Latihan Soal 9 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

 $\begin{array}{l}\\ 81.& \text{Nilai}\cos {\displaystyle \frac{5}{12}\pi -\cos {\displaystyle \frac{1}{12}\pi \text{adalah}....}}\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle \frac{1}{2}\sqrt{6}}& & & \text{d}.& {\displaystyle \frac{1}{2}\sqrt{2}}\\ \\ \text{b}.& -{\displaystyle \frac{1}{2}\sqrt{3}}& \text{c}.& -{\displaystyle \frac{1}{2}\sqrt{2}}& \text{e}.& {\displaystyle \frac{1}{2}\sqrt{6}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \cos {\displaystyle \frac{5}{12}\pi -\cos {\displaystyle \frac{1}{12}\pi }}\\ & =-2\sin \left({\displaystyle \frac{{\displaystyle \frac{5}{12}\pi +{\displaystyle \frac{1}{12}\pi }}}{2}}\right)\sin \left({\displaystyle \frac{{\displaystyle \frac{5}{12}\pi -{\displaystyle \frac{1}{12}\pi }}}{2}}\right)\\ & =-2\sin \left({\displaystyle \frac{{\displaystyle \frac{6}{12}\pi }}{2}}\right)\sin \left({\displaystyle \frac{{\displaystyle \frac{4}{12}\pi }}{2}}\right)\\ & =-2\sin \left({\displaystyle \frac{1}{4}\pi }\right)\sin \left({\displaystyle \frac{1}{6}\pi }\right)\\ & =-2\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)\left({\displaystyle \frac{1}{2}}\right)\\ & ={\displaystyle -\frac{1}{2}\sqrt{2}}\end{array}\end{array}$.

$\begin{array}{l}\\ 82.& \text{Bentuk}\sin (2x-{\displaystyle \frac{3}{2}\pi })-\sin (4x+{\displaystyle \frac{1}{2}\pi })\\ & \text{senilai dengan}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -2\sin 3x.\sin x}& & & \text{d}.& {\displaystyle 2\sin 3x.\sin x}\\ \text{b}.& {\displaystyle -2\cos 3x.\sin x}& & & \text{e}.& {\displaystyle 2\cos 3x.\sin x}\\ \text{c}.& {\displaystyle 2\sin 2(x-\pi )}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \sin (2x-{\displaystyle \frac{3}{2}\pi })-\sin (4x+{\displaystyle \frac{1}{2}\pi })\\ & =2\cos \left({\displaystyle \frac{2x-{\displaystyle \frac{3}{2}\pi +4x+{\displaystyle \frac{1}{2}\pi }}}{2}}\right)\\ & \times \sin \left({\displaystyle \frac{2x-{\displaystyle \frac{3}{2}\pi -(4x+{\displaystyle \frac{1}{2}\pi })}}{2}}\right)\\ & =2\cos (3x-{\displaystyle \frac{1}{2}\pi )\sin (-x-\pi )}\\ & =2\cos \left({\displaystyle \frac{1}{2}\pi -3x}\right)(-\sin (\pi +x))\\ & =2(\sin 3x)(-(-\sin x))\\ & =2\sin 3x.\sin x\end{array}\end{array}$.

$\begin{array}{l}\\ 83.& \text{Nilai}{\displaystyle \frac{\cos {75}^{\circ }-\cos {15}^{\circ }}{\sin {75}^{\circ }-\sin {15}^{\circ }}=....}\\ & \begin{array}{l}\\ \text{a}.-1& & \text{d}.{\displaystyle \frac{1}{3}\sqrt{6}}\\ \\ \text{b}.{\displaystyle \frac{1}{2}\sqrt{2}}& \text{c}.1& \text{e}.{\displaystyle \frac{1}{2}\sqrt{6}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {\displaystyle \frac{\cos {75}^{\circ }-\cos {15}^{\circ }}{\sin {75}^{\circ }-\sin {15}^{\circ }}}\\ & ={\displaystyle \frac{-2\sin {\displaystyle \frac{1}{2}({75}^{\circ }+{15}^{\circ })\sin \frac{1}{2}({75}^{\circ }-{15}^{\circ })}}{2\cos {\displaystyle \frac{1}{2}({75}^{\circ }+{15}^{\circ })\sin \frac{1}{2}({75}^{\circ }-{15}^{\circ })}}}\\ & =-{\displaystyle \frac{2\sin {45}^{\circ }\times \sin {30}^{\circ }}{2\cos {45}^{\circ }\times \sin {30}^{\circ }}}\\ & =-\tan {45}^{\circ }\\ & =-1\end{array}\end{array}$

$\begin{array}{l}\\ 84.& \text{Bentuk}{\displaystyle \frac{\cos 3x-\sin 6x-\cos 9x}{\sin 9x-\cos 6x-\sin 3x}}\\ & \text{senilai dengan}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -\tan 6x}& & & \text{d}.& 6\cot x\\ \text{b}.& {\displaystyle -\cot 6x}& & & \text{e}.& {\displaystyle \tan 6x}\\ \text{c}.& {\displaystyle 6\tan x}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \frac{\cos 3x-\sin 6x-\cos 9x}{\sin 9x-\cos 6x-\sin 3x}}\\ & ={\displaystyle \frac{\cos 3x-\cos 9x-\sin 6x}{\sin 9x-\sin 3x-\cos 6x}}\\ & ={\displaystyle \frac{-2\sin 6x\sin (-3x)-\sin 6x}{2\cos 6x\sin 3x-\cos 6x}}\\ & ={\displaystyle \frac{2\sin 6x\sin 3x-\sin 6x}{2\cos 6x\sin 3x-\cos 6x}}\\ & ={\displaystyle \frac{\sin 6x(2\sin 3x-1)}{\cos 6x(2\sin 3x-1)}}\\ & =\tan 6x\end{array}\end{array}$.

$\begin{array}{l}\\ 85.& \text{Nilai dari}\\ & {\displaystyle \frac{\sin x+\sin 3x+\sin 5x+\sin 7x}{\cos x+\cos 3x+\cos 5x+\cos 7x}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \tan 2x}& & \text{d}.& {\displaystyle \tan 8x}\\ \text{b}.& {\displaystyle \tan 4x}& & \text{e}.& {\displaystyle \tan 16x}\\ \text{c}.& {\displaystyle {\displaystyle \tan 6x}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \frac{\sin x+\sin 3x+\sin 5x+\sin 7x}{\cos x+\cos 3x+\cos 5x+\cos 7x}}\\ & ={\displaystyle \frac{\sin 7x+\sin x+\sin 5x+\sin 3x}{\cos 7x+\cos x+\cos 5x+\cos 3x}}\\ & ={\displaystyle \frac{2\sin 4x\cos 3x+2\sin 4x\cos x}{2\cos 4x\cos 3x+2\cos 4x\cos x}}\\ & ={\displaystyle \frac{2\sin 4x(\cos 3x+\cos x)}{2\cos 4x(\cos 3x+\cos x)}}\\ & =\tan 4x\end{array}\end{array}$.

$\begin{array}{l}\\ 86.& \text{Bentuk sederhana dari}\\ & {\displaystyle \frac{\cos A+\sin A}{\cos A-\sin A}-\frac{\cos A-\sin A}{\cos A+\sin A}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \tan A}& & & \text{d}.& 2\cos 2A\\ \text{b}.& {\displaystyle 2\tan A}& & & \text{e}.& {\displaystyle 2\tan 2A}\\ \text{c}.& {\displaystyle 2\sin 2A}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \frac{\cos A+\sin A}{\cos A-\sin A}-\frac{\cos A-\sin A}{\cos A+\sin A}}\\ & ={\displaystyle \frac{(\cos A+\sin A{)}^2-(\cos A-\sin A{)}^2}{(\cos A-\sin A)(\cos A+\sin A)}}\\ & ={\displaystyle \frac{({\cos }^{2}A+2\cos A\sin A+{\sin }^{2}A)-({\cos }^{2}A-2\cos A\sin A+{\sin }^{2}A)}{{\cos }^{2}A-{\sin }^{2}A}}\\ & ={\displaystyle \frac{4\cos A\sin A}{{\cos }^{2}A-{\sin }^{2}A}}\\ & ={\displaystyle \frac{2\sin 2A}{\cos 2A}}\\ & =2\tan 2A\end{array}\end{array}$.

$\begin{array}{l}\\ 87.& \text{Bentuk sederhana dari}\\ & {\displaystyle \frac{\cos 2x-\cos 2y}{\sin 2x+\sin 2y}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -\sin (x-y)}& & & \text{d}.& \cos (x-y)\\ \text{b}.& {\displaystyle -\tan (x-y)}& & & \text{e}.& {\displaystyle \tan (x-y)}\\ \text{c}.& {\displaystyle \sin (x+y)}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \frac{\cos 2x-\cos 2y}{\sin 2x+\sin 2y}}\\ & ={\displaystyle \frac{-2\sin (x+y)\sin (x-y)}{2\sin (x+y)\cos (x-y)}}\\ & =-\tan (x-y)\end{array}\end{array}$.

$\begin{array}{l}\\ 88.& \text{Nilai dari}\cos {80}^{\circ }+\cos {40}^{\circ }-\cos {20}^{\circ }=....\\ & \begin{array}{l}\\ \text{a}.1& & \text{d}.-{\displaystyle \frac{1}{2}}\\ \\ \text{b}.-1& \text{c}.{\displaystyle \frac{1}{2}}& \text{e}.0\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \cos {80}^{\circ }+\cos {40}^{\circ }-\cos {20}^{\circ }\\ & =2\cos {\displaystyle \frac{1}{2}({80}^{\circ }+{40}^{\circ })\cos {\displaystyle \frac{1}{2}({80}^{\circ }-{40}^{\circ })-\cos {20}^{\circ }}}\\ & =2\cos {60}^{\circ }\cos {20}^{\circ }-\cos {20}^{\circ }\\ & =2.{\displaystyle \frac{1}{2}.\cos {20}^{\circ }-\cos {20}^{\circ }}\\ & =\cos {20}^{\circ }-\cos {20}^{\circ }\\ & =0\end{array}\end{array}$.

$\begin{array}{l}\\ 89.& \text{Nilai dari}\\ & 8\cos 82,5^{\circ }\sin 37,5^{\circ }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 4(\sqrt{3}+\sqrt{2})}& & & \text{d}.& 2(\sqrt{3}-\sqrt{2})\\ \text{b}.& {\displaystyle 4(\sqrt{3}-\sqrt{2})}& & & \text{e}.& {\displaystyle \sqrt{3}-\sqrt{2}}\\ \text{c}.& {\displaystyle 2(\sqrt{3}+\sqrt{2})}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& 8\cos 82,5^{\circ }\sin 37,5^{\circ }\\ & =4\times 2\cos 82,5^{\circ }\sin 37,5^{\circ }\\ & =4\times (\sin (82,5^{\circ }+37,5^{\circ })-\sin (82,5^{\circ }-37,5^{\circ }))\\ & =4\times (\sin {120}^{\circ }-\sin {45}^{\circ })\\ & =4\times (\sin ({180}^{\circ }-{60}^{\circ })-\sin {45}^{\circ })\\ & =4\times (\sin {60}^{\circ }-\sin {45}^{\circ })\\ & =4\times \left({\displaystyle \frac{1}{2}\sqrt{3}-{\displaystyle \frac{1}{2}\sqrt{2}}}\right)\\ & =2(\sqrt{3}-\sqrt{2})\end{array}\end{array}$.

$\begin{array}{l}\\ 90.& \text{Bentuk lain dari}\\ & -2\cos 5A.\cos 7A\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -\cos 6A-\cos A}& & & \\ \text{b}.& {\displaystyle -\cos 6A+\cos A}& & & \\ \text{c}.& {\displaystyle \cos 12A-\cos 2A}\\ \text{d}.& -\cos 12A+\cos 2A\\ \text{e}.& {\displaystyle -\cos 12A-\cos 2A}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& -2\cos 5A.\cos 7A\\ & =-(2\cos 5A.\cos 7A)\\ & =-(\cos 12A+\cos (-2A))\\ & =-(\cos 12A+\cos 2A)\\ & =-\cos 12A-\cos 2A\end{array}\end{array}$.