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

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

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

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

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

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

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

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