Lanjutan Soal Persiapan Penilaian Tengah Semester Gasal Kelas XI Pendalaman Matematika (Peminatan)

$\begin{array}{ll}\\ 6.&\textrm{Nilai}\: \: x \: \: \textrm{positif terkecil yang memenuhi}\\ &\sin x=-\displaystyle \frac{1}{2}\sqrt{3}\: \: \textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \displaystyle 30^{\circ} \\\\ &\textrm{b}.\quad \displaystyle 60^{\circ} \\\\ &\textrm{c}.\quad \displaystyle 120^{\circ} \\\\ &\textrm{d}.\quad \color{red}\displaystyle 240^{\circ} \\\\ &\textrm{e}.\quad \displaystyle 300^{\circ} \\\\ &\textbf{Jawab}:\\ &\begin{aligned}\sin x&=-\frac{1}{2}\sqrt{3}\\ \textrm{Gun}&\textrm{akan rumus persamaan}\\ &\textrm{sederhana, yaitu}:\\ \sin x&=-\sin 60^{\circ}\\ &=\sin \left ( 180^{\circ}+60^{\circ} \right )\\ &=\sin 240^{\circ}\\ x&=\color{red}240^{\circ} \end{aligned} \end{array}$.

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

$\begin{array}{ll}\\ 8.&\textrm{Periode dari fungsi}\: \: f(x)=-2\cos 3x\: \: \textrm{adalah}\: ....\\\\ &\textrm{a}.\quad 90^{\circ} \\\\ &\textrm{b}.\quad 100^{\circ} \\\\ &\textrm{c}.\quad \color{red}120^{\circ} \\\\ &\textrm{d}.\quad 150^{\circ} \\\\ &\textrm{e}.\quad 180^{\circ} \\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Periode dari}\: :\: f(x)=-2\cos 3x\\ &\textrm{adalah}\\ &=\displaystyle \frac{360^{\circ}}{3}\\ &=\color{red}120^{\circ}\\ &\\ &\textrm{Ingat bahwa}\\ &f(x)=a\cos bx,\: \: \textrm{maka periodenya}\\ &=\displaystyle \frac{360^{\circ}}{b} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Perhatikanlah grafik berikut} \end{array}$.

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

$\begin{array}{ll}\\ 10.&\textrm{Nilai dari}\: \: \displaystyle \frac{\sin \displaystyle \frac{3}{4}\pi +\tan \pi +\cos \pi }{\sin \displaystyle \frac{1}{2}\pi +\cos 2\pi -3\cos \displaystyle \frac{1}{3}\pi }=\: ....\\\\ &\textrm{a}.\quad 4 \\\\ &\textrm{b}.\quad 2-\sqrt{2} \\\\ &\textrm{c}.\quad \sqrt{2}-2 \\\\ &\textrm{d}.\quad -4 \\\\ &\textrm{e}.\quad \color{red}-1 \\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\sin \displaystyle \frac{3}{4}\pi +\tan \pi +\cos \pi }{\sin \displaystyle \frac{1}{2}\pi +\cos 2\pi -3\cos \displaystyle \frac{1}{3}\pi }\\ &=\displaystyle \frac{\sin 150^{\circ} +\tan 180^{\circ} +\cos 180^{\circ} }{\sin \displaystyle 90^{\circ} +\cos 360^{\circ} -3\cos \displaystyle 60^{\circ} }\\ &=\displaystyle \frac{\displaystyle \frac{1}{2}+0+(-1)}{1+1-3\left (\displaystyle \frac{1}{2} \right ) }\\ &=\displaystyle \frac{-\displaystyle \frac{1}{2}}{2-\displaystyle \frac{3}{2}}\\ &=-\displaystyle \frac{\displaystyle \frac{1}{2}}{\displaystyle \frac{1}{2}}\\ &=\color{red}-1 \end{aligned} \end{array}$