }\end{array}\\ x& =\{\begin{array}{ll}{30}^{\circ }& +k{.180}^{\circ }\\ {60}^{\circ }& +k{.180}^{\circ }\end{array}\\ \text{saat}& k=0\\ x& =\{\begin{array}{ll}{30}^{\circ }& \\ {60}^{\circ }& \end{array}\\ \text{saat}& k=1\\ x& =\{\begin{array}{ll}{30}^{\circ }& +{1.180}^{\circ }={210}^{\circ }\\ {60}^{\circ }& +{1.180}^{\circ }={240}^{\circ }\end{array}\\ \text{saat}& k=2\\ x& =\{\begin{array}{ll}{30}^{\circ }& +{2.180}^{\circ }={390}^{\circ }\\ {60}^{\circ }& +{2.180}^{\circ }={420}^{\circ }\end{array}\\ \text{kedua}& \text{nnya tidak memenuhi}\end{array}\end{array}$

$\begin{array}{l}\\ 26.& \text{Himpunan penyelesaian dari}\\ & \tan 2x-\sqrt{3}=0\text{untuk}{0}^{\circ }\le x\le {360}^{\circ }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{{15}^{\circ },{105}^{\circ },{195}^{\circ },{285}^{\circ }\}\\ \text{b}.& \{{30}^{\circ },{120}^{\circ },{210}^{\circ },{300}^{\circ }\}\\ \text{c}.& \{{45}^{\circ },{135}^{\circ },{225}^{\circ },{315}^{\circ }\}\\ \text{d}.& \{{15}^{\circ },{105}^{\circ },{195}^{\circ },{285}^{\circ }\}\\ \text{e}.& \{{15}^{\circ },{30}^{\circ },{45}^{\circ },{60}^{\circ },{75}^{\circ }\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\tan 2x& -\sqrt{3}=0\\ \tan 2x& =\sqrt{3}\\ \tan 2x& =\tan {60}^{\circ }\\ 2x& =60+k{.180}^{\circ }\\ x& ={30}^{\circ }+k{.90}^{\circ }\\ \text{saat}& k=0\\ x& ={30}^{\circ }\\ \text{saat}& k=1\\ x& ={30}^{\circ }+{90}^{\circ }={120}^{\circ }\\ \text{saat}& k=2\\ x& ={30}^{\circ }+{180}^{\circ }={210}^{\circ }\\ \text{saat}& k=3\\ x& ={30}^{\circ }+{270}^{\circ }={300}^{\circ }\\ \text{saat}& k=4\\ x& ={30}^{\circ }+{360}^{\circ }={390}^{\circ }\\ \text{tidak}& \text{memenuhi}\end{array}\end{array}$

$\begin{array}{l}\\ 27.& \text{Himpunan penyelesaian dari}\\ & \cos 3x=-{\displaystyle \frac{1}{2}\sqrt{3}\text{untuk}{0}^{\circ }\le x\le {180}^{\circ }}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{{40}^{\circ },{80}^{\circ }\}\\ \text{b}.& \{{50}^{\circ },{70}^{\circ }\}\\ \text{c}.& \{{40}^{\circ },{70}^{\circ },{80}^{\circ }\}\\ \text{d}.& \{{50}^{\circ },{70}^{\circ },{170}^{\circ }\}\\ \text{e}.& \{{50}^{\circ },{80}^{\circ },{170}^{\circ }\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\cos 3x& =-{\displaystyle \frac{1}{2}\sqrt{3}}\\ \cos 3x& =-\cos {30}^{\circ }\\ \cos 3x& =\cos ({180}^{\circ }-{30}^{\circ })=\cos {150}^{\circ }\\ 3x& =\pm {150}^{\circ }+k{.360}^{\circ }\\ x& =\pm {50}^{\circ }+k{.120}^{\circ }\\ \text{saat}& k=0\\ x& =\pm {50}^{\circ }\to x={50}^{\circ }(\text{mm})\\ \text{saat}& k=1\\ x& =\pm {50}^{\circ }+{120}^{\circ }=\{\begin{array}{ll}{170}^{\circ }& (\text{mm})\\ {70}^{\circ }& (\text{mm})\end{array}\end{array}\end{array}$.

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

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

$\begin{array}{l}\\ 30.& \text{Jika diketahui}\\ & f\left({\displaystyle \frac{6}{\sqrt{{\sin }^{2}x+4}}}\right)=\tan x,\pi \le x\le 2\pi ,\\ & \text{maka nilai}f(3)=....\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& 1\\ \text{c}.& {\displaystyle \frac{\pi }{2}}\\ \text{d}.& \pi \\ \text{e}.& 2\pi \end{array}\\ & \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Diketahui bahwa}& \\ f\left({\displaystyle \frac{6}{\sqrt{{\sin }^{2}x+4}}}\right)& =\tan x,\pi \le x\le 2\pi \\ f(3)& =....\\ \text{maka selanjutnya}& \\ {\displaystyle \frac{6}{\sqrt{{\sin }^{2}x+4}}}& =3\\ \sqrt{{\sin }^{2}x+4}& =2\\ {\sin }^{2}x+4& =4\\ {\sin }^{2}x& =0\\ \sin x& =\{\begin{array}{ll}{x}_1=0^{\circ }& +k.2\pi \\ x_2={180}^{\circ }& +k.2\pi \end{array}\\ \tan \pi =\tan 2\pi & =0\\ \therefore f(3)& =0\end{array}\end{array}$

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

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

$\begin{array}{l}\\ 12.& \text{Perhatikanlah grafik berikut}\end{array}$.

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

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

$\begin{array}{l}\\ 14.& \text{Grafik fungsi trigonometri pada gambar}\\ & \text{berikut adalah}....\end{array}$.

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

$\begin{array}{l}\\ 15.& \text{Grafik fungsi trigonometri pada gambar}\\ & \text{berikut adalah}....\end{array}$.

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

$\begin{array}{l}\\ 16.& \text{Grafik fungsi trigonometri pada gambar}\\ & \text{berikut adalah}....\end{array}$.

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

$\begin{array}{l}\\ 17.& \text{Himpunan penyelesaian dari persamaan}\\ & \sin x=\sin {\displaystyle \frac{2}{10}\pi \text{untuk}{0}^{\circ }\le x\le {360}^{\circ }}\\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \left\{{\displaystyle \frac{22}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}}\\ \\ & \text{b}.{\displaystyle \left\{{\displaystyle \frac{2}{10}\pi ,{\displaystyle \frac{28}{10}\pi }}\right\}}\\ \\ & \text{c}.{\displaystyle \left\{{\displaystyle \frac{2}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}}\\ \\ & \text{d}.{\displaystyle \left\{{\displaystyle \frac{22}{10}\pi ,{\displaystyle \frac{28}{10}\pi }}\right\}}\\ \\ & \text{e}.{\displaystyle \left\{{\displaystyle \frac{12}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \sin x=\sin {\displaystyle \frac{2}{10}\pi }\\ & \iff x_1={\displaystyle \frac{2}{10}\pi +k.2\pi \text{atau}}\\ & \iff x_2=(\pi -{\displaystyle \frac{2}{10}\pi })+k.2\pi ={\displaystyle \frac{8}{10}\pi +k.2\pi }\\ & k=0\Rightarrow x_1={\displaystyle \frac{2}{10}\pi (\text{mm})\text{atau}}\\ & x_2={\displaystyle \frac{8}{10}\pi (\text{mm})}\\ & k=1\Rightarrow x_{1,2}=....+2\pi (\text{tidak memenuhi})\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle \frac{2}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}\end{array}\end{array}$.

$\begin{array}{l}\\ 18.& \text{Himpunan penyelesaian dari persamaan}\\ & \tan (2x-{\displaystyle \frac{1}{4}\pi })=\tan {\displaystyle \frac{1}{4}\pi \text{untuk}{0}^{\circ }\le x\le {360}^{\circ }}\\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \left\{{\displaystyle \frac{1}{3}\pi ,\pi ,{\displaystyle \frac{5}{3}\pi ,{\displaystyle \frac{7}{3}\pi }}}\right\}}\\ \\ & \text{b}.{\displaystyle \left\{{\displaystyle \frac{1}{4}\pi ,{\displaystyle \frac{3}{5}\pi ,\frac{5}{4}\pi ,\frac{8}{5}\pi }}\right\}}\\ \\ & \text{c}.{\displaystyle \left\{{\displaystyle \frac{1}{4}\pi ,{\displaystyle \frac{3}{4}\pi ,\frac{6}{4}\pi ,{\displaystyle \frac{7}{4}\pi }}}\right\}}\\ \\ & \text{d}.{\displaystyle \left\{{\displaystyle \frac{2}{4}\pi ,{\displaystyle \frac{3}{4}\pi ,\pi ,{\displaystyle \frac{7}{4}\pi }}}\right\}}\\ \\ & \text{e}.{\displaystyle \left\{{\displaystyle \frac{1}{4}\pi ,\frac{3}{4}\pi ,\frac{5}{4}\pi ,{\displaystyle \frac{7}{4}\pi }}\right\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \tan (2x-{\displaystyle \frac{1}{4}\pi })=\tan {\displaystyle \frac{1}{4}\pi }\\ & \iff 2x-{\displaystyle \frac{1}{4}\pi ={\displaystyle \frac{1}{4}\pi +k.\pi }}\\ & \iff 2x={\displaystyle \frac{2}{4}\pi +k.\pi }\\ & \iff x={\displaystyle \frac{1}{4}\pi +k.\frac{\pi }{2}}\\ & k=0\Rightarrow x={\displaystyle \frac{1}{4}\pi (\text{mm})}\\ & k=1\Rightarrow x={\displaystyle \frac{1}{4}\pi +\frac{\pi }{2}={\displaystyle \frac{3}{4}\pi (\text{mm})}}\\ & k=2\Rightarrow x={\displaystyle \frac{1}{4}\pi +\pi ={\displaystyle \frac{5}{4}\pi (\text{mm})}}\\ & k=3\Rightarrow x={\displaystyle \frac{1}{4}\pi +\frac{3\pi }{2}={\displaystyle \frac{7}{4}\pi (\text{mm})}}\\ & k=4\Rightarrow x={\displaystyle \frac{1}{4}\pi +2\pi ={\displaystyle \frac{9}{4}\pi (\text{tidak memenuhi})}}\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle \frac{1}{4}\pi ,{\displaystyle \frac{3}{4}\pi ,\frac{5}{4}\pi ,\frac{7}{4}\pi }}\right\}\end{array}\end{array}$.

$\begin{array}{l}\\ 19.& \text{Himpunan penyelesaian dari persamaan}\\ & \cos (x-30{)}^{\circ }={\displaystyle \frac{1}{2}\sqrt{2}\text{untuk}{0}^{\circ }\le x\le {360}^{\circ }}\\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \{{75}^{\circ },{285}^{\circ }\}}\\ \\ & \text{b}.{\displaystyle \{{75}^{\circ },{343}^{\circ }\}}\\ \\ & \text{c}.{\displaystyle \{{75}^{\circ },{344}^{\circ }\}}\\ \\ & \text{d}.{\displaystyle \{{75}^{\circ },{345}^{\circ }\}}\\ \\ & \text{e}.{\displaystyle \{{75}^{\circ },{346}^{\circ }\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \cos (x-30{)}^{\circ }={\displaystyle \frac{1}{2}\sqrt{2}}\\ & \cos (x-30{)}^{\circ }=\cos {45}^{\circ }\\ & \iff x-{30}^{\circ }=\pm {45}^{\circ }+k{.360}^{\circ }\\ & \iff x={30}^{\circ }\pm {45}^{\circ }+k{.360}^{\circ }\\ & k=0\Rightarrow x_1={75}^{\circ }(\text{mm})\\ & \text{atau}{x}_2=-{15}^{\circ }(tm)\\ & k=1\Rightarrow x_1={75}^{\circ }+{360}^{\circ }(\text{tm})\\ & \text{atau}{x}_2=-{15}^{\circ }+{360}^{\circ }={345}^{\circ }(mm)\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle {75}^{\circ },{345}^{\circ }}\right\}\end{array}\end{array}$.

$\begin{array}{l}\\ 20.& \text{Himpunan penyelesaian dari persamaan}\\ & \cos (x-30{)}^{\circ }={\displaystyle \frac{1}{2}\sqrt{3}\text{untuk}{0}^{\circ }<x<{360}^{\circ }}\\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \{{100}^{\circ },{330}^{\circ }\}}\\ \\ & \text{b}.{\displaystyle \{{30}^{\circ },{330}^{\circ }\}}\\ \\ & \text{c}.{\displaystyle \{{120}^{\circ },{300}^{\circ }\}}\\ \\ & \text{d}.{\displaystyle \{{60}^{\circ },{120}^{\circ }\}}\\ \\ & \text{e}.{\displaystyle \{{50}^{\circ },{300}^{\circ }\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \cos x^{\circ }={\displaystyle \frac{1}{2}\sqrt{3}}\\ & \cos x^{\circ }=\cos {30}^{\circ }\\ & \iff x=\pm {30}^{\circ }+k{.360}^{\circ }\\ & k=0\Rightarrow x_1={30}^{\circ }(\text{mm})\\ & \text{atau}{x}_2=-{30}^{\circ }(tm)\\ & k=1\Rightarrow x_1={30}^{\circ }+{360}^{\circ }={390}^{\circ }(\text{tm})\\ & \text{atau}{x}_2=-{30}^{\circ }+{360}^{\circ }={330}^{\circ }(mm)\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle {30}^{\circ },{330}^{\circ }}\right\}\end{array}\end{array}$

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

 $\begin{array}{l}\\ 1.& \text{Nilai}{75}^{\circ }\text{jika dinyatakan ke radian}\\ & \text{adalah}....\text{radian}\\ \\ & \text{a}.{\displaystyle \frac{1}{3}\pi }\\ \\ & \text{b}.{\displaystyle \frac{5}{6}\pi }\\ \\ & \text{c}.{\displaystyle \frac{5}{12}\pi }\\ \\ & \text{d}.{\displaystyle \frac{7}{12}\pi }\\ \\ & \text{e}.{\displaystyle \frac{9}{12}\pi }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Diketah}& \text{ui bahwa}\\ {180}^{\circ }& =\pi radian\\ 1^{\circ }& ={\displaystyle \frac{\pi }{180}radian}\\ 75\times 1^{\circ }& =75\times {\displaystyle \frac{\pi }{180}radian}\\ {75}^{\circ }& ={\displaystyle \frac{5}{12}\pi radian}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Jika}\tan \theta ={\displaystyle \frac{5}{12}\text{untuk}{0}^{\circ }\le \theta \le {90}^{\circ }}\\ & \text{maka}\cos \theta \text{adalah}....\\ \\ & \text{a}.{\displaystyle \frac{5}{13}}\\ \\ & \text{b}.{\displaystyle \frac{12}{13}}\\ \\ & \text{c}.{\displaystyle \frac{13}{5}}\\ \\ & \text{d}.{\displaystyle \frac{13}{12}}\\ \\ & \text{e}.{\displaystyle \frac{12}{5}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikanlah gambar segitiga berikut}\end{array}$.

$.\begin{array}{rl}\text{Diketah}& \text{ui bahwa}\\ \tan \theta & ={\displaystyle \frac{5}{12},\text{untuk}{0}^{\circ }\le \theta \le {90}^{\circ }}\\ & \text{lihat gambar di atas}\\ & \text{dengan dalil Pythagoras akan}\\ & \text{didapatkan sisimiringnya}=13\\ \text{jadi}& ,\text{nilai dari}\\ \cos \theta & ={\displaystyle \frac{12}{13}}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Perhatikanlah gambar berikut}\end{array}$.

$.\begin{array}{l}\\ & \text{Panjang BC adalah}....\\ & \text{a}.{\displaystyle 20\sin {36}^{\circ }}\\ & \text{b}.{\displaystyle 20\cos {36}^{\circ }}\\ & \text{c}.{\displaystyle 20\tan {36}^{\circ }}\\ & \text{d}.{\displaystyle 15}\\ & \text{e}.{\displaystyle 16}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Diketah}& \text{ui bahwa}\\ \tan {36}^{\circ }& ={\displaystyle \frac{BC}{20}}\\ \iff & BC=20\tan {36}^{\circ }\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Nilai}\tan {300}^{\circ }\text{adalah}....\\ \\ & \text{a}.{\displaystyle \sqrt{3}}\\ \\ & \text{b}.{\displaystyle \frac{1}{3}\sqrt{3}}\\ \\ & \text{c}.{\displaystyle -\frac{1}{3}\sqrt{3}}\\ \\ & \text{d}.{\displaystyle \frac{1}{2}\sqrt{3}}\\ \\ & \text{e}.{\displaystyle -\sqrt{3}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\tan {300}^{\circ }& =\tan ({360}^{\circ }-{60}^{\circ })\\ & =-\tan {60}^{\circ }\\ & =-\sqrt{3}\\ \mathbf{\text{catatan}}& :\text{ingat sudut berelasi}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Nilai}\tan {60}^{\circ }-\sin {120}^{\circ }-\tan {210}^{\circ }\text{adalah}....\\ \\ & \text{a}.{\displaystyle \frac{1}{6}\sqrt{6}}\\ \\ & \text{b}.{\displaystyle \frac{1}{3}\sqrt{6}}\\ \\ & \text{c}.{\displaystyle -\frac{1}{2}\sqrt{6}}\\ \\ & \text{d}.{\displaystyle \frac{1}{3}\sqrt{3}}\\ \\ & \text{e}.{\displaystyle \frac{1}{6}\sqrt{3}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \tan {60}^{\circ }-\sin {120}^{\circ }-\tan {210}^{\circ }\\ & =\tan {60}^{\circ }-\sin ({180}^{\circ }-{60}^{\circ })-\tan ({180}^{\circ }+{30}^{\circ })\\ & =\tan {60}^{\circ }-\sin {60}^{\circ }-\tan {30}^{\circ }\\ & =\sqrt{3}-{\displaystyle \frac{1}{2}\sqrt{3}-{\displaystyle \frac{1}{3}\sqrt{3}}}\\ & =(1-{\displaystyle \frac{1}{2}-\frac{1}{3}})\sqrt{3}\\ & ={\displaystyle \frac{1}{6}\sqrt{3}}\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Nilai}x\text{positif terkecil yang memenuhi}\\ & \sin x=-{\displaystyle \frac{1}{2}\sqrt{3}\text{adalah}....}\\ \\ & \text{a}.{\displaystyle {30}^{\circ }}\\ \\ & \text{b}.{\displaystyle {60}^{\circ }}\\ \\ & \text{c}.{\displaystyle {120}^{\circ }}\\ \\ & \text{d}.{\displaystyle {240}^{\circ }}\\ \\ & \text{e}.{\displaystyle {300}^{\circ }}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\sin x& =-\frac{1}{2}\sqrt{3}\\ \text{Gun}& \text{akan rumus persamaan}\\ & \text{sederhana, yaitu}:\\ \sin x& =-\sin {60}^{\circ }\\ & =\sin ({180}^{\circ }+{60}^{\circ })\\ & =\sin {240}^{\circ }\\ x& ={240}^{\circ }\end{array}\end{array}$.

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

$\begin{array}{l}\\ 8.& \text{Pada setiap}\alpha \text{berlaku}\\ & \tan \alpha +\cos +\tan (-\alpha )+\cos (-\alpha )=....\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& 2\tan \alpha \\ \text{c}.& 2\cos \alpha \\ \text{d}.& 2(\tan \alpha +\cos \alpha )\\ \text{e}.& 2\\ \\ & & \mathbf{\text{SAT Subjeck Test}}\end{array}\\ & \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\tan \alpha & +\cos +\tan (-\alpha )+\cos (-\alpha )\\ & =\tan \alpha +\cos -\tan \alpha +\cos \alpha \\ & =2\cos \alpha \end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Jika}\sin x=-\sin {35}^{\circ }\text{untuk}{90}^{\circ }\le x\le {270}^{\circ }\\ & \text{maka nilai}x\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {215}^{\circ }\\ \text{b}.& {235}^{\circ }\\ \text{c}.& {240}^{\circ }\\ \text{d}.& {255}^{\circ }\\ \text{e}.& {270}^{\circ }\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Diketahui}\text{bahwa}\sin x=-\sin {35}^{\circ }\\ & \text{untuk}{90}^{\circ }\le x\le {270}^{\circ },\text{dengan}\\ & \sin x=-\sin {35}^{\circ }\\ & \text{(hanya terjadi dikuadran III dan IV)}\\ & \text{karena ba}\text{tasnya}\text{hanya untuk kuadran III saja, }\\ & \text{maka}\\ & =\sin ({180}^{\circ }+{35}^{\circ })\\ & =\sin {215}^{\circ }\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Jika}\tan y=\tan {83}^{\circ }\text{untuk}{90}^{\circ }<y<{270}^{\circ }\\ & \text{maka nilai}x\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {173}^{\circ }\\ \text{b}.& {187}^{\circ }\\ \text{c}.& {263}^{\circ }\\ \text{d}.& {268}^{\circ }\\ \text{e}.& {293}^{\circ }\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa nilai}\\ \tan y& =\tan {83}^{\circ }\text{untuk}{90}^{\circ }<y<{270}^{\circ }\\ \tan y& =\tan ({180}^{\circ }+{83}^{\circ }),\\ & \text{karena berada dikuadran III}\\ & =\tan {263}^{\circ }\end{array}\end{array}$.