Contoh Soal 3 (Segitiga dan Trigonometri)

 

$\begin{array}{l}\\ 11.& \text{Nilai dari}\\ & \sin {\displaystyle \frac{\pi }{14}\sin \frac{3\pi }{14}\sin \frac{9\pi }{14}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle {\displaystyle \frac{1}{16}}}& & & \text{d}.& {\displaystyle \frac{1}{2}}\\ \\ \text{b}.& {\displaystyle \frac{1}{8}}& \text{c}.& {\displaystyle \frac{1}{4}}& \text{e}.& {\displaystyle 1}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Perhat}& \text{ikan bahwa}\\ \sin {\displaystyle {\displaystyle \frac{\pi }{14}}}& =\sin \left({\displaystyle \frac{7\pi }{14}-\frac{6\pi }{14}}\right)=\sin \left({\displaystyle \frac{1}{2}\pi -\frac{6\pi }{14}}\right)\\ & =\cos {\displaystyle \frac{6\pi }{14}}\\ \sin {\displaystyle \frac{3\pi }{14}}& =...=\cos {\displaystyle \frac{4\pi }{14}}\\ \sin {\displaystyle \frac{9\pi }{14}}& =...=\sin {\displaystyle \frac{5\pi }{14}=\cos {\displaystyle \frac{2\pi }{14}}}\end{array}\\ & ...\\ & \sin {\displaystyle \frac{\pi }{14}\sin \frac{3\pi }{14}\sin \frac{9\pi }{14}}\\ & =\cos {\displaystyle \frac{6\pi }{14}\cos {\displaystyle \frac{4\pi }{14}\cos {\displaystyle \frac{2\pi }{14}\times {\displaystyle \frac{2\sin {\displaystyle \frac{2\pi }{14}}}{2\sin {\displaystyle \frac{2\pi }{14}}}}}}}\\ & ={\displaystyle \frac{\cos {\displaystyle \frac{6\pi }{14}\cos {\displaystyle \frac{4\pi }{14}\sin {\displaystyle \frac{4\pi }{14}}}}}{2\sin {\displaystyle \frac{2\pi }{14}}}}\\ & \text{silahkan dilanjutkan}\\ & ...\\ & ={\displaystyle \frac{1}{8}}\end{array}$.

$\begin{array}{l}\\ 12.& \text{Nilai dari}\\ & \cos {\displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\cos {\displaystyle \frac{8\pi }{5}}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -{\displaystyle \frac{1}{16}}}& & & \text{d}.& {\displaystyle \frac{1}{16}}\\ \\ \text{b}.& {\displaystyle \frac{1}{8}}& \text{c}.& {\displaystyle 0}& \text{e}.& {\displaystyle \frac{1}{8}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \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 (\pi +{\displaystyle \frac{3\pi }{5}})}\\ & =\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{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}(\sin \pi -\sin {\displaystyle \frac{3\pi }{5}})}}{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}(\sin \pi -\sin (-{\displaystyle \frac{\pi }{5}}))}}}{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}}}}\\ & =-{\displaystyle \frac{1}{16}}\end{array}\end{array}$.

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

$\begin{array}{l}\\ 14.& \text{Nilai dari}\\ & \sin {18}^{\circ }\cos {36}^{\circ }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{6}}& & & \text{d}.& {\displaystyle \frac{1}{3}}\\ \\ \text{b}.& {\displaystyle \frac{1}{5}}& \text{c}.& {\displaystyle {\displaystyle \frac{1}{4}}}& \text{e}.& {\displaystyle \frac{1}{2}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \sin {18}^{\circ }\cos {36}^{\circ }\\ & =\sin {18}^{\circ }\cos {36}^{\circ }\times {\displaystyle \frac{2\cos {18}^{\circ }}{2\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{\cos {36}^{\circ }(\sin {36}^{\circ }+\sin 0^{\circ })}{4\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{\cos {36}^{\circ }\sin {36}^{\circ }}{4\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{\sin {72}^{\circ }}{4\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{\sin ({90}^{\circ }-{18}^{\circ })}{4\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{\cos {18}^{\circ }}{4\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{1}{4}}\end{array}\end{array}$.

$\begin{array}{l}\\ 15.& \text{Nilai eksak dari}\sin {36}^{\circ }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle {\displaystyle \frac{1}{4}\sqrt{10+2\sqrt{5}}}}& & & \text{d}.& {\displaystyle \frac{\sqrt{5}-1}{4}}\\ \text{b}.& {\displaystyle \frac{1}{4}\sqrt{10-2\sqrt{5}}}& & & \text{e}.& {\displaystyle \frac{\sqrt{5}-1}{2}}\\ \text{c}.& {\displaystyle {\displaystyle \frac{\sqrt{5}+1}{4}}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikanlah ilustrasi gambar berikut}\end{array}$.

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

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)

$\begin{array}{l}\\ 6.& \text{Bentuk}{\displaystyle \frac{\cos 3x-\sin 6x-\cos 9x}{\sin 9x-\cos 6x-\sin 3x}}\\ & \text{senilai dengan}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -\tan 6x}& & & \text{d}.& 6\cot x\\ \text{b}.& {\displaystyle -\cot 6x}& & & \text{e}.& {\displaystyle \tan 6x}\\ \text{c}.& {\displaystyle 6\tan x}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \frac{\cos 3x-\sin 6x-\cos 9x}{\sin 9x-\cos 6x-\sin 3x}}\\ & ={\displaystyle \frac{\cos 3x-\cos 9x-\sin 6x}{\sin 9x-\sin 3x-\cos 6x}}\\ & ={\displaystyle \frac{-2\sin 6x\sin (-3x)-\sin 6x}{2\cos 6x\sin 3x-\cos 6x}}\\ & ={\displaystyle \frac{2\sin 6x\sin 3x-\sin 6x}{2\cos 6x\sin 3x-\cos 6x}}\\ & ={\displaystyle \frac{\sin 6x(2\sin 3x-1)}{\cos 6x(2\sin 3x-1)}}\\ & =\tan 6x\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Nilai dari}\\ & {\displaystyle \frac{\sin x+\sin 3x+\sin 5x+\sin 7x}{\cos x+\cos 3x+\cos 5x+\cos 7x}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \tan 2x}& & \text{d}.& {\displaystyle \tan 8x}\\ \text{b}.& {\displaystyle \tan 4x}& & \text{e}.& {\displaystyle \tan 16x}\\ \text{c}.& {\displaystyle {\displaystyle \tan 6x}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \frac{\sin x+\sin 3x+\sin 5x+\sin 7x}{\cos x+\cos 3x+\cos 5x+\cos 7x}}\\ & ={\displaystyle \frac{\sin 7x+\sin x+\sin 5x+\sin 3x}{\cos 7x+\cos x+\cos 5x+\cos 3x}}\\ & ={\displaystyle \frac{2\sin 4x\cos 3x+2\sin 4x\cos x}{2\cos 4x\cos 3x+2\cos 4x\cos x}}\\ & ={\displaystyle \frac{2\sin 4x(\cos 3x+\cos x)}{2\cos 4x(\cos 3x+\cos x)}}\\ & =\tan 4x\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Nilai dari}\cos {80}^{\circ }+\cos {40}^{\circ }-\cos {20}^{\circ }=....\\ & \begin{array}{l}\\ \text{a}.1& & \text{d}.-{\displaystyle \frac{1}{2}}\\ \\ \text{b}.-1& \text{c}.{\displaystyle \frac{1}{2}}& \text{e}.0\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \cos {80}^{\circ }+\cos {40}^{\circ }-\cos {20}^{\circ }\\ & =2\cos {\displaystyle \frac{1}{2}({80}^{\circ }+{40}^{\circ })\cos {\displaystyle \frac{1}{2}({80}^{\circ }-{40}^{\circ })-\cos {20}^{\circ }}}\\ & =2\cos {60}^{\circ }\cos {20}^{\circ }-\cos {20}^{\circ }\\ & =2.{\displaystyle \frac{1}{2}.\cos {20}^{\circ }-\cos {20}^{\circ }}\\ & =\cos {20}^{\circ }-\cos {20}^{\circ }\\ & =0\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Nilai dari}\\ & \sqrt{3}\sin {80}^{\circ }\sin {160}^{\circ }\sin {320}^{\circ }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -{\displaystyle \frac{3}{8}}}& & & \text{d}.& {\displaystyle \frac{3}{8}}\\ \\ \text{b}.& {\displaystyle -\frac{1}{8}}& & & \text{e}.& {\displaystyle \frac{5}{8}}\\ \text{c}.& {\displaystyle \frac{1}{8}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \sqrt{3}\sin {80}^{\circ }\sin {160}^{\circ }\sin {320}^{\circ }\\ & =\sqrt{3}\sin {80}^{\circ }\sin {20}^{\circ }(-\sin {40}^{\circ })\\ & =-\sqrt{3}\sin {80}^{\circ }\sin {40}^{\circ }\sin {20}^{\circ }\\ & =-\sqrt{3}\sin {80}^{\circ }(-{\displaystyle \frac{1}{2}(\cos {60}^{\circ }-\cos {20}^{\circ })})\\ & =-\sqrt{3}\sin {80}^{\circ }(-{\displaystyle \frac{1}{4}+{\displaystyle \frac{\cos {20}^{\circ }}{2}}})\\ & ={\displaystyle \frac{1}{4}\sqrt{3}\sin {80}^{\circ }-{\displaystyle \frac{1}{2}\sqrt{3}\sin {80}^{\circ }\cos {20}^{\circ }}}\\ & ={\displaystyle \frac{1}{4}\sqrt{3}\sin {80}^{\circ }-{\displaystyle \frac{1}{4}\sqrt{3}(\sin {100}^{\circ }+\sin {60}^{\circ })}}\\ & ={\displaystyle \frac{1}{4}\sqrt{3}\sin {80}^{\circ }-{\displaystyle \frac{1}{4}\sqrt{3}(\sin {80}^{\circ }+{\displaystyle \frac{1}{2}\sqrt{3}})}}\\ & ={\displaystyle \frac{1}{4}\sqrt{3}\sin {80}^{\circ }-{\displaystyle \frac{1}{4}\sqrt{3}\sin {80}^{\circ }+{\displaystyle \frac{1}{8}\sqrt{9}}}}\\ & ={\displaystyle \frac{3}{8}}\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Nilai dari}\\ & \cos {\displaystyle \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -{\displaystyle \frac{1}{8}}}& & & \text{d}.& {\displaystyle \frac{1}{2}}\\ \\ \text{b}.& {\displaystyle -\frac{1}{4}}& \text{c}.& 0& \text{e}.& {\displaystyle \frac{1}{3}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \mathbf{\text{Alternatif 1}}\\ & \begin{array}{rl}& \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 (-{\displaystyle \frac{\pi }{7}})}}}}{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{array}\\ & \mathbf{\text{Alternatif 2}}\\ & \begin{array}{rl}& \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}(\cos (\pi -{\displaystyle \frac{\pi }{7}})+\cos {\displaystyle \frac{2\pi }{7}})\cos {\displaystyle \frac{\pi }{7}}}\\ & ={\displaystyle \frac{1}{2}(-\cos {\displaystyle \frac{\pi }{7}+\cos {\displaystyle \frac{2\pi }{7}}})\cos {\displaystyle \frac{\pi }{7}}}\\ & ={\displaystyle \frac{1}{2}(-{\cos }^2{\displaystyle \frac{\pi }{7}+\cos {\displaystyle \frac{2\pi }{7}\cos {\displaystyle \frac{\pi }{7}}}})}\\ & ={\displaystyle \frac{1}{4}(-\cos {\displaystyle \frac{2\pi }{7}-\cos 0+\cos {\displaystyle \frac{3\pi }{7}+\cos {\displaystyle \frac{\pi }{7}}}})}\\ & ={\displaystyle \frac{1}{4}(-\cos 0+\cos {\displaystyle \frac{\pi }{7}-\cos {\displaystyle \frac{2\pi }{7}+\cos {\displaystyle \frac{3\pi }{7}}}})}\\ & ={\displaystyle \frac{1}{4}(-1+{\displaystyle \frac{1}{2}})}\\ & ={\displaystyle \frac{1}{4}\times (-\frac{1}{2})}\\ & =-{\displaystyle \frac{1}{8}}\end{array}\end{array}$.

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

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

Contoh Soal 1 (Segitiga dan Trigonometri)

$\begin{array}{l}\\ 1.& \text{Tunjukkan bahwa}\\ & \cot 7{\displaystyle {\frac{1}{2}}^0=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}\cot \alpha & ={\displaystyle \frac{\cos \alpha }{\sin \alpha }={\displaystyle \frac{2{\cos }^2\alpha }{2\sin \alpha \cos \alpha }}}\\ & ={\displaystyle \frac{1+\cos 2\alpha }{\sin 2\alpha }}\\ & ={\displaystyle \frac{1+\cos {15}^{\circ }}{\sin {15}^{\circ }}}\\ \cot 7{\displaystyle {\frac{1}{2}}^0}& ={\displaystyle \frac{1+\cos (45-30)}{\sin (45-30)}}\\ \\ & =\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}\\ & ={\displaystyle \frac{1+{\displaystyle \frac{1}{4}\sqrt{6}+\frac{1}{4}\sqrt{2}}}{{\displaystyle \frac{1}{4}\sqrt{6}-\frac{1}{4}\sqrt{2}}}}\\ & ={\displaystyle \frac{4+\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}}\\ & ={\displaystyle \frac{(4+\sqrt{6}+\sqrt{2})(\sqrt{6}+\sqrt{2})}{6-2}}\\ & ={\displaystyle \frac{4\sqrt{6}+4\sqrt{2}+6+2\sqrt{3}+2\sqrt{3}+2}{4}}\\ & ={\displaystyle \frac{8+4\sqrt{2}+4\sqrt{3}+4\sqrt{6}}{4}}\\ & =2+\sqrt{2}+\sqrt{3}+\sqrt{6}\\ & =\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}◼\end{array}\end{array}$.

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

$\begin{array}{l}\\ 3.& \text{Tunjukkan bahwa}\\ & \tan 11{\displaystyle {\frac{1}{4}}^{\circ }=\sqrt{4+2\sqrt{2}}-\sqrt{2}-1}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Langkah awal}\\ & \tan 22{\frac{1}{2}}^{\circ }={\displaystyle \frac{\sin 22{\displaystyle {\frac{1}{2}}^{\circ }}}{\cos 22{\displaystyle {\frac{1}{2}}^{\circ }}}={\displaystyle \frac{2\sin 22{\displaystyle {\frac{1}{2}}^{\circ }\sin 22{\displaystyle {\frac{1}{2}}^{\circ }}}}{2\sin 22{\displaystyle {\frac{1}{2}}^{\circ }\cos 22{\displaystyle {\frac{1}{2}}^{\circ }}}}}}\\ & ={\displaystyle \frac{1-\cos {45}^{\circ }}{\sin {45}^{\circ }}={\displaystyle \frac{1-{\displaystyle \frac{1}{2}\sqrt{2}}}{{\displaystyle \frac{1}{2}\sqrt{2}}}={\displaystyle \frac{2-\sqrt{2}}{\sqrt{2}}}}}\\ & ={\displaystyle \frac{2-\sqrt{2}}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}={\displaystyle \frac{2\sqrt{2}-2}{2}}}\\ & =\sqrt{2}-1\\ & \text{Langkah berikutnya}\\ & \text{Misalkan}\tan 11{\frac{1}{4}}^{\circ }=x,\text{maka}\\ & \tan 22{\displaystyle {\frac{1}{2}}^{\circ }={\displaystyle \frac{2\tan 11{\displaystyle {\frac{1}{4}}^{\circ }}}{1-{\tan }^{2}11{\displaystyle {\frac{1}{4}}^{\circ }}}}}\\ & \iff \sqrt{2}-1={\displaystyle \frac{2x}{1-x^2}}\\ & \iff {\displaystyle \frac{1}{\sqrt{2}-1}={\displaystyle \frac{1-x^2}{2x}}}\\ & \iff x^2+2(\sqrt{2}+1)x-1=0\\ & \iff x_{{}_{1,2}}={\displaystyle \frac{-2(\sqrt{2}+1)\pm \sqrt{4{(\sqrt{2}+1)}^2+4}}{2}}\\ & \iff x_{{}_{1,2}}=-(\sqrt{2}+1)\pm \sqrt{3+2\sqrt{2}+1}\\ & \iff x_{{}_{1,2}}=-\sqrt{2}-1\pm \sqrt{4+2\sqrt{2}}\\ & \text{Ambil yang nilai positif saja}\\ & \text{Sehingga}\\ & \tan 11{\displaystyle {\frac{1}{4}}^{\circ }=-\sqrt{2}-1+\sqrt{4+2\sqrt{2}}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Jika}\alpha ,\beta ,\text{dan}\gamma \text{adalah sudut}\\ & \text{pada segitiga ABC, buktikan bahwa}\\ & \tan {\displaystyle \frac{\alpha }{2}\tan {\displaystyle \frac{\beta }{2}+\tan {\displaystyle \frac{\beta }{2}\tan {\displaystyle \frac{\gamma }{2}+\tan {\displaystyle \frac{\gamma }{2}\tan {\displaystyle \frac{\alpha }{2}=1}}}}}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Dikatahui bahwa}\\ & \alpha +\beta ={180}^{\circ }-\gamma \\ & \text{atau}{\displaystyle \frac{1}{2}(\alpha +\beta )={\displaystyle \frac{1}{2}({180}^{\circ }-\gamma )={90}^{\circ }-{\displaystyle \frac{\gamma }{2}}}}\\ & \text{Maka}\\ & \tan \left({\displaystyle \frac{\alpha }{2}+\frac{\beta }{2}}\right)=\tan ({90}^{\circ }-{\displaystyle \frac{\gamma }{2}})=\cot {\displaystyle \frac{\gamma }{2}}\\ & \iff \tan \left({\displaystyle \frac{\alpha }{2}+\frac{\beta }{2}}\right)={\displaystyle \frac{1}{\tan {\displaystyle \frac{\gamma }{2}}}}\\ & \iff {\displaystyle \frac{\tan {\displaystyle \frac{\alpha }{2}+\tan {\displaystyle \frac{\beta }{2}}}}{1-\tan {\displaystyle \frac{\alpha }{2}\tan {\displaystyle \frac{\beta }{2}}}}={\displaystyle \frac{1}{\tan {\displaystyle \frac{\gamma }{2}}}}}\\ & \iff \tan {\displaystyle \frac{\alpha }{2}\tan {\displaystyle \frac{\gamma }{2}+\tan {\displaystyle \frac{\beta }{2}\tan {\displaystyle \frac{\gamma }{2}=1-\tan {\displaystyle \frac{\alpha }{2}\tan {\displaystyle \frac{\beta }{2}}}}}}}\\ & \iff \tan {\displaystyle \frac{\alpha }{2}\tan {\displaystyle \frac{\beta }{2}+\tan {\displaystyle \frac{\beta }{2}\tan {\displaystyle \frac{\gamma }{2}+\tan {\displaystyle \frac{\gamma }{2}\tan {\displaystyle \frac{\alpha }{2}=1◼}}}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Jika}\alpha ,\beta ,\text{dan}\gamma \text{adalah sudut}\\ & \text{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}}}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Dikatahui bahwa}\\ & \alpha +\beta ={180}^{\circ }-\gamma \\ & \text{atau}{\displaystyle \frac{1}{2}(\alpha +\beta )={\displaystyle \frac{1}{2}({180}^{\circ }-\gamma )={90}^{\circ }-{\displaystyle \frac{\gamma }{2}}}}\\ & \text{Maka}\\ & \cos \alpha +\cos \beta +\cos \gamma =2\cos \left({\displaystyle \frac{\alpha +\beta }{2}}\right)\cos \left({\displaystyle \frac{\alpha -\beta }{2}}\right)\\ & +\cos ({180}^{\circ }-(\alpha +\beta ))\\ & \iff \cos \alpha +\cos \beta +\cos \gamma =2\cos \left({\displaystyle \frac{\alpha +\beta }{2}}\right)\cos \left({\displaystyle \frac{\alpha -\beta }{2}}\right)\\ & -\cos (\alpha +\beta )\\ & \iff \cos \alpha +\cos \beta +\cos \gamma =2\cos \left({\displaystyle \frac{\alpha +\beta }{2}}\right)\cos \left({\displaystyle \frac{\alpha -\beta }{2}}\right)\\ & -2{\cos }^2\left({\displaystyle \frac{\alpha +\beta }{2}}\right)+1\\ & \iff \cos \alpha +\cos \beta +\cos \gamma \\ & =2\cos \left({\displaystyle \frac{\alpha +\beta }{2}}\right)(\cos \left({\displaystyle \frac{\alpha -\beta }{2}}\right)-\cos \left({\displaystyle \frac{\alpha +\beta }{2}}\right))+1\\ & \iff \cos \alpha +\cos \beta +\cos \gamma \\ & =2\cos \left({\displaystyle \frac{\alpha +\beta }{2}}\right)(-2\sin {\displaystyle \frac{\alpha }{2}\sin {\displaystyle \frac{-\beta }{2}}})+1\\ & \iff \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\\ & \iff \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\\ & \iff \cos \alpha +\cos \beta +\cos \gamma =2\cos ({90}^{\circ }-{\displaystyle \frac{\gamma }{2}})\left(2\sin {\displaystyle \frac{\alpha }{2}\sin {\displaystyle \frac{\beta }{2}}}\right)+1\\ & \iff \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}\\ & \iff \cos \alpha +\cos \beta +\cos \gamma =1+4\sin {\displaystyle \frac{\alpha }{2}\sin {\displaystyle \frac{\beta }{2}\sin {\displaystyle \frac{\gamma }{2}◼}}}\end{array}\end{array}$

DAFTAR PUSTAKA

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