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

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

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

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

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

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

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

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

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

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

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

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

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