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

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

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

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

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

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

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

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

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

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

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

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

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