$\begin{array}{ll}\\ 81.&\textrm{Nilai}\: \: \cos \displaystyle \frac{5}{12}\pi -\cos \displaystyle \frac{1}{12}\pi \: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&-\displaystyle \frac{1}{2}\sqrt{6}&&&\textrm{d}.&\displaystyle \frac{1}{2}\sqrt{2}\\\\ \textrm{b}.&-\displaystyle \frac{1}{2}\sqrt{3}&\textrm{c}.&\color{red}-\displaystyle \frac{1}{2}\sqrt{2}&\textrm{e}.&\displaystyle \frac{1}{2}\sqrt{6} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\cos \displaystyle \frac{5}{12}\pi -\cos \displaystyle \frac{1}{12}\pi \\ &=-2\sin \left ( \displaystyle \frac{\displaystyle \frac{5}{12}\pi +\displaystyle \frac{1}{12}\pi }{2} \right )\sin \left ( \displaystyle \frac{\displaystyle \frac{5}{12}\pi -\displaystyle \frac{1}{12}\pi }{2} \right )\\ &=-2\sin \left (\displaystyle \frac{\displaystyle \frac{6}{12}\pi }{2} \right )\sin \left (\displaystyle \frac{\displaystyle \frac{4}{12}\pi }{2} \right ) \\ &=-2\sin \left (\displaystyle \frac{1 }{4}\pi \right )\sin \left (\displaystyle \frac{1 }{6}\pi \right ) \\ &=-2\left ( \displaystyle \frac{1}{2}\sqrt{2} \right )\left ( \displaystyle \frac{1}{2} \right )\\ &=\color{red}\displaystyle -\frac{1}{2}\sqrt{2} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 82.&\textrm{Bentuk}\: \: \sin \left ( 2x-\displaystyle \frac{3}{2}\pi \right )-\sin \left ( 4x+\displaystyle \frac{1}{2}\pi \right )\\ & \textrm{senilai dengan}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle -2\sin 3x.\sin x &&&\textrm{d}.&\color{red}\displaystyle 2\sin 3x.\sin x\\ \textrm{b}.&\displaystyle -2\cos 3x.\sin x&&&\textrm{e}.&\displaystyle 2\cos 3x.\sin x\\ \textrm{c}.&\displaystyle 2\sin 2\left ( x-\pi \right ) \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\sin \left ( 2x-\displaystyle \frac{3}{2}\pi \right )-\sin \left ( 4x+\displaystyle \frac{1}{2}\pi \right )\\ &=2\cos \left ( \displaystyle \frac{2x-\displaystyle \frac{3}{2}\pi+4x+\displaystyle \frac{1}{2}\pi}{2} \right )\\ &\qquad \times \sin \left ( \displaystyle \frac{2x-\displaystyle \frac{3}{2}\pi-\left (4x+\displaystyle \frac{1}{2}\pi \right )}{2} \right )\\ &=2\cos (3x-\displaystyle \frac{1}{2}\pi )\sin\left ( -x-\pi \right )\\ &=2\cos \left ( \displaystyle \frac{1}{2}\pi -3x \right )\left ( -\sin (\pi +x) \right )\\ &=2\left ( \sin 3x \right )\left ( -(-\sin x) \right )\\ &=2\sin 3x.\sin x \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 83.&\textrm{Nilai}\: \: \displaystyle \frac{\cos 75^{\circ}-\cos 15^{\circ}}{\sin 75^{\circ}-\sin 15^{\circ}}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}-1&&\textrm{d}.\quad \displaystyle \frac{1}{3}\sqrt{6}\\\\ \textrm{b}.\quad \displaystyle \frac{1}{2}\sqrt{2}&\textrm{c}.\quad 1&\textrm{e}.\quad \displaystyle \frac{1}{2}\sqrt{6} \end{array}\\\\ &\textbf{Jawab}:\: \: \color{red}\textbf{a}\\ &\begin{aligned}&\displaystyle \frac{\cos 75^{\circ}-\cos 15^{\circ}}{\sin 75^{\circ}-\sin 15^{\circ}}\\ &=\displaystyle \frac{-2\sin \displaystyle \frac{1}{2}\left ( 75^{\circ}+15^{\circ} \right )\sin \frac{1}{2}\left ( 75^{\circ}-15^{\circ} \right )}{2\cos \displaystyle \frac{1}{2}\left ( 75^{\circ}+15^{\circ} \right )\sin \frac{1}{2}\left ( 75^{\circ}-15^{\circ} \right )}\\ &=-\displaystyle \frac{2\sin 45^{\circ}\times \sin 30^{\circ}}{2\cos 45^{\circ}\times \sin 30^{\circ}}\\ &=-\tan 45^{\circ}\\ &=\color{red}-1 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 84.&\textrm{Bentuk}\: \: \displaystyle \frac{\cos 3x-\sin 6x-\cos 9x}{\sin 9x-\cos 6x-\sin 3x}\\ & \textrm{senilai dengan}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle -\tan 6x &&&\textrm{d}.&6\cot x\\ \textrm{b}.&\displaystyle -\cot 6x&&&\textrm{e}.&\displaystyle \color{red}\tan 6x\\ \textrm{c}.&\displaystyle 6\tan x \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\cos 3x-\sin 6x-\cos 9x}{\sin 9x-\cos 6x-\sin 3x}\\ &=\displaystyle \frac{\cos 3x-\cos 9x-\sin 6x}{\sin 9x-\sin 3x-\cos 6x}\\ &=\displaystyle \frac{-2\sin 6x\sin (-3x)-\sin 6x}{2\cos 6x\sin 3x-\cos 6x}\\ &=\displaystyle \frac{2\sin 6x\sin 3x-\sin 6x}{2\cos 6x\sin 3x-\cos 6x}\\ &=\displaystyle \frac{\sin 6x(2\sin 3x-1)}{\cos 6x(2\sin 3x-1)}\\ &=\tan 6x \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 85.&\textrm{Nilai dari}\\ &\displaystyle \frac{\sin x+\sin 3x+\sin 5x+\sin 7x}{\cos x+\cos 3x+\cos 5x+\cos 7x}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \tan 2x &&\textrm{d}.&\displaystyle \tan 8x\\ \textrm{b}.&\color{red}\displaystyle \tan 4x&\quad&\textrm{e}.&\displaystyle \tan 16x\\ \textrm{c}.&\displaystyle \displaystyle \tan 6x \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\sin x+\sin 3x+\sin 5x+\sin 7x}{\cos x+\cos 3x+\cos 5x+\cos 7x}\\ &=\displaystyle \frac{\sin 7x+\sin x+\sin 5x+\sin 3x}{\cos 7x+\cos x+\cos 5x+\cos 3x}\\ &=\displaystyle \frac{2\sin 4x\cos 3x+2\sin 4x\cos x}{2\cos 4x\cos 3x+2\cos 4x\cos x}\\ &=\displaystyle \frac{2\sin 4x\left ( \cos 3x+\cos x \right )}{2\cos 4x\left ( \cos 3x+\cos x \right )}\\ &=\tan 4x \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 86.&\textrm{Bentuk sederhana dari}\\ &\quad\quad \displaystyle \frac{\cos A+\sin A}{\cos A-\sin A}-\frac{\cos A-\sin A}{\cos A+\sin A}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \tan A &&&\textrm{d}.&2\cos 2A\\ \textrm{b}.&\displaystyle 2\tan A&&&\textrm{e}.&\displaystyle \color{red}2\tan 2A\\ \textrm{c}.&\displaystyle 2\sin 2A \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\cos A+\sin A}{\cos A-\sin A}-\frac{\cos A-\sin A}{\cos A+\sin A}\\ &=\displaystyle \frac{(\cos A+\sin A)^{2}-(\cos A-\sin A)^{2}}{(\cos A-\sin A)(\cos A+\sin A)}\\ &=\displaystyle \frac{(\cos ^{2}A+2\cos A\sin A+\sin ^{2}A)-(\cos ^{2}A-2\cos A\sin A+\sin ^{2}A)}{\cos ^{2}A-\sin ^{2}A}\\ &=\displaystyle \frac{4\cos A\sin A}{\cos ^{2}A-\sin ^{2}A}\\ &=\displaystyle \frac{2\sin 2A}{\cos 2A}\\ &=2\tan 2A \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 87.&\textrm{Bentuk sederhana dari}\\ &\qquad\quad \displaystyle \frac{\cos 2x-\cos 2y}{\sin 2x+\sin 2y}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle -\sin (x-y) &&&\textrm{d}.&\cos (x-y)\\ \textrm{b}.&\displaystyle \color{red}-\tan (x-y)&&&\textrm{e}.&\displaystyle \tan (x-y)\\ \textrm{c}.&\displaystyle \sin (x+y) \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\cos 2x-\cos 2y}{\sin 2x+\sin 2y}\\ &=\displaystyle \frac{-2\sin (x+y)\sin (x-y)}{2\sin (x+y)\cos (x-y)}\\ &=-\tan (x-y) \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 88.&\textrm{Nilai dari}\: \: \cos 80^{\circ}+\cos 40^{\circ}-\cos 20^{\circ}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 1&&\textrm{d}.\quad -\displaystyle \frac{1}{2}\\\\ \textrm{b}.\quad -1&\textrm{c}.\quad \displaystyle \frac{1}{2}&\textrm{e}.\quad \color{red}0 \end{array}\\\\ &\textbf{Jawab}:\: \: \color{red}\textbf{e}\\ &\begin{aligned}&\cos 80^{\circ}+\cos 40^{\circ}-\cos 20^{\circ}\\ &=2\cos \displaystyle \frac{1}{2}\left ( 80^{\circ}+40^{\circ} \right )\cos \displaystyle \frac{1}{2}\left ( 80^{\circ}-40^{\circ} \right )-\cos 20^{\circ}\\ &=2\cos 60^{\circ}\cos 20^{\circ}-\cos 20^{\circ}\\ &=2.\displaystyle \frac{1}{2}.\cos 20^{\circ}-\cos 20^{\circ}\\ &=\cos 20^{\circ}-\cos 20^{\circ}\\ &=\color{red}0 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 89.&\textrm{Nilai dari}\\ &\quad\quad 8\cos 82,5^{\circ}\sin 37,5^{\circ}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle 4(\sqrt{3}+\sqrt{2}) &&&\textrm{d}.&\color{red}2(\sqrt{3}-\sqrt{2})\\ \textrm{b}.&\displaystyle 4(\sqrt{3}-\sqrt{2})&&&\textrm{e}.&\displaystyle \sqrt{3}-\sqrt{2}\\ \textrm{c}.&\displaystyle 2(\sqrt{3}+\sqrt{2}) \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&8\cos 82,5^{\circ}\sin 37,5^{\circ}\\ &=4\times 2\cos 82,5^{\circ}\sin 37,5^{\circ}\\ &=4\times \left ( \sin (82,5^{\circ}+37,5^{\circ})-\sin (82,5^{\circ}-37,5^{\circ}) \right )\\ &=4\times \left (\sin 120^{\circ}-\sin 45^{\circ} \right )\\ &=4\times \left ( \sin \left ( 180^{\circ}-60^{\circ} \right )-\sin 45^{\circ} \right )\\ &=4\times \left ( \sin 60^{\circ}-\sin 45^{\circ} \right )\\ &=4\times \left ( \displaystyle \frac{1}{2}\sqrt{3}-\displaystyle \frac{1}{2}\sqrt{2} \right )\\ &=2\left ( \sqrt{3}-\sqrt{2} \right ) \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 90.&\textrm{Bentuk lain dari}\\ &\quad\quad -2\cos 5A.\cos 7A\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle -\cos 6A-\cos A &&&\\ \textrm{b}.&\displaystyle -\cos 6A+\cos A&&&\\ \textrm{c}.&\displaystyle \cos 12A-\cos 2A\\ \textrm{d}.&-\cos 12A+\cos 2A\\ \textrm{e}.&\color{red}\displaystyle -\cos 12A-\cos 2A \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&-2\cos 5A.\cos 7A\\ &=-\left ( 2\cos 5A.\cos 7A \right )\\ &=-\left ( \cos 12A+\cos (-2A) \right )\\ &=-\left ( \cos 12A+\cos 2A \right )\\ &=-\cos 12A-\cos 2A \\ \end{aligned} \end{array}$.
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