Contoh Soal 3 (Segitiga dan Trigonometri)

 

$\begin{array}{ll}\\ 11.&\textrm{Nilai dari}\\ &\quad\quad \sin \displaystyle \frac{\pi }{14}\sin \frac{3\pi }{14}\sin \frac{9\pi }{14}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \displaystyle \frac{1}{16} &&&\textrm{d}.&\displaystyle \frac{1}{2}\\\\ \textrm{b}.&\displaystyle \color{red}\frac{1}{8}&\quad \textrm{c}.&\displaystyle \frac{1}{4}\quad &\textrm{e}.&\displaystyle 1 \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\textrm{Perhat}&\textrm{ikan bahwa}\\ \sin \displaystyle \displaystyle \frac{\pi }{14}&=\sin \left (\displaystyle \frac{7\pi }{14}-\frac{6\pi }{14} \right )=\sin \left ( \displaystyle \frac{1}{2}\pi -\frac{6\pi }{14} \right )\\ &=\cos \displaystyle \frac{6\pi }{14} \\ \sin \displaystyle \frac{3\pi }{14}&=...=\cos \displaystyle \frac{4\pi }{14}\\ \sin \displaystyle \frac{9\pi }{14}&=...=\sin \displaystyle \frac{5\pi }{14}=\cos \displaystyle \frac{2\pi }{14} \end{aligned}\\ &...\\ &\sin \displaystyle \frac{\pi }{14}\sin \frac{3\pi }{14}\sin \frac{9\pi }{14}\\ &=\cos \displaystyle \frac{6\pi }{14}\cos \displaystyle \frac{4\pi }{14}\cos \displaystyle \frac{2\pi }{14}\times \displaystyle \frac{2\sin \displaystyle \frac{2\pi }{14}}{2\sin \displaystyle \frac{2\pi }{14}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{6\pi }{14}\cos \displaystyle \frac{4\pi }{14}\sin \displaystyle \frac{4\pi }{14}}{2\sin \displaystyle \frac{2\pi }{14}}\\ &\textrm{silahkan dilanjutkan}\\ &...\\ &=\displaystyle \frac{1}{8} \end{array}$.

$\begin{array}{ll}\\ 12.&\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}$.

$\begin{array}{ll}\\ 13.&\textrm{Nilai dari}\: \: \sin \displaystyle \frac{\pi }{24}.\sin \frac{5\pi }{24}.\sin \frac{7\pi }{24}.\sin \frac{11\pi }{24}\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{5}{16}&&\textrm{d}.\quad \displaystyle \frac{2}{16}\\\\ \textrm{b}.\quad \displaystyle \frac{4}{16}&\textrm{c}.\quad \displaystyle \frac{3}{16}&\textrm{e}.\quad \color{red}\displaystyle \frac{1}{16} \end{array}\\\\ &\textbf{(Olimpiade Sains PORSEMA NU 2012)}\\\\ &\textbf{Jawab}:\: \: \color{red}\textbf{e}\\ &\begin{aligned}&\sin \displaystyle \frac{\pi }{24}.\sin \frac{5\pi }{24}.\sin \frac{7\pi }{24}.\sin \frac{11\pi }{24}\\ &=\displaystyle \frac{1}{4}\left ( 2\sin \displaystyle \frac{11\pi }{24}.\sin \frac{\pi }{24}.2\sin \frac{7\pi }{24}.\sin \frac{5\pi }{24} \right )\\ &=\displaystyle \frac{1}{4}\left [ \left ( \cos \left ( \frac{10\pi }{24} \right )-\cos \left ( \frac{12\pi }{24} \right ) \right )\times \left ( \cos \left ( \frac{2\pi }{24} \right )-\cos \left ( \frac{12\pi }{24} \right ) \right ) \right ]\\ &=\displaystyle \frac{1}{4}\left [ \left ( \cos 75^{\circ}-\cos 90^{\circ} \right )\times \left ( \cos 15^{\circ}-\cos 90^{\circ} \right ) \right ]\\ &=\displaystyle \frac{1}{4}\left [ \cos 75^{\circ}.\cos 15^{\circ} \right ]\\ &=\displaystyle \frac{1}{8}\left [ \cos 90^{\circ}+\cos 60^{\circ} \right ]\\ &=\displaystyle \frac{1}{8}\left ( 0+\frac{1}{2} \right )\\ &=\color{red}\displaystyle \frac{1}{16} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 14.&\textrm{Nilai dari}\\ & \qquad\qquad\sin 18^{\circ}\cos 36^{\circ}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \frac{1}{6} &&&\textrm{d}.&\displaystyle \frac{1}{3}\\\\ \textrm{b}.&\displaystyle \frac{1}{5}&\quad \textrm{c}.&\displaystyle \color{red}\displaystyle \frac{1}{4}\quad &\textrm{e}.&\displaystyle \frac{1}{2} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\sin 18^{\circ}\cos 36^{\circ}\\ &=\sin 18^{\circ}\cos 36^{\circ}\times \displaystyle \frac{2\cos 18^{\circ}}{2\cos 18^{\circ}}\\ &=\displaystyle \frac{\cos 36^{\circ}\left ( \sin 36^{\circ}+\sin 0^{\circ} \right )}{4\cos 18^{\circ}}\\ &=\displaystyle \frac{\cos 36^{\circ}\sin 36^{\circ}}{4\cos 18^{\circ}}\\ &=\displaystyle \frac{\sin 72^{\circ}}{4\cos 18^{\circ}}\\ &=\displaystyle \frac{\sin \left ( 90^{\circ}-18^{\circ} \right )}{4\cos 18^{\circ}}\\ &=\displaystyle \frac{\cos 18^{\circ}}{4\cos 18^{\circ}}\\ &=\color{red}\displaystyle \frac{1}{4} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 15.&\textrm{Nilai eksak dari}\: \: \sin 36^{\circ}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \displaystyle \frac{1}{4}\sqrt{10+2\sqrt{5}}&&&\textrm{d}.&\displaystyle \frac{\sqrt{5}-1}{4}\\ \textrm{b}.&\color{red}\displaystyle \frac{1}{4}\sqrt{10-2\sqrt{5}}&&\quad &\textrm{e}.&\displaystyle \frac{\sqrt{5}-1}{2}\\ \textrm{c}.&\displaystyle \displaystyle \frac{\sqrt{5}+1}{4} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\textrm{Perhatikanlah ilustrasi gambar berikut} \end{array}$.

$.\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}$.

DAFTAR PUSTAKA

1. Budhi, W.S. 2014. Matematika 4: Bahan Ajar Persiapan Menuju Olimpiade Matematika Sain Nasional/Internasional SMA. Jakarta: TRISULA ADISAKTI.
2. Sembiring, S., Zulkifli, M., Marsito, Rusdi, I. 2017. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SRIKANDI EMPAT WIDYA UTAMA.
3. Sukino. 2016. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.

Contoh Soal 2 (Segitiga dan Trigonometri)

$\begin{array}{ll}\\ 6.&\textrm{Bentuk}\: \: \displaystyle \frac{\cos 3x-\sin 6x-\cos 9x}{\sin 9x-\cos 6x-\sin 3x}\\ & \textrm{senilai dengan}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle -\tan 6x &&&\textrm{d}.&6\cot x\\ \textrm{b}.&\displaystyle -\cot 6x&&&\textrm{e}.&\displaystyle \color{red}\tan 6x\\ \textrm{c}.&\displaystyle 6\tan x \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\cos 3x-\sin 6x-\cos 9x}{\sin 9x-\cos 6x-\sin 3x}\\ &=\displaystyle \frac{\cos 3x-\cos 9x-\sin 6x}{\sin 9x-\sin 3x-\cos 6x}\\ &=\displaystyle \frac{-2\sin 6x\sin (-3x)-\sin 6x}{2\cos 6x\sin 3x-\cos 6x}\\ &=\displaystyle \frac{2\sin 6x\sin 3x-\sin 6x}{2\cos 6x\sin 3x-\cos 6x}\\ &=\displaystyle \frac{\sin 6x(2\sin 3x-1)}{\cos 6x(2\sin 3x-1)}\\ &=\tan 6x \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Nilai dari}\\ &\displaystyle \frac{\sin x+\sin 3x+\sin 5x+\sin 7x}{\cos x+\cos 3x+\cos 5x+\cos 7x}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \tan 2x &&\textrm{d}.&\displaystyle \tan 8x\\ \textrm{b}.&\color{red}\displaystyle \tan 4x&\quad&\textrm{e}.&\displaystyle \tan 16x\\ \textrm{c}.&\displaystyle \displaystyle \tan 6x \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\sin x+\sin 3x+\sin 5x+\sin 7x}{\cos x+\cos 3x+\cos 5x+\cos 7x}\\ &=\displaystyle \frac{\sin 7x+\sin x+\sin 5x+\sin 3x}{\cos 7x+\cos x+\cos 5x+\cos 3x}\\ &=\displaystyle \frac{2\sin 4x\cos 3x+2\sin 4x\cos x}{2\cos 4x\cos 3x+2\cos 4x\cos x}\\ &=\displaystyle \frac{2\sin 4x\left ( \cos 3x+\cos x \right )}{2\cos 4x\left ( \cos 3x+\cos x \right )}\\ &=\tan 4x \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Nilai dari}\: \: \cos 80^{\circ}+\cos 40^{\circ}-\cos 20^{\circ}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 1&&\textrm{d}.\quad -\displaystyle \frac{1}{2}\\\\ \textrm{b}.\quad -1&\textrm{c}.\quad \displaystyle \frac{1}{2}&\textrm{e}.\quad \color{red}0 \end{array}\\\\ &\textbf{Jawab}:\: \: \color{red}\textbf{e}\\ &\begin{aligned}&\cos 80^{\circ}+\cos 40^{\circ}-\cos 20^{\circ}\\ &=2\cos \displaystyle \frac{1}{2}\left ( 80^{\circ}+40^{\circ} \right )\cos \displaystyle \frac{1}{2}\left ( 80^{\circ}-40^{\circ} \right )-\cos 20^{\circ}\\ &=2\cos 60^{\circ}\cos 20^{\circ}-\cos 20^{\circ}\\ &=2.\displaystyle \frac{1}{2}.\cos 20^{\circ}-\cos 20^{\circ}\\ &=\cos 20^{\circ}-\cos 20^{\circ}\\ &=\color{red}0 \end{aligned} \end{array}$.

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

$\begin{array}{ll}\\ 10.&\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  $\cos \displaystyle \frac{\pi }{7}-\cos \displaystyle \frac{2\pi }{7}+\cos \displaystyle \frac{3\pi }{7}=\color{red}\displaystyle \frac{1}{2}$.

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

Contoh Soal 1 (Segitiga dan Trigonometri)

$\begin{array}{ll}\\ 1.&\textrm{Tunjukkan bahwa}\\ &\cot 7\displaystyle \frac{1}{2}^{0}=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}\\\\ &\textbf{Bukti}\\ &\begin{aligned}\cot \alpha &=\displaystyle \frac{\cos \alpha }{\sin \alpha }=\displaystyle \frac{2\cos ^{2}\alpha }{2\sin \alpha \cos \alpha }\\ &=\displaystyle \frac{1+\cos 2\alpha }{\sin 2\alpha }\\ &=\displaystyle \frac{1+\cos 15^{\circ} }{\sin 15^{\circ} }\\ \cot 7\displaystyle \frac{1}{2}^{0}&=\displaystyle \frac{1+\cos (45-30) }{\sin (45-30) }\\\\ &=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}\\ &=\displaystyle \frac{1+\displaystyle \frac{1}{4}\sqrt{6}+\frac{1}{4}\sqrt{2}}{\displaystyle \frac{1}{4}\sqrt{6}-\frac{1}{4}\sqrt{2}}\\ &=\displaystyle \frac{4+\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}\\ &=\displaystyle \frac{\left ( 4+\sqrt{6}+\sqrt{2} \right )\left ( \sqrt{6}+\sqrt{2} \right )}{6-2}\\ &=\displaystyle \frac{4\sqrt{6}+4\sqrt{2}+6+2\sqrt{3}+2\sqrt{3}+2}{4}\\ &=\displaystyle \frac{8+4\sqrt{2}+4\sqrt{3}+4\sqrt{6}}{4}\\ &=2+\sqrt{2}+\sqrt{3}+\sqrt{6}\\ &=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}\qquad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tentukan nilai eksak dari}\: \: \sin 18^{\circ}\\\\ &\textbf{Jawab}\\ &\begin{aligned}&\textrm{Diketahui}\: \: 4(18^{\circ})=72^{\circ}=90^{\circ}-18^{\circ}\\ &\textrm{maka kita pilih}\: \: x=18^{\circ}.\: \textrm{Selanjutnya}\\ &\sin 4x=\sin (90^{\circ}-x)=\cos x\\ &\Leftrightarrow 2\sin 2x\cos 2x=\cos x\\ &\Leftrightarrow 2(2\sin x\cos x)(1-2\sin ^{2}x)=\cos x\\ &\Leftrightarrow 4\sin x(1-2\sin ^{2}x)=1\\ &\Leftrightarrow 4\sin x-8\sin ^{3}x-1=0\\ &\Leftrightarrow 8\sin ^{3}x-4\sin x+1=0\\ &\Leftrightarrow \left ( \sin x-1 \right )\left (4\sin ^{2}+2\sin x-1 \right )=0\\ &\Leftrightarrow 2\sin x=1\: \: \textrm{atau}\: \: 2\sin x=\displaystyle \frac{-1\pm \sqrt{5}}{2}\\ &\textrm{Nilai yang mungkin untuk}\: \: 2\sin x\: \: \textrm{dari}\\ &\textrm{ketiga nilai di atas adalah}\: \: \displaystyle \frac{-1+\sqrt{5}}{2}\\ &\textrm{Sehinga nilai dari}\\ &2\sin x=\displaystyle \frac{-1+\sqrt{5}}{2}\\ &\Leftrightarrow \sin x=\displaystyle \frac{-1+\sqrt{5}}{4}\\ &\textrm{karena}\: \: x=18^{\circ},\: \: \textrm{akan didapatkan}\\ &\textrm{nilai}\: \: \sin 18^{\circ}=\displaystyle \frac{-1+\sqrt{5}}{4}\\ &\textrm{Jadi, nilai eksak dari}\: \: \sin 18^{\circ}=\displaystyle \frac{\sqrt{5}-1}{4} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Tunjukkan bahwa}\\ & \tan 11\displaystyle \frac{1}{4}^{\circ}=\sqrt{4+2\sqrt{2}}-\sqrt{2}-1\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\textrm{Langkah awal}\\ &\tan 22\frac{1}{2}^{\circ}=\displaystyle \frac{\sin 22\displaystyle \frac{1}{2}^{\circ}}{\cos 22\displaystyle \frac{1}{2}^{\circ}}=\displaystyle \frac{2\sin 22\displaystyle \frac{1}{2}^{\circ}\sin 22\displaystyle \frac{1}{2}^{\circ}}{2\sin 22\displaystyle \frac{1}{2}^{\circ}\cos 22\displaystyle \frac{1}{2}^{\circ}}\\ &\qquad =\displaystyle \frac{1-\cos 45^{\circ}}{\sin 45^{\circ}}=\displaystyle \frac{1-\displaystyle \frac{1}{2}\sqrt{2}}{\displaystyle \frac{1}{2}\sqrt{2}}=\displaystyle \frac{2-\sqrt{2}}{\sqrt{2}}\\ &\qquad =\displaystyle \frac{2-\sqrt{2}}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\displaystyle \frac{2\sqrt{2}-2}{2}\\ &\qquad =\sqrt{2}-1\\ &\textrm{Langkah berikutnya}\\ &\textrm{Misalkan}\: \: \tan 11\frac{1}{4}^{\circ}=x,\: \: \textrm{maka}\\ &\tan 22\displaystyle \frac{1}{2}^{\circ}=\displaystyle \frac{2\tan 11\displaystyle \frac{1}{4}^{\circ}}{1-\tan ^{2}11\displaystyle \frac{1}{4}^{\circ}}\\ &\Leftrightarrow \sqrt{2}-1=\displaystyle \frac{2x}{1-x^{2}}\\ &\Leftrightarrow \displaystyle \frac{1}{\sqrt{2}-1}=\displaystyle \frac{1-x^{2}}{2x}\\ &\Leftrightarrow x^{2}+2\left ( \sqrt{2}+1 \right )x-1=0\\ &\Leftrightarrow \: x_{_{1,2}}=\displaystyle \frac{-2\left ( \sqrt{2}+1 \right )\pm \sqrt{4\left ( \sqrt{2}+1 \right )^{2}+4}}{2}\\ &\Leftrightarrow \: x_{_{1,2}}= -\left ( \sqrt{2}+1 \right )\pm \sqrt{3+2\sqrt{2}+1}\\ &\Leftrightarrow \: x_{_{1,2}}=-\sqrt{2}-1\pm \sqrt{4+2\sqrt{2}}\\ &\textrm{Ambil yang nilai positif saja}\\ &\textrm{Sehingga}\\ & \tan 11\displaystyle \frac{1}{4}^{\circ}=-\sqrt{2}-1+\sqrt{4+2\sqrt{2}}\quad \blacksquare \\ \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Jika}\: \: \alpha ,\: \beta ,\: \: \textrm{dan}\: \: \gamma \: \: \textrm{adalah sudut}\\ & \textrm{pada segitiga ABC, buktikan bahwa}\\ &\tan \displaystyle \frac{\alpha }{2}\tan \displaystyle \frac{\beta }{2}+\tan \displaystyle \frac{\beta }{2}\tan \displaystyle \frac{\gamma }{2}+\tan \displaystyle \frac{\gamma }{2}\tan \displaystyle \frac{\alpha }{2}=1\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\textrm{Dikatahui bahwa}\\ &\alpha +\beta =180^{\circ}-\gamma \\ & \textrm{atau}\: \: \displaystyle \frac{1}{2}(\alpha +\beta )=\displaystyle \frac{1}{2}\left ( 180^{\circ}-\gamma \right )=90^{\circ}-\displaystyle \frac{\gamma }{2}\\ &\textrm{Maka}\\ &\tan \left ( \displaystyle \frac{\alpha }{2}+\frac{\beta }{2} \right )=\tan \left ( 90^{\circ}-\displaystyle \frac{\gamma }{2} \right )=\cot \displaystyle \frac{\gamma }{2}\\ &\Leftrightarrow \tan \left ( \displaystyle \frac{\alpha }{2}+\frac{\beta }{2} \right )=\displaystyle \frac{1}{\tan \displaystyle \frac{\gamma }{2}}\\ &\Leftrightarrow \displaystyle \frac{\tan \displaystyle \frac{\alpha }{2}+\tan \displaystyle \frac{\beta }{2}}{1-\tan \displaystyle \frac{\alpha }{2}\tan \displaystyle \frac{\beta }{2}}=\displaystyle \frac{1}{\tan \displaystyle \frac{\gamma }{2}}\\ &\Leftrightarrow \tan \displaystyle \frac{\alpha }{2}\tan \displaystyle \frac{\gamma }{2}+\tan \displaystyle \frac{\beta }{2}\tan \displaystyle \frac{\gamma }{2}=1-\tan \displaystyle \frac{\alpha }{2}\tan \displaystyle \frac{\beta }{2}\\ &\Leftrightarrow \tan \displaystyle \frac{\alpha }{2}\tan \displaystyle \frac{\beta }{2}+\tan \displaystyle \frac{\beta }{2}\tan \displaystyle \frac{\gamma }{2}+\tan \displaystyle \frac{\gamma }{2}\tan \displaystyle \frac{\alpha }{2}=1\quad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Jika}\: \: \alpha ,\: \beta ,\: \: \textrm{dan}\: \: \gamma \: \: \textrm{adalah sudut}\\ & \textrm{pada segitiga ABC, buktikan bahwa}\\ &\cos \alpha +\cos \beta +\cos \gamma =1+4\sin \displaystyle \frac{\alpha }{2}\sin \displaystyle \frac{\beta }{2}\sin \displaystyle \frac{\gamma }{2}\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\textrm{Dikatahui bahwa}\\ &\alpha +\beta =180^{\circ}-\gamma \\ & \textrm{atau}\: \: \displaystyle \frac{1}{2}(\alpha +\beta )=\displaystyle \frac{1}{2}\left ( 180^{\circ}-\gamma \right )=90^{\circ}-\displaystyle \frac{\gamma }{2}\\ &\textrm{Maka}\\ &\cos \alpha +\cos \beta +\cos \gamma =2\cos \left ( \displaystyle \frac{\alpha +\beta }{2} \right )\cos \left ( \displaystyle \frac{\alpha -\beta }{2} \right )\\ &\qquad\qquad\qquad\qquad\qquad +\cos \left ( 180^{\circ}-\left ( \alpha +\beta \right ) \right )\\ &\Leftrightarrow \cos \alpha +\cos \beta +\cos \gamma =2\cos \left ( \displaystyle \frac{\alpha +\beta }{2} \right )\cos \left ( \displaystyle \frac{\alpha -\beta }{2} \right )\\ &\qquad\qquad\qquad\qquad\qquad -\cos (\alpha +\beta )\\ &\Leftrightarrow \cos \alpha +\cos \beta +\cos \gamma =2\cos \left ( \displaystyle \frac{\alpha +\beta }{2} \right )\cos \left ( \displaystyle \frac{\alpha -\beta }{2} \right )\\ &\qquad\qquad\qquad\qquad\qquad -2\cos ^{2}\left ( \displaystyle \frac{\alpha +\beta }{2} \right )+1\\ &\Leftrightarrow \cos \alpha +\cos \beta +\cos \gamma \\ &\quad\quad =2\cos \left ( \displaystyle \frac{\alpha +\beta }{2} \right )\left (\cos \left ( \displaystyle \frac{\alpha -\beta }{2} \right )-\cos \left ( \displaystyle \frac{\alpha +\beta }{2} \right ) \right )+1\\ &\Leftrightarrow \cos \alpha +\cos \beta +\cos \gamma \\ &\quad\quad =2\cos \left ( \displaystyle \frac{\alpha +\beta }{2} \right )\left ( -2\sin \displaystyle \frac{\alpha }{2}\sin \displaystyle \frac{-\beta }{2} \right )+1\\ &\Leftrightarrow \cos \alpha +\cos \beta +\cos \gamma \\ &\quad\quad =2\cos \left ( \displaystyle \frac{\alpha +\beta }{2} \right )\left ( 2\sin \displaystyle \frac{\alpha }{2}\sin \displaystyle \frac{\beta }{2} \right )+1\\ &\Leftrightarrow \cos \alpha +\cos \beta +\cos \gamma =2\cos \left ( \displaystyle \frac{\alpha +\beta }{2} \right )\left ( 2\sin \displaystyle \frac{\alpha }{2}\sin \displaystyle \frac{\beta }{2} \right )+1\\ &\Leftrightarrow \cos \alpha +\cos \beta +\cos \gamma =2\cos \left ( 90^{\circ}-\displaystyle \frac{\gamma }{2} \right )\left ( 2\sin \displaystyle \frac{\alpha }{2}\sin \displaystyle \frac{\beta }{2} \right )+1\\ &\Leftrightarrow \cos \alpha +\cos \beta +\cos \gamma =4\sin \displaystyle \frac{\gamma }{2}\left ( \sin \displaystyle \frac{\alpha }{2}\sin \displaystyle \frac{\beta }{2} \right )+1\\ &\Leftrightarrow \cos \alpha +\cos \beta +\cos \gamma =1+4\sin \displaystyle \frac{\alpha }{2}\sin \displaystyle \frac{\beta }{2}\sin \displaystyle \frac{\gamma }{2}\qquad \blacksquare \end{aligned} \end{array}$

DAFTAR PUSTAKA

1. Bambang, S. 2012. Materi, Soal dan Penyelesaian Olimpiade Matematika Tingkat SMA/MA. Jakarta: BINA PRESTASI INSANI