Latihan Soal 10 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

$\begin{array}{ll}\\ 91.&\textrm{Bentuk sederhana dari}\\ &\quad\quad 4\sin \left ( \displaystyle \frac{1}{4}\pi +x \right )\cos \left ( \displaystyle \frac{1}{4}\pi -x \right )\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \color{red}2+2\sin 2x &&&\textrm{d}.&2+2\sin x\\ \textrm{b}.&\displaystyle 2+\sin 2x&&&\textrm{e}.&2+\sin x\\ \textrm{c}.&\displaystyle 2\sin 2x \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&4\sin \left ( \displaystyle \frac{1}{4}\pi +x \right )\cos \left ( \displaystyle \frac{1}{4}\pi -x \right )\\ &=2\left ( \sin \left ( \displaystyle \frac{1}{2}\pi \right )+\sin (2x) \right )\\ &=2\left (1+\sin 2x \right )\\ &=2+2\sin 2x \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 92.&\textrm{Bentuk sederhana dari}\\ & 2\sin \left ( \displaystyle \frac{3}{4}\pi +x \right ).\cos \left ( \displaystyle \frac{\pi }{4}+x \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}1-\sin 2x&&\textrm{d}.\quad \cos 2x\\\\ \textrm{b}.\quad 1+\sin 2x&\textrm{c}.\quad 1-\cos 2x&\textrm{e}.\quad -\cos 2x\end{array}\\\\ &\textbf{Jawab}:\: \: \color{red}\textbf{a}\\ &\begin{aligned}&2\sin \left ( \displaystyle \frac{3}{4}\pi +x \right ).\cos \left ( \displaystyle \frac{\pi }{4}+x \right )\\ &=2\sin \left ( 135^{\circ}+x \right ).\cos \left ( 45^{\circ}+x \right )\\ &=2\sin \left ( 90^{\circ}+45^{\circ}+x \right ).\cos \left ( 45^{\circ}+x \right )\\ &=2\cos \left ( 45^{\circ}+x \right ).\cos \left ( 45^{\circ}+x \right )\\ &=2\cos ^{2}\left ( 45^{\circ}+x \right )\\ &=2\left ( \cos 45^{\circ}\cos x-\sin 45^{\circ}\sin x \right )^{2}\\ &=2\left ( \frac{1}{2}\sqrt{2}.\cos x-\displaystyle \frac{1}{2}\sqrt{2}.\sin x \right )^{2}\\ &=\displaystyle \left ( \cos x-\sin x \right )^{2}\\ &=\cos ^{2}x+\sin ^{2}x-2\sin x\cos x\\ &=\color{red}1-\sin 2x \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 93.&\textrm{Bentuk sederhana dari}\\ & 2\cos \left ( x+\displaystyle \frac{\pi }{4} \right ). \cos \left ( \displaystyle \frac{3}{4}\pi -x \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \cos (2x+\pi )&&\textrm{d}.\quad -\cos \left ( 2x-\displaystyle \frac{\pi }{2} \right )\\ \textrm{b}.\quad \cos \left ( 2x+\displaystyle \frac{\pi }{2} \right )&&\textrm{e}.\quad \color{red}\sin 2x-1\\ \textrm{c}.\quad \cos 2x \end{array}\\\\ &\textbf{Jawab}:\: \: \color{red}\textbf{e}\\ &\begin{aligned}&2\cos \left ( x+\displaystyle \frac{\pi }{4} \right ). \cos \left ( \displaystyle \frac{3}{4}\pi -x \right )\\ &=\cos \left ( x+\displaystyle \frac{\pi }{4}+\displaystyle \frac{3}{4}\pi -x \right )+\cos \left (x+\displaystyle \frac{\pi }{4} -\left ( \displaystyle \frac{3}{4}\pi -x \right ) \right )\\ &=\cos \pi +\cos \left ( 2x-\frac{2}{4}\pi \right )\\ &=-1+\cos \left ( \displaystyle \frac{1}{2}\pi -2x \right ),\\ &\quad\quad \color{blue}\textrm{ingat bahwa}\: \cos (-\alpha )=\cos \alpha \\ &=-1+\sin 2x\\ &=\color{red}\sin 2x-1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 94.&\textrm{Nilai dari}\\ &\quad\quad \sqrt{3}\sin 80^{\circ}\sin 160^{\circ}\sin 320^{\circ}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle -\displaystyle \frac{3}{8} &&&\textrm{d}.&\displaystyle \color{red}\frac{3}{8}\\\\ \textrm{b}.&\displaystyle -\frac{1}{8}&&&\textrm{e}.&\displaystyle \frac{5}{8}\\ \textrm{c}.&\displaystyle \frac{1}{8} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\sqrt{3}\sin 80^{\circ}\sin 160^{\circ}\sin 320^{\circ}\\ &=\sqrt{3}\sin 80^{\circ}\sin 20^{\circ}\left (-\sin 40^{\circ} \right )\\ &=-\sqrt{3}\sin 80^{\circ}\sin 40^{\circ}\sin 20^{\circ}\\ &=-\sqrt{3}\sin 80^{\circ}\left ( -\displaystyle \frac{1}{2}\left ( \cos 60^{\circ}-\cos 20^{\circ} \right ) \right )\\ &=-\sqrt{3}\sin 80^{\circ}\left ( -\displaystyle \frac{1}{4}+\displaystyle \frac{\cos 20^{\circ}}{2} \right )\\ &=\displaystyle \frac{1}{4}\sqrt{3}\sin 80^{\circ}-\displaystyle \frac{1}{2}\sqrt{3}\sin 80^{\circ}\cos 20^{\circ}\\ &=\displaystyle \frac{1}{4}\sqrt{3}\sin 80^{\circ}-\displaystyle \frac{1}{4}\sqrt{3}\left ( \sin 100^{\circ}+\sin 60^{\circ} \right )\\ &=\displaystyle \frac{1}{4}\sqrt{3}\sin 80^{\circ}-\displaystyle \frac{1}{4}\sqrt{3}\left ( \sin 80^{\circ}+\displaystyle \frac{1}{2}\sqrt{3} \right )\\ &=\displaystyle \frac{1}{4}\sqrt{3}\sin 80^{\circ}-\displaystyle \frac{1}{4}\sqrt{3}\sin 80^{\circ}+\displaystyle \frac{1}{8}\sqrt{9}\\ &=\displaystyle \frac{3}{8} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 95.&\textrm{Nilai dari}\\ &\quad\quad \cos \displaystyle \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \color{red}-\displaystyle \frac{1}{8} &&&\textrm{d}.&\displaystyle \frac{1}{2}\\\\ \textrm{b}.&\displaystyle -\frac{1}{4}&\quad \textrm{c}.&0\quad &\textrm{e}.&\displaystyle \frac{1}{3} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\textbf{Alternatif 1}\\ &\begin{aligned}&\cos \displaystyle \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}\times \displaystyle \frac{2\sin \displaystyle \frac{2\pi }{7}}{2\sin \displaystyle \frac{2\pi }{7}}\\ &=\left (\sin \displaystyle \frac{4\pi }{7}-\sin 0 \right )\frac{\displaystyle \cos \frac{\pi }{7}\cos \displaystyle \frac{4\pi }{7}}{2\sin \displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin \displaystyle \frac{4\pi }{7}\displaystyle \cos \frac{\pi }{7}\cos \displaystyle \frac{4\pi }{7}}{2\sin \displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\left ( \sin \displaystyle \frac{5\pi }{7}+\sin \displaystyle \frac{3\pi }{7} \right )\cos \displaystyle \frac{4\pi }{7}}{4\sin \displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin \displaystyle \frac{5\pi }{7}\cos \displaystyle \frac{4\pi }{7}+\sin \displaystyle \frac{3\pi }{7}\cos \displaystyle \frac{4\pi }{7}}{4\sin \displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin \displaystyle \frac{9\pi }{7}+\sin \displaystyle \frac{\pi }{7}+\sin \displaystyle \frac{7\pi }{7}+\sin \left (-\displaystyle \frac{\pi }{7} \right )}{8\sin \displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{-\sin \displaystyle \frac{2\pi }{7}+\sin \displaystyle \frac{\pi }{7}+0-\sin \displaystyle \frac{\pi }{7}}{8\sin \displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{-\sin \displaystyle \frac{2\pi }{7}}{8\sin \displaystyle \frac{2\pi }{7}}\\ &=-\displaystyle \frac{1}{8} \end{aligned}\\ &\textbf{Alternatif 2}\\ &\begin{aligned}&\cos \displaystyle \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}\\ &=\cos \displaystyle \frac{4\pi }{7}\cos \frac{2\pi }{7}\cos \frac{\pi }{7}\\ &=\displaystyle \frac{1}{2}\left ( \cos \displaystyle \frac{6\pi }{7}+\cos \displaystyle \frac{2\pi }{7} \right )\cos \displaystyle \frac{\pi }{7}\\ &=\displaystyle \frac{1}{2}\left ( \cos \left ( \pi -\displaystyle \frac{\pi }{7} \right )+\cos \displaystyle \frac{2\pi }{7} \right )\cos \displaystyle \frac{\pi }{7}\\ &=\displaystyle \frac{1}{2}\left ( -\cos \displaystyle \frac{\pi }{7}+\cos \displaystyle \frac{2\pi }{7} \right )\cos \displaystyle \frac{\pi }{7}\\ &= \displaystyle \frac{1}{2}\left (-\cos ^{2}\displaystyle \frac{\pi }{7}+\cos \displaystyle \frac{2\pi }{7}\cos \displaystyle \frac{\pi }{7} \right )\\ &=\displaystyle \frac{1}{4}\left ( -\cos \displaystyle \frac{2\pi }{7}-\cos 0+\cos \displaystyle \frac{3\pi }{7}+\cos \displaystyle \frac{\pi }{7} \right )\\ &=\displaystyle \frac{1}{4}\left ( -\cos 0+\color{red}\cos \displaystyle \frac{\pi }{7}-\cos \displaystyle \frac{2\pi }{7}+\cos \displaystyle \frac{3\pi }{7} \color{black}\right )\\ &=\displaystyle \frac{1}{4}\left ( -1+\color{red}\displaystyle \frac{1}{2}\color{black} \right )\\ &=\displaystyle \frac{1}{4}\times \left (-\frac{1}{2} \right )\\ &=-\displaystyle \frac{1}{8} \end{aligned} \end{array}$.

Berikut penjelasan untuk  $\cos \displaystyle \frac{\pi }{7}-\cos \displaystyle \frac{2\pi }{7}+\cos \displaystyle \frac{3\pi }{7}=\color{red}\displaystyle \frac{1}{2}$.

$\begin{aligned}&\cos \displaystyle \frac{\pi }{7}-\cos \displaystyle \frac{2\pi }{7}+\cos \displaystyle \frac{3\pi }{7}\\ &=\cos \displaystyle \frac{\pi }{7}-\cos \displaystyle \frac{2\pi }{7}+\cos \displaystyle \frac{3\pi }{7}\times \displaystyle \frac{\left (2\sin\displaystyle \frac{2\pi }{7} \right ) }{\left (2\sin\displaystyle \frac{2\pi }{7} \right )}\\ &=\displaystyle \frac{2\cos\displaystyle \frac{\pi }{7}\sin\displaystyle \frac{2\pi }{7}-2\cos\displaystyle \frac{2\pi }{7}\sin\displaystyle \frac{2\pi }{7}+2\cos\displaystyle \frac{3\pi }{7}\sin\displaystyle \frac{2\pi }{7}}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin\displaystyle \frac{3\pi }{7}-\sin\left (-\displaystyle \frac{\pi }{7} \right )-\left ( \sin\displaystyle \frac{4\pi }{7}-\sin\displaystyle \frac{0\pi }{7} \right )+\sin\displaystyle \frac{5\pi }{7}-\sin\displaystyle \frac{\pi }{7}}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin\displaystyle \frac{3\pi }{7}+\sin\displaystyle \frac{\pi }{7}-\sin\displaystyle \frac{4\pi }{7}+\sin\displaystyle \frac{5\pi }{7}-\sin\displaystyle \frac{\pi }{7}}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin\displaystyle \frac{3\pi }{7}-\sin\displaystyle \frac{4\pi }{7}+\sin\displaystyle \frac{5\pi }{7}}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin\left (\pi -\displaystyle \frac{4\pi }{7} \right )-\sin\displaystyle \frac{4\pi }{7}+\sin\left (\pi -\displaystyle \frac{2\pi }{7} \right )}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin\displaystyle \frac{4\pi }{7}-\sin\displaystyle \frac{4\pi }{7}+\sin\displaystyle \frac{2\pi }{7}}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin\displaystyle \frac{2\pi }{7}}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{1}{2}\qquad \blacksquare \end{aligned}$.

$\begin{array}{ll}\\ 96.&\textrm{Nilai dari}\\ &\quad\quad \sin \displaystyle \frac{\pi }{14}\sin \frac{3\pi }{14}\sin \frac{9\pi }{14}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \displaystyle \frac{1}{16} &&&\textrm{d}.&\displaystyle \frac{1}{2}\\\\ \textrm{b}.&\displaystyle \color{red}\frac{1}{8}&\quad \textrm{c}.&\displaystyle \frac{1}{4}\quad &\textrm{e}.&\displaystyle 1 \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\textrm{Perhat}&\textrm{ikan bahwa}\\ \sin \displaystyle \displaystyle \frac{\pi }{14}&=\sin \left (\displaystyle \frac{7\pi }{14}-\frac{6\pi }{14} \right )=\sin \left ( \displaystyle \frac{1}{2}\pi -\frac{6\pi }{14} \right )\\ &=\cos \displaystyle \frac{6\pi }{14} \\ \sin \displaystyle \frac{3\pi }{14}&=...=\cos \displaystyle \frac{4\pi }{14}\\ \sin \displaystyle \frac{9\pi }{14}&=...=\sin \displaystyle \frac{5\pi }{14}=\cos \displaystyle \frac{2\pi }{14} \end{aligned}\\ &...\\ &\sin \displaystyle \frac{\pi }{14}\sin \frac{3\pi }{14}\sin \frac{9\pi }{14}\\ &=\cos \displaystyle \frac{6\pi }{14}\cos \displaystyle \frac{4\pi }{14}\cos \displaystyle \frac{2\pi }{14}\times \displaystyle \frac{2\sin \displaystyle \frac{2\pi }{14}}{2\sin \displaystyle \frac{2\pi }{14}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{6\pi }{14}\cos \displaystyle \frac{4\pi }{14}\sin \displaystyle \frac{4\pi }{14}}{2\sin \displaystyle \frac{2\pi }{14}}\\ &\textrm{silahkan dilanjutkan}\\ &...\\ &=\displaystyle \frac{1}{8} \end{array}$.

$\begin{array}{ll}\\ 97.&\textrm{Nilai dari}\\ & \cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\cos \displaystyle \frac{8\pi }{5}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \color{red}-\displaystyle \frac{1}{16} &&&\textrm{d}.&\displaystyle \frac{1}{16}\\\\ \textrm{b}.&\displaystyle \frac{1}{8}&\quad \textrm{c}.&\displaystyle 0\quad &\textrm{e}.&\displaystyle \frac{1}{8} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\cos \displaystyle \frac{8\pi }{5}\\ &=\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\cos \left (\pi +\displaystyle \frac{3\pi }{5} \right )\\ &=\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\left (-\cos \displaystyle \frac{3\pi }{5} \right )\\ &=-\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\cos \displaystyle \frac{3\pi }{5}\\ &=-\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{3\pi }{5}\cos \displaystyle \frac{4\pi }{5}\\ &=-\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{3\pi }{5}\cos \displaystyle \frac{4\pi }{5}\times \displaystyle \frac{2\sin \displaystyle \frac{\pi }{5}}{2\sin \displaystyle \frac{\pi }{5}}\\ &=\frac{-\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{3\pi }{5}\left ( \sin \pi -\sin \displaystyle \frac{3\pi }{5} \right )}{2\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{3\pi }{5} \sin \frac{3\pi }{5}}{2\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{\pi }{5}\cos \displaystyle \frac{3\pi }{5}\left ( \cos \displaystyle \frac{2\pi }{5}\sin \displaystyle \frac{3\pi }{5} \right )}{2\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{\pi }{5}\cos \displaystyle \frac{3\pi }{5}\left ( \sin \pi -\sin \left ( -\displaystyle \frac{\pi }{5} \right ) \right )}{4\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{\pi }{5}\cos \displaystyle \frac{3\pi }{5}\sin \displaystyle \frac{\pi }{5}}{4\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{3\pi }{5}\cos \displaystyle \frac{\pi }{5}\sin \displaystyle \frac{\pi }{5}}{4\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{3\pi }{5}\left (\cos \displaystyle \frac{\pi }{5}\sin \displaystyle \frac{\pi }{5} \right )}{4\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{3\pi }{5}\left (\sin \displaystyle \frac{2\pi }{5}-\sin 0 \right )}{8\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{3\pi }{5}\sin \displaystyle \frac{2\pi }{5}}{8\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\sin \pi -\sin \displaystyle \frac{\pi }{5}}{16\sin \displaystyle \frac{\pi }{5}}\\ &=-\displaystyle \frac{\sin \displaystyle \frac{\pi }{5}}{16\sin \displaystyle \frac{\pi }{5}}\\ &=\color{red}-\displaystyle \frac{1}{16} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 98.&\textrm{Nilai dari}\: \: \sin \displaystyle \frac{\pi }{24}.\sin \frac{5\pi }{24}.\sin \frac{7\pi }{24}.\sin \frac{11\pi }{24}\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{5}{16}&&\textrm{d}.\quad \displaystyle \frac{2}{16}\\\\ \textrm{b}.\quad \displaystyle \frac{4}{16}&\textrm{c}.\quad \displaystyle \frac{3}{16}&\textrm{e}.\quad \color{red}\displaystyle \frac{1}{16} \end{array}\\\\ &\textbf{(Olimpiade Sains PORSEMA NU 2012)}\\\\ &\textbf{Jawab}:\: \: \color{red}\textbf{e}\\ &\begin{aligned}&\sin \displaystyle \frac{\pi }{24}.\sin \frac{5\pi }{24}.\sin \frac{7\pi }{24}.\sin \frac{11\pi }{24}\\ &=\displaystyle \frac{1}{4}\left ( 2\sin \displaystyle \frac{11\pi }{24}.\sin \frac{\pi }{24}.2\sin \frac{7\pi }{24}.\sin \frac{5\pi }{24} \right )\\ &=\displaystyle \frac{1}{4}\left [ \left ( \cos \left ( \frac{10\pi }{24} \right )-\cos \left ( \frac{12\pi }{24} \right ) \right )\times \left ( \cos \left ( \frac{2\pi }{24} \right )-\cos \left ( \frac{12\pi }{24} \right ) \right ) \right ]\\ &=\displaystyle \frac{1}{4}\left [ \left ( \cos 75^{\circ}-\cos 90^{\circ} \right )\times \left ( \cos 15^{\circ}-\cos 90^{\circ} \right ) \right ]\\ &=\displaystyle \frac{1}{4}\left [ \cos 75^{\circ}.\cos 15^{\circ} \right ]\\ &=\displaystyle \frac{1}{8}\left [ \cos 90^{\circ}+\cos 60^{\circ} \right ]\\ &=\displaystyle \frac{1}{8}\left ( 0+\frac{1}{2} \right )\\ &=\color{red}\displaystyle \frac{1}{16} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 99.&\textrm{Nilai dari}\\ & \qquad\qquad\sin 18^{\circ}\cos 36^{\circ}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \frac{1}{6} &&&\textrm{d}.&\displaystyle \frac{1}{3}\\\\ \textrm{b}.&\displaystyle \frac{1}{5}&\quad \textrm{c}.&\displaystyle \color{red}\displaystyle \frac{1}{4}\quad &\textrm{e}.&\displaystyle \frac{1}{2} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\sin 18^{\circ}\cos 36^{\circ}\\ &=\sin 18^{\circ}\cos 36^{\circ}\times \displaystyle \frac{2\cos 18^{\circ}}{2\cos 18^{\circ}}\\ &=\displaystyle \frac{\cos 36^{\circ}\left ( \sin 36^{\circ}+\sin 0^{\circ} \right )}{4\cos 18^{\circ}}\\ &=\displaystyle \frac{\cos 36^{\circ}\sin 36^{\circ}}{4\cos 18^{\circ}}\\ &=\displaystyle \frac{\sin 72^{\circ}}{4\cos 18^{\circ}}\\ &=\displaystyle \frac{\sin \left ( 90^{\circ}-18^{\circ} \right )}{4\cos 18^{\circ}}\\ &=\displaystyle \frac{\cos 18^{\circ}}{4\cos 18^{\circ}}\\ &=\color{red}\displaystyle \frac{1}{4} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 100.&\textrm{Nilai eksak dari}\: \: \sin 36^{\circ}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \displaystyle \frac{1}{4}\sqrt{10+2\sqrt{5}}&&&\textrm{d}.&\displaystyle \frac{\sqrt{5}-1}{4}\\ \textrm{b}.&\color{red}\displaystyle \frac{1}{4}\sqrt{10-2\sqrt{5}}&&\quad &\textrm{e}.&\displaystyle \frac{\sqrt{5}-1}{2}\\ \textrm{c}.&\displaystyle \displaystyle \frac{\sqrt{5}+1}{4} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\textrm{Perhatikanlah ilustrasi gambar berikut} \end{array}$.

$.\qquad\begin{aligned}&\textrm{Perhatikan bahwa}\: \: \color{red}\bigtriangleup ABC\: \: \color{black}\textrm{sama kaki}\\ &\textrm{dengan}\: \: AD=DC=CB=1,\: AC=x\\ &\textrm{Diketahui pula}\: \: CD\: \: \textrm{adalah garis bagi}\\ &\textrm{serta}\: \: ABC\: \: \textrm{sebangun}\: \: \bigtriangleup BCD\\ &\textrm{akibatnya}:\\ &\color{red}\textrm{perbandingan sisi yang bersesuaian}\\ &\color{red}\textrm{akan sama},\: \: \color{black}\textrm{maka}\\ &\displaystyle \frac{AB}{BC}=\displaystyle \frac{BC}{AB-AD}\\ &\Leftrightarrow \displaystyle \frac{x}{1}=\frac{1}{x-1}\\ &\Leftrightarrow x(x-1)=1\\ &\Leftrightarrow x^{2}-x-1=0\\ &\Leftrightarrow x=\displaystyle \frac{1\pm \sqrt{5}}{2}\\ &\textrm{akibatnya}\: \: AB=AC=\displaystyle \frac{1+\sqrt{5}}{2}\\ &\textrm{Selanjutnya gunakan}\: \: \color{blue}\textrm{aturan sinus}\\ &\displaystyle \frac{AB}{\sin \angle C}=\frac{BC}{\sin \angle A}\\ &\Leftrightarrow \displaystyle \frac{AB}{BC}=\frac{\sin \angle C}{\sin \angle A}\\ &\Leftrightarrow \displaystyle \frac{ \left (\displaystyle \frac{1+\sqrt{5} }{2} \right )}{1}=\frac{\sin 72^{\circ}}{\sin 36^{\circ}}\\ &\Leftrightarrow \displaystyle \frac{1+\sqrt{5}}{2}=\displaystyle \frac{2\sin 36^{\circ}\cos 36^{\circ}}{\sin 36^{\circ}}\\ &\Leftrightarrow \displaystyle \frac{1+\sqrt{5}}{2}=2\cos 36^{\circ}\\ &\Leftrightarrow \cos 36^{\circ}=\Leftrightarrow \displaystyle \frac{1+\sqrt{5}}{4}\\ &\textrm{Dari fakta di atas kita akan dengan}\\ &\textrm{mudah menentukan nilai sinusnya}\\ &\textrm{yaitu dengan menggunakan}\\ &\textrm{identitas trigonometri berikut}:\\ &\sin ^{2}36^{\circ}+\cos ^{2}36^{\circ}=1\\ &\Leftrightarrow \sin ^{2}36^{\circ}=1-\cos ^{2}36^{\circ}\\ &\Leftrightarrow \sin 36^{\circ}=\sqrt{1-\cos ^{2}36^{\circ}}\\ &\Leftrightarrow \: \: \quad\quad\quad =\sqrt{1-\left ( \displaystyle \frac{1+\sqrt{5}}{4} \right )^{2}}\\ &\Leftrightarrow \: \: \quad\quad\quad =\sqrt{1-\displaystyle \frac{6+2\sqrt{5}}{16}}\\ &\Leftrightarrow \: \: \quad\quad\quad =\sqrt{\displaystyle \frac{10-2\sqrt{5}}{16}}\\ &\Leftrightarrow \: \: \quad\quad\quad =\color{red}\displaystyle \frac{1}{4}\sqrt{10-2\sqrt{5}} \end{aligned}$.