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Soal Trigonometri Sudut Ganda dan Sudut Pertengahan dan Pembahasan

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Hallo...! Pengunjung setia Catatan Matematika, kali ini Bang RP (Reikson Panjaitan, S.Pd) berbagi Kumpulan Soal Trigonometri Sudut Rangkap dan Sudut Pertengahan berserta pembahasannya. Ayo... manfaatkan website Catatan Matematika ini untuk belajar matematika secara online.
Tata Cara Belajar:
Cobalah mengerjakan soal-soal yang tersedia secara mandiri. Setelah itu cek jawaban kamu dengan pembahasan yang telah disediakan, dengan cara:
klik "LIHAT/TUTUP:".
Soal No. 1
Jika $1-\cot \alpha =-\frac{1}{3}$ maka nilai $\sin 2\alpha +\cos 2\alpha $ = ….
A. $\frac{17}{25}$
B. 1
C. $\frac{6}{5}$
D. $\frac{31}{25}$
E. $\frac{7}{5}$
Penyelesaian: Lihat/Tutup $\begin{align}1-\cot \alpha &= -\frac{1}{3} \\ 1+\frac{1}{3} &= \cot \alpha \\ \cot \alpha =\frac{4}{3} = \frac{sa}{de} \end{align}$
$\begin{align}mi &= \sqrt{de^2+sa^2} \\ &= \sqrt{3^2+4^2} \\ mi &= 5 \end{align}$
$\sin \alpha =\frac{de}{mi}\Leftrightarrow \sin \alpha =\frac{3}{5}$
$\cos \alpha =\frac{sa}{mi}\Leftrightarrow \cos \alpha =\frac{4}{5}$
maka:
$\sin 2\alpha +\cos 2\alpha $
= $2\sin \alpha \cos \alpha +2{{\cos }^{2}}\alpha -1$
= $2.\frac{3}{5}.\frac{4}{5}+2{\left( \frac{4}{5} \right)^2}-1$
= $\frac{24}{25}+\frac{32}{25}-\frac{25}{25}$
= $\frac{31}{25}$
Jawaban: D

Soal No. 2
Jika $\sin A=\frac{4}{5}$ untuk A sudut tumpul, maka $\cos 2A$ adalah ….
A. $\frac{12}{25}$
B. $\frac{7}{25}$
C. $\frac{3}{25}$
D. $-\frac{7}{25}$
E. $-\frac{3}{25}$
Penyelesaian: Lihat/Tutup $\sin A=\frac{4}{5}$ maka:
$\begin{align}\cos 2A &= 1-2{\sin }^2A \\ &= 1-2{\left( \frac{4}{5} \right)^2} \\ &= 1-\frac{32}{25} \\ \cos 2A &= -\frac{7}{25} \end{align}$
Jawaban: D

Soal No. 3
Bentuk $\frac{\sin 4p}{\sin 2p}$ identik dengan ….
A. $1-2{\cos }^2p$
B. $2{\sin }^2p-1$
C. $1-2{\sin }^2p$
D. $2(1-2{\sin }^2p)$
E. $2{\cos }^2p-1$
Penyelesaian: Lihat/Tutup $\begin{align}\frac{\sin 4p}{\sin 2p} &= \frac{\sin 2(2p)}{\sin 2p} \\ &= \frac{2\sin 2p\cos 2p}{\sin 2p} \\ &= 2\cos 2p \\ \frac{\sin 4p}{\sin 2p} &= 2(1-2{\sin }^2p) \end{align}$
Jawaban: D

Soal No. 4
Diketahui $\cos 2A=\frac{1}{3}$ dengan A adalah sudut lancip maka nilai $\tan A$ = ….
A. $\frac{1}{3}\sqrt{3}$
B. $\frac{1}{2}\sqrt{2}$
C. $\frac{1}{3}\sqrt{6}$
D. $\frac{2}{3}\sqrt{6}$
E. $\frac{2}{5}\sqrt{5}$
Penyelesaian: Lihat/Tutup Cara 1. Menggunakan rumus trigonometri sudut ganda.
$\begin{align}\cos 2A &= \frac{1}{3} \\ 2{\cos }^2A-1 &= \frac{1}{3} \\ 2{\cos }^2A &= \frac{1}{3}+1 \\ 2{\cos }^2A &= \frac{4}{3} \\ {\cos }^2A &= \frac{2}{3} \\ \cos A &= \frac{\sqrt{2}}{\sqrt{3}}= \frac{sa}{mi} \end{align}$
$\begin{align}de &= \sqrt{mi^2-sa^2} \\ &= \sqrt{(\sqrt{3})^2-(\sqrt{2})^2} \\ de &= 1 \end{align}$
$\tan A=\frac{de}{sa}\Leftrightarrow \tan A=\frac{1}{\sqrt{2}}=\frac{1}{2}\sqrt{2}$

Cara 2. Menggunakan rumus trigonometri pertengahan sudut: $\tan \frac{1}{2}\alpha =\pm \sqrt{\frac{1-\cos \alpha }{1+\cos \alpha }}$ maka:
$\begin{align}\tan A &= \tan \frac{1}{2}(2A) \\ &= \sqrt{\frac{1-\cos 2A}{1+\cos 2A}} \\ &= \sqrt{\frac{1-\frac{1}{3}}{1+\frac{1}{3}}} \\ &= \sqrt{\frac{\frac{2}{3}}{\frac{4}{3}}} \\ &= \sqrt{\frac{2}{3}\times \frac{3}{4}} \\ &= \sqrt{\frac{2}{4}} \\ \tan A &= \frac{\sqrt{2}}{2} \end{align}$
Jawaban: B

Soal No. 5
Nilai $\cos \left( 22\frac{1}{2} \right)^\circ $ = ….
A. $\sqrt{2+\sqrt{2}}$
B. $\sqrt{2-\sqrt{2}}$
C. $\frac{1}{2}\sqrt{2+\sqrt{2}}$
D. $\frac{1}{2}\sqrt{2-\sqrt{2}}$
E. $2\sqrt{2+\sqrt{2}}$
Penyelesaian: Lihat/Tutup $\begin{align}\cos \left( 22\frac{1}{2} \right)^\circ &= \cos \left( \frac{45}{2} \right)^\circ \\ &= \cos \frac{1}{2}.45^\circ \\ &= \sqrt{\frac{1+\cos 45^\circ }{2}} \\ &= \sqrt{\frac{1+\frac{\sqrt{2}}{2}}{2}} \\ &= \sqrt{\frac{\frac{2+\sqrt{2}}{2}}{2}} \\ &= \sqrt{\frac{2+\sqrt{2}}{4}} \\ \cos \left( 22\frac{1}{2} \right)^\circ &= \frac{1}{2}\sqrt{2+\sqrt{2}} \end{align}$
Jawaban: C

Soal No. 6
Nilai $\sin 112,5^\circ $ = ….
A. $\sqrt{2+\sqrt{2}}$
B. $\sqrt{2-\sqrt{2}}$
C. $\frac{1}{2}\sqrt{2+\sqrt{2}}$
D. $\frac{1}{2}\sqrt{2-\sqrt{2}}$
E. $2\sqrt{2+\sqrt{2}}$
Penyelesaian: Lihat/Tutup $\begin{align}\sin 112,5^\circ &= \sin \frac{1}{2}.225^\circ \\ &= \sqrt{\frac{1-\cos 225^\circ }{2}} \\ &= \sqrt{\frac{1-\left( -\frac{\sqrt{2}}{2} \right)}{2}} \\ &= \sqrt{\frac{1+\frac{\sqrt{2}}{2}}{2}} \\ &= \sqrt{\frac{2+\sqrt{2}}{4}} \\ \sin 112,5^\circ &= \frac{1}{2}\sqrt{2+\sqrt{2}} \end{align}$
Jawaban: C

Soal No. 7
Jika $\tan C=\frac{1}{2}$ dan C sudut pada kuadran III maka nilai $\cos 2C$ = ….
A. $\frac{3}{5}$
B. $\frac{4}{5}$
C. $\frac{2}{5}$
D. $\frac{1}{3}\sqrt{5}$
E. $\frac{1}{2}\sqrt{5}$
Penyelesaian: Lihat/Tutup C sudut pada kuadran III maka $\sin C$ negatif.
$\tan C=\frac{1}{2}=\frac{de}{sa}$
$\begin{align}mi &= \sqrt{de^2+sa^2} \\ &= \sqrt{1^2+2^2} \\ mi &= \sqrt{5} \end{align}$
$\sin C=-\frac{de}{mi}\Leftrightarrow \sin C=-\frac{1}{\sqrt{5}}$
$\begin{align}\cos 2C &= 1-2{\sin }^2C \\ &= 1-2.{\left( -\frac{1}{\sqrt{5}} \right)^2} \\ &= 1-\frac{2}{5} \\ \cos 2C &= \frac{3}{5} \end{align}$
Jawaban: A

Soal No. 8
Diketahui nilai $\tan x=\frac{2}{3}$ dan $0^\circ < x <90^\circ $ maka $\tan 2x$ = ….
A. $\frac{3}{2}$
B. $\frac{5}{12}$
C. $\frac{12}{5}$
D. $\frac{13}{5}$
E. $\frac{36}{5}$
Penyelesaian: Lihat/Tutup $\begin{align}\tan 2x &= \frac{2\tan x}{1-{\tan }^2x} \\ &= \frac{2.\frac{2}{3}}{1-{\left( \frac{2}{3} \right)^2}} \\ &= \frac{\frac{4}{3}}{\frac{5}{9}} \\ &= \frac{4}{3}\times \frac{9}{5} \\ \tan 2x &= \frac{12}{5} \end{align}$
Jawaban: C

Soal No. 9
Bentuk $\frac{2\tan x}{1+{\tan }^2x}$ ekuivalen dengan ….
A. $2\sin x$
B. $\sin 2x$
C. $2\cos x$
D. $\cos 2x$
E. $\tan 2x$
Penyelesaian: Lihat/Tutup $\begin{align}\frac{2\tan x}{1+{\tan }^2x} &= \frac{2\tan x}{{{\sec }^2}x} \\ &= 2\tan x{\cos }^2x \\ &= 2\frac{\sin x}{\cos x}.{\cos }^2x \\ &= 2\sin x\cos x \\ \frac{2\tan x}{1+{\tan }^2x} &= \sin 2x \end{align}$
Jawaban: B

Soal No. 10
Jika $\tan x=a$, maka $\sin 2x$ sama dengan ….
A. $\frac{2a}{1+a^2}$
B. $\frac{1+a^2}{2a}$
C. $\frac{1-a^2}{1+a^2}$
D. $\frac{1+a^2}{1-a^2}$
E. $\frac{a}{a+a^2}$
Penyelesaian: Lihat/Tutup $\tan x=a\Leftrightarrow \tan x=\frac{a}{1}=\frac{de}{sa}$
$\begin{align}mi &= \sqrt{de^2+sa^2} \\ &= \sqrt{a^2+1^2} \\ mi &= \sqrt{a^2+1} \end{align}$
$\sin x=\frac{de}{mi}\Leftrightarrow \sin x=\frac{a}{\sqrt{a^2+1}}$
$\cos x=\frac{sa}{mi}\Leftrightarrow \cos x=\frac{1}{\sqrt{a^2+1}}$
$\begin{align}\sin 2x &= 2\sin x\cos x \\ &= 2.\frac{a}{\sqrt{a^2+1}}.\frac{1}{\sqrt{a^2+1}} \\ \sin 2x &= \frac{2a}{a^2+1} \end{align}$
Jawaban: A

Soal No. 11
Nilai dari $12\sin \left( 22\frac{1}{2} \right)^\circ .\cos \left( 22\frac{1}{2} \right)^\circ $ = ….
A. $3\sqrt{2}$
B. $4\sqrt{2}$
C. $3\sqrt{3}$
D. $2\sqrt{3}$
E. $2\sqrt{2}$
Penyelesaian: Lihat/Tutup $12\sin \left( 22\frac{1}{2} \right)^\circ .\cos \left( 22\frac{1}{2} \right)^\circ $
= $6\times 2\sin \left( 22\frac{1}{2} \right)^\circ .\cos \left( 22\frac{1}{2} \right)^\circ $
= $6\sin 2\times \left( 22\frac{1}{2} \right)^\circ $
= $6\sin 45^\circ $
= $6\times \frac{1}{2}\sqrt{2}$
= $3\sqrt{2}$
Jawaban: A

Soal No. 12
Jika diketahui bahwa $\cos \frac{1}{2}\theta =\sqrt{\frac{x+1}{2x}}$ maka ${{x}^2}-\frac{1}{{{x}^2}}$ = ….
A. ${\tan }^2\theta +{\sin }^2\theta $
B. ${\tan }^2\theta -{\sin }^2\theta $
C. ${\sin }^2\theta -{\cos }^2\theta $
D. ${\cos }^2\frac{1}{2}\theta +{\tan }^2\frac{1}{2}\theta $
E. ${\sin }^2\frac{1}{2}\theta +{\tan }^2\frac{1}{2}\theta $
Penyelesaian: Lihat/Tutup $\begin{align}\cos \frac{1}{2}\theta &= \sqrt{\frac{x+1}{2x}} \\ \sqrt{\frac{1+\cos \theta }{2}} &= \sqrt{\frac{x+1}{2x}} \\ \frac{1+\cos \theta }{2} &= \frac{x+1}{2x} \\ 2x+2x\cos \theta &= 2x+2 \\ 2x\cos \theta &= 2 \\ x &= \frac{2}{2\cos \theta } \\ x &= \frac{1}{\cos \theta } \end{align}$
$\begin{align}x^2-\frac{1}{x^2} &= {\left( \frac{1}{\cos \theta } \right)^2}-\frac{1}{{\left( \frac{1}{\cos \theta } \right)^2}} \\ &= \frac{1}{{\cos }^2\theta }-{\cos }^2\theta \\ &= {{\sec }^2}\theta -{\cos }^2\theta \\ &= {\tan }^2\theta +1-{\cos }^2\theta \\ x^2-\frac{1}{{{x}^2}} &= {\tan }^2\theta +{\sin }^2\theta \end{align}$
Jawaban: A

Soal No. 13
Jika $\tan 2\alpha =4\sin \alpha \cos \alpha $ untuk $\frac{\pi }{2} ≪ a < \pi $ maka $\cos \alpha $ = ….
A. $\frac{1}{2}\sqrt{3}$
B. $\frac{1}{2}$
C. 0
D. $-\frac{1}{2}\sqrt{3}$
E. $-\frac{1}{2}$
Penyelesaian: Lihat/Tutup $\frac{\pi }{2}≪ a < \pi $ maka $\cos \alpha $ negatif.
$\begin{align}\tan 2\alpha &= 4\sin \alpha \cos \alpha \\ \tan 2\alpha &= 2.2\sin \alpha \cos \alpha \\ \frac{\sin 2\alpha }{\cos 2\alpha } &= 2\sin 2\alpha \\ \sin 2\alpha &= 2\sin 2\alpha \cos 2\alpha \\ 1 &= 2\cos 2\alpha \\ \cos 2\alpha &= \frac{1}{2} \\ 2{\cos }^2\alpha -1 &= \frac{1}{2} \\ 2{\cos }^2\alpha &= \frac{1}{2}+1 \\ 2{\cos }^2\alpha &= \frac{3}{2} \\ {\cos }^2\alpha &= \frac{3}{4} \\ \cos \alpha &= -\frac{\sqrt{3}}{2} \end{align}$
Jawaban: D

Soal No. 14
Diketahui sudut lancip A dengan $\cos 2A=\frac{1}{3}$ nilai $\sin A$ = ….
A. $\frac{1}{3}\sqrt{3}$
B. $\frac{1}{2}\sqrt{2}$
C. $\frac{1}{3}\sqrt{6}$
D. $\frac{2}{3}\sqrt{5}$
E. $\frac{2}{3}\sqrt{6}$
Penyelesaian: Lihat/Tutup A sudut lancip maka $\sin A$ positif.
Cara 1. Menggunakan rumus trigonometri sudut ganda
$\begin{align}\cos 2A &= \frac{1}{3} \\ 1-2{\sin }^2A &= \frac{1}{3} \\ -2{\sin }^2A &= \frac{1}{3}-1 \\ -2{\sin }^2A &= -\frac{2}{3} \\ {\sin }^2A &= \frac{1}{3} \\ \sin A &= \sqrt{\frac{1}{3}} \\ \sin A &= \frac{1}{\sqrt{3}} \\ \sin A &= \frac{1}{3}\sqrt{3} \end{align}$
Cara 2. Menggunakan rumus trigonometri sudut pertengahan.
$\begin{align}\sin A &= \sin \frac{1}{2}(2A) \\ &= \sqrt{\frac{1-\cos 2A}{2}} \\ &= \sqrt{\frac{1-\frac{1}{3}}{2}} \\ &= \sqrt{\frac{1}{3}} \\ &= \frac{1}{\sqrt{3}} \\ \sin A &= \frac{1}{3}\sqrt{3} \end{align}$
Jawaban: A

Soal No. 15
Diketahui ${\sin }^2x=\frac{3}{5}$ untuk $\frac{\pi }{2}< x <\pi $, nilai $\tan 2x$ = ….
A. $2\sqrt{6}$
B. $\frac{2}{5}\sqrt{6}$
C. $\frac{2}{5\sqrt{6}}$
D. $-\frac{2}{5}\sqrt{6}$
E. $-2\sqrt{6}$
Penyelesaian: Lihat/Tutup $\frac{\pi }{2}< x <\pi $ maka $\sin x$ positif dan $\tan x$ negatif.
$\begin{align}{\sin }^2x &= \frac{3}{5} \\ \sin x &= \sqrt{\frac{3}{5}} \\ \sin x &= \frac{\sqrt{3}}{\sqrt{5}}=\frac{de}{mi} \end{align}$
$\begin{align}sa &= \sqrt{mi^2-de^2} \\ &= \sqrt{{{(\sqrt{5})}^2}-(\sqrt{3})^2} \\ sa &= \sqrt{2} \end{align}$
$\tan x=-\frac{de}{sa}\Leftrightarrow \tan x=-\frac{\sqrt{3}}{\sqrt{2}}$
$\begin{align}\tan 2x &= \frac{2\tan x}{1-{\tan }^2x} \\ &= \frac{2\left( -\frac{\sqrt{3}}{\sqrt{2}} \right)}{1-{\left( -\frac{\sqrt{3}}{\sqrt{2}} \right)^2}} \\ &= \frac{-\frac{2\sqrt{3}}{\sqrt{2}}}{1-\frac{3}{2}} \\ &= \frac{-\frac{2\sqrt{3}}{\sqrt{2}}}{-\frac{1}{2}} \\ &= -\frac{2\sqrt{3}}{\sqrt{2}}\times \left( -\frac{2}{1} \right) \\ &= \frac{4\sqrt{3}}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}} \\ \tan 2x &= 2\sqrt{6} \end{align}$
Jawaban: A

Soal No. 16
Nilai dari $6-12{\sin }^2\frac{\pi }{12}$ = ….
A. $3\sqrt{2}$
B. $4\sqrt{2}$
C. $3\sqrt{3}$
D. $2\sqrt{3}$
E. $2\sqrt{2}$
Penyelesaian: Lihat/Tutup Ingat, $1-2{\sin }^2A=\cos 2A$ maka:
$\begin{align}& 6-12{\sin }^2\frac{\pi }{12} &= 6\left( 1-2{\sin }^2\frac{\pi }{12} \right) \\ &= 6\left( \cos 2.\frac{\pi }{12} \right) \\ &= 6\cos \frac{\pi }{6} \\ &= 6\times \frac{1}{2}\sqrt{3} \\ 6-12{\sin }^2\frac{\pi }{12} &= 3\sqrt{3} \end{align}$
Jawaban: C

Soal No. 17
Nilai dari $4-8{\cos }^2\frac{3\pi }{8}$ = ….
A. $3\sqrt{2}$
B. $4\sqrt{2}$
C. $3\sqrt{3}$
D. $2\sqrt{3}$
E. $2\sqrt{2}$
Penyelesaian: Lihat/Tutup $\begin{align}4-8{\cos }^2\frac{3\pi }{8} &= 4-8\left( 1-{\sin }^2\frac{3\pi }{8} \right) \\ &= -4+8{\sin }^2\frac{3\pi }{8} \\ &= -4\left( 1-2{\sin }^2\frac{3\pi }{8} \right) \\ &= -4\cos \left( 2.\frac{3\pi }{8} \right) \\ &= -4\cos \frac{3\pi }{4} \\ &= -4\cos 135^\circ \\ &= -4\times \left( -\frac{1}{2}\sqrt{2} \right) \\ 4-8{\cos }^2\frac{3\pi }{8} &= 2\sqrt{2} \end{align}$
Jawaban: E

Soal No. 18
Diketahui $\sin p=\frac{2}{\sqrt{5}}$ dan $0^\circ < p < 90^\circ $, nilai dari $\tan 2p$ = ….
A. $-2$
B. $-\frac{4}{3}$
C. $-\frac{4}{5}$
D. $\frac{4}{3}$
E. 2
Penyelesaian: Lihat/Tutup $0^\circ < p < 90^\circ $ maka $\tan p$ positif.
$\sin p=\frac{2}{\sqrt{5}}=\frac{de}{mi}$
$\begin{align}sa &= \sqrt{mi^2-de^2} \\ &= \sqrt{{{(\sqrt{5})}^2}-2^2} \\ sa &= 1 \end{align}$
$\tan p=\frac{de}{sa}\Leftrightarrow \tan p=\frac{2}{1}=2$
$\begin{align}\tan 2p &= \frac{2\tan p}{1-{\tan }^2p} \\ &= \frac{2.2}{1-2^2} \\ \tan 2p &= -\frac{4}{3} \end{align}$
Jawaban: B

Soal No. 19
Diketahui $\sin A=\frac{7}{25}$ dan sudut A lancip. Nilai dari $\sin 2A$ adalah ….
A. $\frac{14}{625}$
B. $\frac{24}{625}$
C. $\frac{48}{625}$
D. $\frac{168}{625}$
E. $\frac{336}{625}$
Penyelesaian: Lihat/Tutup $\sin A=\frac{7}{25}=\frac{de}{mi}$
$\begin{align}sa &= \sqrt{mi^2-de^2} \\ &= \sqrt{{{25}^2}-{{7}^2}} \\ sa &= 24 \end{align}$
$\cos A=\frac{sa}{mi}\Leftrightarrow \cos A=\frac{24}{25}$
$\begin{align}\sin 2A &= 2\sin A\cos A \\ &= 2.\frac{7}{25}.\frac{24}{25} \\ \sin 2A &= \frac{336}{625} \end{align}$
Jawaban: E

Soal No. 20
Nilai dari $\frac{2\tan 112,5^\circ }{1-{\tan }^2112,5^\circ }$ = ….
A. 1
B. $\frac{3}{2}$
C. $\frac{1}{2}$
D. $\sqrt{2}$
E. $2\sqrt{3}$
Penyelesaian: Lihat/Tutup Ingat: $\frac{2\tan x}{1-{\tan }^2x}=\tan 2x$
$\begin{align}\frac{2\tan 112,5^\circ }{1-{\tan }^2112,5^\circ } &= \tan (2\times 112,5^\circ ) \\ &= \tan 225^\circ \\ &= 1 \end{align}$
Jawaban: A

Soal No. 21
Ditentukan $\sin A=\frac{7}{25}$ maka $\cos 2A$ = ….
A. $\frac{576}{675}$
B. $\frac{572}{675}$
C. $\frac{563}{625}$
D. $\frac{527}{625}$
E. $\frac{513}{576}$
Penyelesaian: Lihat/Tutup $\begin{align}\cos 2A &= 1-2{\sin }^2A \\ &= 1-2{\left( \frac{7}{25} \right)^2} \\ &= 1-\frac{98}{625} \\ \cos 2A &= \frac{527}{625} \end{align}$
Jawaban: D

Soal No. 22
Jika $\cos P=\frac{1}{3}$ dan P sudut lancip maka nilai $\cos 2P$ = ….
A. $\frac{4\sqrt{2}}{9}$
B. $\frac{7}{9}$
C. $-\frac{7}{9}$
D. $-\frac{4}{7}$
E. $\frac{4\sqrt{7}}{7}$
Penyelesaian: Lihat/Tutup $\begin{align}\cos 2P &= 2{\cos }^2P-1 \\ &= 2{\left( \frac{1}{3} \right)^2}-1 \\ &= \frac{2}{9}-1 \\ \cos 2P &= -\frac{7}{9} \end{align}$
Jawaban: C

Soal No. 23
Diketahui $\sin \alpha =\frac{1}{4}$ maka $\cos 2\alpha $ = ….
A. $\frac{1}{16}$
B. $\frac{7}{16}$
C. $\frac{1}{4}$
D. $\frac{7}{8}$
E. 1
Penyelesaian: Lihat/Tutup $\begin{align}\cos 2\alpha &= 1-2{\sin }^2\alpha \\ &= 1-2{\left( \frac{1}{4} \right)^2} \\ &= 1-\frac{1}{8} \\ \cos 2\alpha &= \frac{7}{8} \end{align}$
Jawaban: D

Soal No. 24
Jika $\tan \alpha =\frac{1}{2}$ dan $\alpha $ sudut lancip maka nilai $\sin 2\alpha $ = ….
A. $\frac{2}{\sqrt{5}}$
B. $\frac{2}{5}$
C. $\frac{4}{5}$
D. $\frac{3}{5}$
E. $\frac{2}{5}\sqrt{5}$
Penyelesaian: Lihat/Tutup $\alpha $ sudut lancip maka $\sin \alpha $ positif dan $\cos \alpha $ positif.
$\tan \alpha =\frac{1}{2}=\frac{de}{sa}$
$\begin{align}mi &= \sqrt{de^2+sa^2} \\ &= \sqrt{1^2+2^2} \\ mi &= \sqrt{5} \end{align}$
$\sin \alpha =\frac{de}{mi}\Leftrightarrow \sin \alpha =\frac{1}{\sqrt{5}}$
$\cos \alpha =\frac{sa}{mi}\Leftrightarrow \cos \alpha =\frac{2}{\sqrt{5}}$
$\begin{align}\sin 2\alpha &= 2\sin \alpha \cos \alpha \\ &= 2.\frac{1}{\sqrt{5}}.\frac{2}{\sqrt{5}} \\ \sin 2\alpha &= \frac{4}{5} \end{align}$
Jawaban: C

Soal No. 25
Diketahui $\sin \alpha =\frac{1}{5}\sqrt{13}$, $\alpha $ sudut lancip. Nilai $\cos 2\alpha $ = ….
A. $-1$
B. $-\frac{1}{2}$
C. $-\frac{1}{5}$
D. $-\frac{1}{25}$
E. 1
Penyelesaian: Lihat/Tutup $\begin{align}\cos 2\alpha &= 1-2{\sin }^2\alpha \\ &= 1-2{\left( \frac{\sqrt{13}}{5} \right)^2} \\ &= 1-\frac{26}{25} \\ \cos 2\alpha &= -\frac{1}{25} \end{align}$
Jawaban: D

Soal No. 26
Himpunan penyelesaian persamaan $\sin 2x+2\cos x=0$ untuk $0\le x\le 2\pi $ adalah ….
A. $\{0,\pi \}$
B. $\left\{ \frac{\pi }{2},\pi \right\}$
C. $\left\{ \frac{3\pi }{2},\pi \right\}$
D. $\left\{ \frac{\pi }{2},\frac{3\pi }{2} \right\}$
E. $\left\{ 0,\frac{3\pi }{2} \right\}$
Penyelesaian: Lihat/Tutup Ingat: Rumus Trigonometri Sudut Ganda, $\sin 2\alpha =2\sin \alpha \cos \alpha $.
$\begin{align}\sin 2x+2\cos x&= 0 \\ 2\sin x\cos x+2\cos x &= 0 \\ \sin x\cos x+\cos x &= 0 \\ \cos x(\sin x+1) &= 0 \end{align}$
$\cos x=0\to x=90^\circ \,\text{atau}\,x=270^\circ $
$\sin x+1=0\Leftrightarrow \sin x=-1\to x=270^\circ $
HP = $\left\{ 90^\circ ,270^\circ \right\}$ atau
HP = $\left\{ \frac{90^\circ \pi }{180^\circ },\frac{270^\circ \pi }{180^\circ } \right\}$ = $\left\{ \frac{\pi }{2},\frac{3\pi }{2} \right\}$
Jawaban: D

Soal No. 27
Nilai $x$ yang memenuhi persamaan $2\sin 2x+4\cos x=0$ dan $0^\circ \le x\le 360^\circ $ adalah ….
A. $\{30^\circ ,60^\circ \}$
B. $\{60^\circ ,90^\circ \}$
C. $\{90^\circ ,270^\circ \}$
D. $\{150^\circ ,300^\circ \}$
E. $\{270^\circ ,360^\circ \}$
Penyelesaian: Lihat/Tutup Ingat: Rumus Trigonometri Sudut Ganda, $\sin 2\alpha =2\sin \alpha \cos \alpha $.
$\begin{align}2\sin 2x+4\cos x &= 0 \\ \sin 2x+2\cos x &= 0 \\ 2\sin x\cos x+2\cos x &= 0 \\ \sin x\cos x+\cos x &= 0 \\ \cos x(\sin x+1) &= 0 \end{align}$
$\cos x=0\to x=90^\circ \,\text{atau}\,x=270^\circ $
$\sin x+1=0\Leftrightarrow \sin x=-1\to x=270^\circ $
HP = $\left\{ 90^\circ ,270^\circ \right\}$
Jawaban: C

Soal No. 28
Nilai dari $\cos 72^\circ +\sin 72^\circ .\tan 36^\circ $ = ….
A. 3
B. $3\sqrt{2}$
C. $\frac{1}{2}\sqrt{3}$
D. $\frac{1}{2}$
E. 1
Penyelesaian: Lihat/Tutup $\cos 72^\circ +\sin 72^\circ .\tan 36^\circ $
= $\cos 72^\circ +\sin (2.36)^\circ .\tan 36^\circ $
= $\cos 72^\circ +2\sin 36^\circ \cos 36^\circ .\frac{\sin 36^\circ }{\cos 36^\circ }$
= $\cos (2.36)^\circ +2{\sin }^236^\circ $
= $1-2{\sin }^236^\circ +2{\sin }^236^\circ $
= 1
Jawaban: E

Soal No. 29
Bentuk $\frac{\sin 2A}{\sin A}-\frac{\cos 2A}{\cos A}$ = ….
A. $\sec A$
B. $\csc A$
C. ${{\sec }^2}A$
D. $2+\csc A$
E. $2\sec A$
Penyelesaian: Lihat/Tutup $\frac{\sin 2A}{\sin A}-\frac{\cos 2A}{\cos A}$
= $\frac{2\sin A\cos A}{\sin A}-\frac{2{\cos }^2A-1}{\cos A}$
= $2\cos A-2\cos A+\frac{1}{\cos A}$
= $\frac{1}{\cos A}$
= $\sec A$
Jawaban: A

Soal No. 30
Jika $\sin A=\frac{3}{5}$ dan A sudut tumpul maka nilai $\tan 2A$ = ….
A. $\frac{12}{7}$
B. $-\frac{24}{7}$
C. $-\frac{12}{7}$
D. $\frac{24}{7}$
E. $-\frac{15}{7}$
Penyelesaian: Lihat/Tutup A sudut tumpul maka $\tan A$ negatif.
$\sin A=\frac{3}{5}=\frac{de}{mi}$
$\begin{align}sa &= \sqrt{mi^2-de^2} \\ &= \sqrt{5^2-3^2} \\ sa &= 4 \end{align}$
$\tan A=-\frac{de}{sa}\Leftrightarrow \tan A=-\frac{3}{4}$
$\begin{align}\tan 2A &= \frac{2\tan A}{1-{\tan }^2A} \\ &= \frac{2\left( -\frac{3}{4} \right)}{1-{\left( -\frac{3}{4} \right)^2}} \\ &= \frac{-\frac{3}{2}}{1-\frac{9}{16}} \\ &= \frac{-\frac{3}{2}}{\frac{7}{16}} \\ &= -\frac{3}{2}\times \frac{16}{7} \\ \tan 2A &= -\frac{24}{7} \end{align}$
Jawaban: B

Soal No. 31
Bentuk sederhana dari $\frac{1+\sin 2A-\cos 2A}{1+\sin 2A+\cos 2A}$ adalah ….
A. $\tan A$
B. $2\tan A$
C. $\cot A$
D. $\cot 2A$
E. $\tan 2A$
Penyelesaian: Lihat/Tutup $\frac{1+\sin 2A-\cos 2A}{1+\sin 2A+\cos 2A}$
= $\frac{1+\sin 2A-(1-2{\sin }^2A)}{1+\sin 2A+(2{\cos }^2A-1)}$
= $\frac{\sin 2A+2{\sin }^2A}{\sin 2A+2{\cos }^2A}$
= $\frac{2\sin A\cos A+2{\sin }^2A}{2\sin A\cos A+2{\cos }^2A}$
= $\frac{2\sin A(\cos A+\sin A)}{2\cos A(\sin A+\cos A)}$
= $\frac{\sin A}{\cos A}$
= $\tan A$
Jawaban: A

Soal No. 32
Diketahui $\tan 3x=n$ maka nilai $\sin 12x$ adalah ….
A. $\frac{4n}{\sqrt{1+n^2}}$
B. $\frac{4(1-n^2)}{1+n^2}$
C. $\frac{4n}{1+n^2\sqrt{1+n^2}}$
D. $\frac{4n(1-n^2)}{1+n^2}$
E. $\frac{4n(1-n^2)}{{{(1+n^2)}^2}}$
Penyelesaian: Lihat/Tutup $\tan 3x=n\Leftrightarrow \tan 3x=\frac{n}{1}=\frac{de}{sa}$ maka:
$mi=\sqrt{de^2+sa^2}=\sqrt{n^2+1}$
$\sin 3x=\frac{de}{mi}\Leftrightarrow \sin 3x=\frac{n}{\sqrt{n^2+1}}$
$\cos 3x=\frac{sa}{mi}\Leftrightarrow \cos 3x=\frac{1}{\sqrt{n^2+1}}$
maka:
$\sin 12x$
= $\sin 2(6x)$
= $2\sin 6x\cos 6x$
= $2\sin 2(3x).\cos 2(3x)$
= $2.2\sin 3x\cos 3x.(2{\cos }^23x-1)$
= $4\sin 3x\cos 3x.(2{\cos }^23x-1)$
= $4.\frac{n}{\sqrt{n^2+1}}.\frac{1}{\sqrt{n^2+1}}\left( 2{\left( \frac{1}{\sqrt{n^2+1}} \right)^2}-1 \right)$
= $\frac{4n}{n^2+1}\left( \frac{2}{n^2+1}-1 \right)$
= $\frac{4n}{n^2+1}\left( \frac{2}{n^2+1}-\frac{n^2+1}{n^2+1} \right)$
= $\frac{4n}{n^2+1}.\frac{1-n^2}{n^2+1}$
= $\frac{4n(1-n^2)}{{{(n^2+1)}^2}}$
Jawaban: E

Soal No. 33
Diketahui $\tan A=\frac{3}{4}$ dan $180^\circ < A < 270^\circ $, maka nilai $\cos \frac{1}{2}A$ = ….
A. $-\frac{1}{2}\sqrt{2}$
B. $-\frac{3}{10}\sqrt{10}$
C. $-\frac{1}{10}\sqrt{10}$
D. $-\frac{1}{10}\sqrt{30}$
E. $-\frac{9}{10}\sqrt{10}$
Penyelesaian: Lihat/Tutup $180^\circ < A < 270^\circ $ maka $\cos A$ negatif.
$\frac{180^\circ }{2} < \frac{A}{2} < \frac{270^\circ }{2}\Leftrightarrow 90^\circ < \frac{1}{2}A < 135^\circ $ maka $\cos \frac{1}{2}A$ negatif.
$\tan A=\frac{3}{4}=\frac{de}{sa}$
$\begin{align}mi &= \sqrt{de^2+sa^2} \\ &= \sqrt{3^2+4^2} \\ mi &= 5 \end{align}$
$\cos A=-\frac{sa}{mi}\Leftrightarrow \cos A=-\frac{4}{5}$
$\begin{align}\cos \frac{1}{2}A &= -\sqrt{\frac{1-\cos A}{2}} \\ &= -\sqrt{\frac{1-\left( -\frac{4}{5} \right)}{2}} \\ &= -\sqrt{\frac{9}{10}} \\ &= -\frac{3}{\sqrt{10}} \\ \cos A &= -\frac{3}{10}\sqrt{10} \end{align}$
Jawaban: B

Soal No. 34
Nilai $\cos \frac{9\pi }{8}$ adalah ….
A. $-\frac{1}{2}\sqrt{4+\sqrt{2}}$
B. $-\frac{1}{2}\sqrt{2+\sqrt{2}}$
C. $-\frac{1}{2}\sqrt{1+\sqrt{2}}$
D. $-\frac{1}{2}\sqrt{1-\sqrt{2}}$
E. $-\frac{1}{2}\sqrt{2-\sqrt{2}}$
Penyelesaian: Lihat/Tutup
Ingat:
$\cos (180^\circ +\alpha )=-\cos \alpha $
$\cos \frac{1}{2}\alpha =\pm \sqrt{\frac{1+\cos \alpha }{2}}$
$\begin{align}\cos \left( \frac{9\pi }{8} \right) &= \cos \left( \frac{9\times 180^\circ }{8} \right) \\ &= \cos 202,5^\circ \\ &= \cos (180^\circ +22,5^\circ ) \\ &= -\cos 22,5^\circ \\ &= -\cos \left( \frac{1}{2}\times 45^\circ \right) \\ &= -\sqrt{\frac{1+\cos 45^\circ }{2}} \\ &= -\sqrt{\frac{1+\frac{1}{2}\sqrt{2}}{2}} \\ &= -\sqrt{\frac{\frac{2+\sqrt{2}}{2}}{2}} \\ &= -\sqrt{\frac{2+\sqrt{2}}{4}} \\ &= -\frac{\sqrt{2+\sqrt{2}}}{2} \\ \cos \left( \frac{9\pi }{8} \right) &= -\frac{1}{2}\sqrt{2+\sqrt{2}} \end{align}$
Jawaban: B

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