Soal dan jawaban Persiapan Semester gasal Kelas XI Matematika Peminatan Bagian Kelima

$\begin{array}{l}\\ 21.& \text{Nilai}\cos {\displaystyle \frac{5}{12}\pi -\cos {\displaystyle \frac{1}{12}\pi \text{adalah}....}}\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle \frac{1}{2}\sqrt{6}}& & & \text{d}.& {\displaystyle \frac{1}{2}\sqrt{2}}\\ \\ \text{b}.& -{\displaystyle \frac{1}{2}\sqrt{3}}& \text{c}.& -{\displaystyle \frac{1}{2}\sqrt{2}}& \text{e}.& {\displaystyle \frac{1}{2}\sqrt{6}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \cos {\displaystyle \frac{5}{12}\pi -\cos {\displaystyle \frac{1}{12}\pi }}\\ & =-2\sin \left({\displaystyle \frac{{\displaystyle \frac{5}{12}\pi +{\displaystyle \frac{1}{12}\pi }}}{2}}\right)\sin \left({\displaystyle \frac{{\displaystyle \frac{5}{12}\pi -{\displaystyle \frac{1}{12}\pi }}}{2}}\right)\\ & =-2\sin \left({\displaystyle \frac{{\displaystyle \frac{6}{12}\pi }}{2}}\right)\sin \left({\displaystyle \frac{{\displaystyle \frac{4}{12}\pi }}{2}}\right)\\ & =-2\sin \left({\displaystyle \frac{1}{4}\pi }\right)\sin \left({\displaystyle \frac{1}{6}\pi }\right)\\ & =-2\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)\left({\displaystyle \frac{1}{2}}\right)\\ & ={\displaystyle -\frac{1}{2}\sqrt{2}}\end{array}\end{array}$.

$\begin{array}{l}\\ 22.& \text{Bentuk}\sin (2x-{\displaystyle \frac{3}{2}\pi })-\sin (4x+{\displaystyle \frac{1}{2}\pi })\\ & \text{senilai dengan}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -2\sin 3x.\sin x}& & & \text{d}.& {\displaystyle 2\sin 3x.\sin x}\\ \text{b}.& {\displaystyle -2\cos 3x.\sin x}& & & \text{e}.& {\displaystyle 2\cos 3x.\sin x}\\ \text{c}.& {\displaystyle 2\sin 2(x-\pi )}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \sin (2x-{\displaystyle \frac{3}{2}\pi })-\sin (4x+{\displaystyle \frac{1}{2}\pi })\\ & =2\cos \left({\displaystyle \frac{2x-{\displaystyle \frac{3}{2}\pi +4x+{\displaystyle \frac{1}{2}\pi }}}{2}}\right)\\ & \times \sin \left({\displaystyle \frac{2x-{\displaystyle \frac{3}{2}\pi -(4x+{\displaystyle \frac{1}{2}\pi })}}{2}}\right)\\ & =2\cos (3x-{\displaystyle \frac{1}{2}\pi )\sin (-x-\pi )}\\ & =2\cos \left({\displaystyle \frac{1}{2}\pi -3x}\right)(-\sin (\pi +x))\\ & =2(\sin 3x)(-(-\sin x))\\ & =2\sin 3x.\sin x\end{array}\end{array}$.

$\begin{array}{l}\\ 23.& \text{Bentuk}{\displaystyle \frac{\cos 3x-\sin 6x-\cos 9x}{\sin 9x-\cos 6x-\sin 3x}}\\ & \text{senilai dengan}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -\tan 6x}& & & \text{d}.& 6\cot x\\ \text{b}.& {\displaystyle -\cot 6x}& & & \text{e}.& {\displaystyle \tan 6x}\\ \text{c}.& {\displaystyle 6\tan x}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \frac{\cos 3x-\sin 6x-\cos 9x}{\sin 9x-\cos 6x-\sin 3x}}\\ & ={\displaystyle \frac{\cos 3x-\cos 9x-\sin 6x}{\sin 9x-\sin 3x-\cos 6x}}\\ & ={\displaystyle \frac{-2\sin 6x\sin (-3x)-\sin 6x}{2\cos 6x\sin 3x-\cos 6x}}\\ & ={\displaystyle \frac{2\sin 6x\sin 3x-\sin 6x}{2\cos 6x\sin 3x-\cos 6x}}\\ & ={\displaystyle \frac{\sin 6x(2\sin 3x-1)}{\cos 6x(2\sin 3x-1)}}\\ & =\tan 6x\end{array}\end{array}$.

$\begin{array}{l}\\ 24.& \text{Nilai dari}\\ & {\displaystyle \frac{\sin x+\sin 3x+\sin 5x+\sin 7x}{\cos x+\cos 3x+\cos 5x+\cos 7x}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \tan 2x}& & \text{d}.& {\displaystyle \tan 8x}\\ \text{b}.& {\displaystyle \tan 4x}& & \text{e}.& {\displaystyle \tan 16x}\\ \text{c}.& {\displaystyle {\displaystyle \tan 6x}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \frac{\sin x+\sin 3x+\sin 5x+\sin 7x}{\cos x+\cos 3x+\cos 5x+\cos 7x}}\\ & ={\displaystyle \frac{\sin 7x+\sin x+\sin 5x+\sin 3x}{\cos 7x+\cos x+\cos 5x+\cos 3x}}\\ & ={\displaystyle \frac{2\sin 4x\cos 3x+2\sin 4x\cos x}{2\cos 4x\cos 3x+2\cos 4x\cos x}}\\ & ={\displaystyle \frac{2\sin 4x(\cos 3x+\cos x)}{2\cos 4x(\cos 3x+\cos x)}}\\ & =\tan 4x\end{array}\end{array}$.

$\begin{array}{l}\\ 25.& \text{Bentuk sederhana dari}\\ & {\displaystyle \frac{\cos A+\sin A}{\cos A-\sin A}-\frac{\cos A-\sin A}{\cos A+\sin A}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \tan A}& & & \text{d}.& 2\cos 2A\\ \text{b}.& {\displaystyle 2\tan A}& & & \text{e}.& {\displaystyle 2\tan 2A}\\ \text{c}.& {\displaystyle 2\sin 2A}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \frac{\cos A+\sin A}{\cos A-\sin A}-\frac{\cos A-\sin A}{\cos A+\sin A}}\\ & ={\displaystyle \frac{(\cos A+\sin A{)}^2-(\cos A-\sin A{)}^2}{(\cos A-\sin A)(\cos A+\sin A)}}\\ & ={\displaystyle \frac{({\cos }^{2}A+2\cos A\sin A+{\sin }^{2}A)-({\cos }^{2}A-2\cos A\sin A+{\sin }^{2}A)}{{\cos }^{2}A-{\sin }^{2}A}}\\ & ={\displaystyle \frac{4\cos A\sin A}{{\cos }^{2}A-{\sin }^{2}A}}\\ & ={\displaystyle \frac{2\sin 2A}{\cos 2A}}\\ & =2\tan 2A\end{array}\end{array}$.