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

$\begin{array}{l}\\ 91.& \text{Bentuk sederhana dari}\\ & 4\sin \left({\displaystyle \frac{1}{4}\pi +x}\right)\cos \left({\displaystyle \frac{1}{4}\pi -x}\right)\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 2+2\sin 2x}& & & \text{d}.& 2+2\sin x\\ \text{b}.& {\displaystyle 2+\sin 2x}& & & \text{e}.& 2+\sin x\\ \text{c}.& {\displaystyle 2\sin 2x}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& 4\sin \left({\displaystyle \frac{1}{4}\pi +x}\right)\cos \left({\displaystyle \frac{1}{4}\pi -x}\right)\\ & =2(\sin \left({\displaystyle \frac{1}{2}\pi }\right)+\sin (2x))\\ & =2(1+\sin 2x)\\ & =2+2\sin 2x\end{array}\end{array}$.

$\begin{array}{l}\\ 92.& \text{Bentuk sederhana dari}\\ & 2\sin \left({\displaystyle \frac{3}{4}\pi +x}\right).\cos \left({\displaystyle \frac{\pi }{4}+x}\right)=....\\ & \begin{array}{l}\\ \text{a}.1-\sin 2x& & \text{d}.\cos 2x\\ \\ \text{b}.1+\sin 2x& \text{c}.1-\cos 2x& \text{e}.-\cos 2x\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}& 2\sin \left({\displaystyle \frac{3}{4}\pi +x}\right).\cos \left({\displaystyle \frac{\pi }{4}+x}\right)\\ & =2\sin ({135}^{\circ }+x).\cos ({45}^{\circ }+x)\\ & =2\sin ({90}^{\circ }+{45}^{\circ }+x).\cos ({45}^{\circ }+x)\\ & =2\cos ({45}^{\circ }+x).\cos ({45}^{\circ }+x)\\ & =2{\cos }^2({45}^{\circ }+x)\\ & =2{(\cos {45}^{\circ }\cos x-\sin {45}^{\circ }\sin x)}^2\\ & =2{(\frac{1}{2}\sqrt{2}.\cos x-{\displaystyle \frac{1}{2}\sqrt{2}.\sin x})}^2\\ & ={\displaystyle {(\cos x-\sin x)}^2}\\ & ={\cos }^{2}x+{\sin }^{2}x-2\sin x\cos x\\ & =1-\sin 2x\end{array}\end{array}$.

$\begin{array}{l}\\ 93.& \text{Bentuk sederhana dari}\\ & 2\cos (x+{\displaystyle \frac{\pi }{4}}).\cos \left({\displaystyle \frac{3}{4}\pi -x}\right)=....\\ & \begin{array}{l}\\ \text{a}.\cos (2x+\pi )& & \text{d}.-\cos (2x-{\displaystyle \frac{\pi }{2}})\\ \text{b}.\cos (2x+{\displaystyle \frac{\pi }{2}})& & \text{e}.\sin 2x-1\\ \text{c}.\cos 2x\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}& 2\cos (x+{\displaystyle \frac{\pi }{4}}).\cos \left({\displaystyle \frac{3}{4}\pi -x}\right)\\ & =\cos (x+{\displaystyle \frac{\pi }{4}+{\displaystyle \frac{3}{4}\pi -x}})+\cos (x+{\displaystyle \frac{\pi }{4}-\left({\displaystyle \frac{3}{4}\pi -x}\right)})\\ & =\cos \pi +\cos (2x-\frac{2}{4}\pi )\\ & =-1+\cos \left({\displaystyle \frac{1}{2}\pi -2x}\right),\\ & \text{ingat bahwa}\cos (-\alpha )=\cos \alpha \\ & =-1+\sin 2x\\ & =\sin 2x-1\end{array}\end{array}$.

$\begin{array}{l}\\ 94.& \text{Nilai dari}\\ & \sqrt{3}\sin {80}^{\circ }\sin {160}^{\circ }\sin {320}^{\circ }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -{\displaystyle \frac{3}{8}}}& & & \text{d}.& {\displaystyle \frac{3}{8}}\\ \\ \text{b}.& {\displaystyle -\frac{1}{8}}& & & \text{e}.& {\displaystyle \frac{5}{8}}\\ \text{c}.& {\displaystyle \frac{1}{8}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \sqrt{3}\sin {80}^{\circ }\sin {160}^{\circ }\sin {320}^{\circ }\\ & =\sqrt{3}\sin {80}^{\circ }\sin {20}^{\circ }(-\sin {40}^{\circ })\\ & =-\sqrt{3}\sin {80}^{\circ }\sin {40}^{\circ }\sin {20}^{\circ }\\ & =-\sqrt{3}\sin {80}^{\circ }(-{\displaystyle \frac{1}{2}(\cos {60}^{\circ }-\cos {20}^{\circ })})\\ & =-\sqrt{3}\sin {80}^{\circ }(-{\displaystyle \frac{1}{4}+{\displaystyle \frac{\cos {20}^{\circ }}{2}}})\\ & ={\displaystyle \frac{1}{4}\sqrt{3}\sin {80}^{\circ }-{\displaystyle \frac{1}{2}\sqrt{3}\sin {80}^{\circ }\cos {20}^{\circ }}}\\ & ={\displaystyle \frac{1}{4}\sqrt{3}\sin {80}^{\circ }-{\displaystyle \frac{1}{4}\sqrt{3}(\sin {100}^{\circ }+\sin {60}^{\circ })}}\\ & ={\displaystyle \frac{1}{4}\sqrt{3}\sin {80}^{\circ }-{\displaystyle \frac{1}{4}\sqrt{3}(\sin {80}^{\circ }+{\displaystyle \frac{1}{2}\sqrt{3}})}}\\ & ={\displaystyle \frac{1}{4}\sqrt{3}\sin {80}^{\circ }-{\displaystyle \frac{1}{4}\sqrt{3}\sin {80}^{\circ }+{\displaystyle \frac{1}{8}\sqrt{9}}}}\\ & ={\displaystyle \frac{3}{8}}\end{array}\end{array}$.

$\begin{array}{l}\\ 95.& \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  $\cos {\displaystyle \frac{\pi }{7}-\cos {\displaystyle \frac{2\pi }{7}+\cos {\displaystyle \frac{3\pi }{7}={\displaystyle \frac{1}{2}}}}}$.

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

$\begin{array}{l}\\ 96.& \text{Nilai dari}\\ & \sin {\displaystyle \frac{\pi }{14}\sin \frac{3\pi }{14}\sin \frac{9\pi }{14}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle {\displaystyle \frac{1}{16}}}& & & \text{d}.& {\displaystyle \frac{1}{2}}\\ \\ \text{b}.& {\displaystyle \frac{1}{8}}& \text{c}.& {\displaystyle \frac{1}{4}}& \text{e}.& {\displaystyle 1}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Perhat}& \text{ikan bahwa}\\ \sin {\displaystyle {\displaystyle \frac{\pi }{14}}}& =\sin \left({\displaystyle \frac{7\pi }{14}-\frac{6\pi }{14}}\right)=\sin \left({\displaystyle \frac{1}{2}\pi -\frac{6\pi }{14}}\right)\\ & =\cos {\displaystyle \frac{6\pi }{14}}\\ \sin {\displaystyle \frac{3\pi }{14}}& =...=\cos {\displaystyle \frac{4\pi }{14}}\\ \sin {\displaystyle \frac{9\pi }{14}}& =...=\sin {\displaystyle \frac{5\pi }{14}=\cos {\displaystyle \frac{2\pi }{14}}}\end{array}\\ & ...\\ & \sin {\displaystyle \frac{\pi }{14}\sin \frac{3\pi }{14}\sin \frac{9\pi }{14}}\\ & =\cos {\displaystyle \frac{6\pi }{14}\cos {\displaystyle \frac{4\pi }{14}\cos {\displaystyle \frac{2\pi }{14}\times {\displaystyle \frac{2\sin {\displaystyle \frac{2\pi }{14}}}{2\sin {\displaystyle \frac{2\pi }{14}}}}}}}\\ & ={\displaystyle \frac{\cos {\displaystyle \frac{6\pi }{14}\cos {\displaystyle \frac{4\pi }{14}\sin {\displaystyle \frac{4\pi }{14}}}}}{2\sin {\displaystyle \frac{2\pi }{14}}}}\\ & \text{silahkan dilanjutkan}\\ & ...\\ & ={\displaystyle \frac{1}{8}}\end{array}$.

$\begin{array}{l}\\ 97.& \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}$.

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

$\begin{array}{l}\\ 99.& \text{Nilai dari}\\ & \sin {18}^{\circ }\cos {36}^{\circ }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{6}}& & & \text{d}.& {\displaystyle \frac{1}{3}}\\ \\ \text{b}.& {\displaystyle \frac{1}{5}}& \text{c}.& {\displaystyle {\displaystyle \frac{1}{4}}}& \text{e}.& {\displaystyle \frac{1}{2}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \sin {18}^{\circ }\cos {36}^{\circ }\\ & =\sin {18}^{\circ }\cos {36}^{\circ }\times {\displaystyle \frac{2\cos {18}^{\circ }}{2\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{\cos {36}^{\circ }(\sin {36}^{\circ }+\sin 0^{\circ })}{4\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{\cos {36}^{\circ }\sin {36}^{\circ }}{4\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{\sin {72}^{\circ }}{4\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{\sin ({90}^{\circ }-{18}^{\circ })}{4\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{\cos {18}^{\circ }}{4\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{1}{4}}\end{array}\end{array}$.

$\begin{array}{l}\\ 100.& \text{Nilai eksak dari}\sin {36}^{\circ }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle {\displaystyle \frac{1}{4}\sqrt{10+2\sqrt{5}}}}& & & \text{d}.& {\displaystyle \frac{\sqrt{5}-1}{4}}\\ \text{b}.& {\displaystyle \frac{1}{4}\sqrt{10-2\sqrt{5}}}& & & \text{e}.& {\displaystyle \frac{\sqrt{5}-1}{2}}\\ \text{c}.& {\displaystyle {\displaystyle \frac{\sqrt{5}+1}{4}}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikanlah ilustrasi gambar berikut}\end{array}$.

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