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

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

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

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

$\begin{array}{l}\\ 9.& \text{Perhatikanlah grafik berikut}\end{array}$.

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

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