Bank Soal: Identitas Trigonometri dan Pembahasan - CATATAN MATEMATIKA

Bank Soal: Identitas Trigonometri dan Pembahasan

Soal Identitas Trigonometri dan Pembahasan
Berikut ini adalah Soal Identitas Trigonometri dan Pembahasan, yaitu salah satu sub materi TRIGONOMETRI pada mata pelajaran Matematika Wajib Kelas 10. Silahkan dipelajari dan jangan lupa share/bagikan ke media sosial kalian, agar manfaat postingan ini dapat dirasakan oleh siswa/i yang lain. Terima kasih.
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Soal No. 1
Hasil dari $\frac{\sin x+\tan x}{\cot x+\csc x}$ = ….
(A) $\sin x\cos x$
(B) $\sin x\cot x$
(C) $\sin x\csc x$
(D) $\sin x\sec x$
(E) $\sin x\tan x$
$\begin{align}\frac{\sin x+\tan x}{\cot x+\csc x} &=\frac{\sin x+\frac{\sin x}{\cos x}}{\frac{\cos x}{\sin x}+\frac{1}{\sin x}} \\ &=\frac{\frac{\sin x.\cos x+\sin x}{\cos x}}{\frac{\cos x+1}{\sin x}} \\ &=\frac{\sin x(\cos x+1)}{\cos x}\times \frac{\sin x}{\cos x+1} \\ \frac{\sin x+\tan x}{\cot x+\csc x} &=\sin x.\tan x \end{align}$
Jawaban: E

Soal No. 2
Hasil dari $\frac{\tan \alpha .\sin \alpha }{\cos \alpha .\sec \alpha }$ = ….
(A) $\frac{{{\sin }^{2}}\alpha }{\cos \alpha }$
(B) $\frac{{{\sin }^{2}}\alpha }{{{\cos }^{3}}\alpha }$
(C) $\frac{{{\sin }^{2}}\alpha +\sin \alpha }{{{\cos }^{2}}\alpha }$
(D) ${{\sin }^{3}}\alpha $
(E) ${{\cos }^{3}}\alpha $
$\begin{align}\frac{\tan \alpha .\sin \alpha }{\cos \alpha .\sec \alpha } &=\frac{\frac{\sin \alpha }{\cos \alpha }.\sin \alpha }{\cos \alpha .\frac{1}{\cos \alpha }} \\ \frac{\tan \alpha .\sin \alpha }{\cos \alpha .\sec \alpha } &=\frac{{{\sin }^{2}}\alpha }{\cos \alpha } \end{align}$
Jawaban: A

Soal No. 3
$\frac{1-\cos x}{\sin x}$ = …
(A) $\frac{-\sin x}{1+\cos x}$
(B) $\frac{-\cos x}{1-\sin x}$
(C) $\frac{\sin x}{1-\cos x}$
(D) $\frac{\cos x}{1+\sin x}$
(E) $\frac{\sin x}{1+\cos x}$
$\begin{align}\frac{1-\cos x}{\sin x} &=\frac{1-\cos x}{\sin x}\times \frac{1+\cos x}{1+\cos x} \\ &=\frac{1-{{\cos }^{2}}x}{\sin x(1+\cos x)} \\ &=\frac{{{\sin }^{2}}x}{\sin x(1+\cos x)} \\ \frac{1-\cos x}{\sin x} &=\frac{\sin x}{1+\cos x} \end{align}$
Jawaban: E

Soal No. 4
$\frac{\tan x+1}{\tan x-1}$ = ….
(A) $\frac{1-\sin x.\cos x}{1-\sin x.\cos x}$
(B) $\frac{1+\sin x.\cos x}{1-\sin x.\cos x}$
(C) $\frac{1-\sin x.\cos x}{1+\sin x.\cos x}$
(D) $\frac{\csc x+\sec x}{\csc x-\sec x}$
(E) $\frac{\csc x+\sec x}{\sec x-\csc x}$
$\begin{align}\frac{\tan x+1}{\tan x-1} &=\frac{\left. \frac{\sin x}{\cos x}+1 \right\}\times \frac{1}{\sin x}}{\left. \frac{\sin x}{\cos x}-1 \right\}\times \frac{1}{\sin x}} \\ &=\frac{\frac{1}{\cos x}+\frac{1}{\sin x}}{\frac{1}{\cos x}-\frac{1}{\sin x}} \\ \frac{\tan x+1}{\tan x-1} &=\frac{\sec x+\csc x}{\sec x-\csc x} \end{align}$
Jawaban: E

Soal No. 5
Bentuk $(1-{{\cos }^{2}}A).{{\cot }^{2}}A$ dapat disederhanakan menjadi …
(A) $2{{\sin }^{2}}A-1$
(B) ${{\sin }^{2}}A+{{\cos }^{2}}A$
(C) $1-{{\cos }^{2}}A$
(D) $1-{{\sin }^{2}}A$
(E) ${{\cos }^{2}}A+2$
$\begin{align}(1-{{\cos }^{2}}A).{{\cot }^{2}}A &={{\sin }^{2}}A.\frac{{{\cos }^{2}}A}{{{\sin }^{2}}A} \\ &={{\cos }^{2}}A \\ (1-{{\cos }^{2}}A).{{\cot }^{2}}A &=1-{{\sin }^{2}}A \end{align}$
Jawaban: D
Soal No. 6
Bentuk sederhana dari $\tan A+\frac{\cos A}{1+\sin A}$ adalah …
(A) $\sec A$
(B) $\cos A$
(C) $\cot A$
(D) $\tan A$
(E) $\csc A$
$\begin{align}\tan A+\frac{\cos A}{1+\sin A} &=\frac{\sin A}{\cos A}+\frac{\cos A}{1+\sin A} \\ &=\frac{\sin A(1+\sin A)+\cos A.\cos A}{\cos A(1+\sin A)} \\ &=\frac{\sin A+{{\sin }^{2}}A+{{\cos }^{2}}A}{\cos A(1+\sin A} \\ &=\frac{(1+\sin A)}{\cos A(1+\sin A)} \\ &=\frac{1}{\cos A} \\ \tan A+\frac{\cos A}{1+\sin A} &=\sec A \end{align}$
Jawaban: A

Soal No. 7
Untuk setiap sudut $\beta $, maka $(1-{{\sin }^{2}}\beta )(1+{{\tan }^{2}}\beta )$ dapat disederhanakan menjadi …
(A) $1+{{\sin }^{2}}\beta $
(B) ${{\sin }^{2}}\beta -{{\cos }^{2}}\beta $
(C) $1+{{\cos }^{2}}\beta $
(D) 1
(E) ${{\sin }^{2}}\beta $
$\begin{align}(1-{{\sin }^{2}}\beta )(1+{{\tan }^{2}}\beta ) &=co{{s}^{2}}\beta .{{\sec }^{2}}\beta \\ &={{\cos }^{2}}\beta .\frac{1}{{{\cos }^{2}}\beta } \\ (1-{{\sin }^{2}}\beta )(1+{{\tan }^{2}}\beta ) &=1 \end{align}$
Jawaban: D

Soal No. 8
Diketahui $p-q=\cos x$ dan $\sqrt{2pq}=\sin x$ maka ${{p}^{2}}+{{q}^{2}}$ = …
(A) $\sin x+\cos x$
(B) ${{\sin }^{2}}x+{{\cos }^{2}}x$
(C) ${{\sin }^{2}}x-{{\cos }^{2}}x$
(D) ${{\sin }^{2}}x+{{\sin }^{2}}x$
(E) ${{\cos }^{2}}x-{{\sin }^{2}}x$
$\sqrt{2pq}=\sin x\Leftrightarrow 2pq={{\sin }^{2}}x$
$\begin{align}p-q &=\cos x \\ {{(p-q)}^{2}} &={{\cos }^{2}}x \\ {{p}^{2}}+{{q}^{2}}-2pq &={{\cos }^{2}}x \\ {{p}^{2}}+{{q}^{2}} &=2pq+{{\cos }^{2}}x \\ {{p}^{2}}+{{q}^{2}} &={{\sin }^{2}}x+{{\cos }^{2}}x \end{align}$
Jawaban: B

Soal No. 9
Bentuk sederhana dari $\frac{1+\sec x}{\tan x+\sin x}$ adalah ….
(A) $\sec x$
(B) $\sin x$
(C) $\tan x$
(D) $\csc x$
(E) $\cos x$
$\begin{align}\frac{1+\sec x}{\tan x+\sin x} &=\frac{1+\frac{1}{\cos x}}{\frac{\sin x}{\cos x}+\sin x} \\ &=\frac{\frac{\cos x+1}{\cos x}}{\frac{\sin x+\sin x.\cos x}{\cos x}} \\ &=\frac{1+\cos x}{\sin x+\sin x.\cos x} \\ &=\frac{(1+\cos x)}{\sin x(1+\cos x)} \\ &=\frac{1}{\sin x} \\ \frac{1+\sec x}{\tan x+\sin x} &=\csc x \end{align}$
Jawaban: D

Soal No. 10
Nilai dari $\frac{2\tan x}{1+{{\tan }^{2}}x}$ adalah ….
(A) $2\sin x\cos x$
(B) $\sin x\cos x$
(C) $1-2\sin x$
(D) $2\sin x$
(E) $2\cos x$
$\begin{align}\frac{2\tan x}{1+{{\tan }^{2}}x} &=\frac{2.\frac{\sin x}{\cos x}}{{{\sec }^{2}}x} \\ &=\frac{\frac{2\sin x}{\cos x}}{\frac{1}{{{\cos }^{2}}x}} \\ &=\frac{2\sin x}{\cos x}\times \frac{{{\cos }^{2}}x}{1} \\ \frac{2\tan x}{1+{{\tan }^{2}}x} &=2\sin x.\cos x \end{align}$
Jawaban: A
Soal No. 11
Bentuk sederhana dari $\frac{{{\sin }^{4}}x-{{\cos }^{4}}x}{\sin x-\cos x}$ = ….
(A) ${{\sin }^{3}}x-{{\cos }^{2}}x$
(B) ${{\sin }^{3}}x+{{\cos }^{3}}x$
(C) ${{\sin }^{2}}x-{{\cos }^{2}}x$
(D) $\sin x-\cos x$
(E) $\sin x+\cos x$
$\begin{align}\frac{{{\sin }^{4}}x-{{\cos }^{4}}x}{\sin x-\cos x} &=\frac{({{\sin }^{2}}x-{{\cos }^{2}}x)({{\sin }^{2}}x+{{\cos }^{2}}x)}{\sin x-\cos x} \\ &=\frac{(\sin x+\cos x)(\sin x-\cos x)(1)}{(\sin x-\cos x)} \\ \frac{{{\sin }^{4}}x-{{\cos }^{4}}x}{\sin x-\cos x} &=\sin x+\cos x \end{align}$
Jawaban: E

Soal No. 12
Bentuk yang senilai dengan $5{{\tan }^{2}}x+3$ adalah ….
(A) $\frac{5}{{{\sin }^{2}}x}-2$
(B) $\frac{5}{{{\cos }^{2}}x}-2$
(C) $\frac{5}{{{\sin }^{2}}x}+3$
(D) $\frac{3}{{{\sin }^{2}}x}+2$
(E) $\frac{2}{{{\cos }^{2}}x}+5$
$\begin{align}5{{\tan }^{2}}x+3 &=5(se{{c}^{2}}x-1)+3 \\ &=5{{\sec }^{2}}x-5+3 \\ 5{{\tan }^{2}}x+3 &=\frac{5}{{{\cos }^{2}}x}-2 \end{align}$
Jawaban: B

Soal No. 13
Bentuk $(1-{{\sin }^{2}}A).{{\tan }^{2}}A$ dapat disederhanakan menjadi …
(A) $2{{\sin }^{2}}A-1$
(B) ${{\sin }^{2}}A+{{\cos }^{2}}A$
(C) $1-{{\cos }^{2}}A$
(D) $1-{{\sin }^{2}}A$
(E) ${{\cos }^{2}}A+2$
$\begin{align}(1-{{\sin }^{2}}A).{{\tan }^{2}}A &={{\cos }^{2}}A.\frac{{{\sin }^{2}}A}{{{\cos }^{2}}A} \\ &={{\sin }^{2}}A \\ (1-{{\sin }^{2}}A).{{\tan }^{2}}A &=1-{{\cos }^{2}}A \end{align}$
Jawaban: C

Soal No. 14
${{\cos }^{2}}20{}^\circ +{{\cos }^{2}}40{}^\circ +{{\cos }^{2}}50{}^\circ +{{\cos }^{2}}70{}^\circ $ = …
(A) 2
(B) $\frac{3}{2}$
(C) 1
(D) $\frac{1}{2}$
(E) 0
${{\cos }^{2}}20{}^\circ +{{\cos }^{2}}40{}^\circ +{{\cos }^{2}}50{}^\circ +{{\cos }^{2}}70{}^\circ $
= ${{\cos }^{2}}20{}^\circ +{{\cos }^{2}}70{}^\circ +{{\cos }^{2}}40{}^\circ +{{\cos }^{2}}50{}^\circ $
= ${{\cos }^{2}}(90{}^\circ -70{}^\circ )+{{\cos }^{2}}70{}^\circ +{{\cos }^{2}}(90{}^\circ -50{}^\circ )+{{\cos }^{2}}50{}^\circ $
= ${{\sin }^{2}}70{}^\circ +{{\cos }^{2}}70{}^\circ +{{\sin }^{2}}50{}^\circ +{{\cos }^{2}}50{}^\circ $
= 1 + 1
= 2
Jawaban: A

Soal No. 15
Bentuk sederhana dari $\frac{1+\sin x}{\cos x}+\frac{\cos x}{1+\sin x}$ = ….
(A) $2\sec x$
(B) $2\cos x$
(C) $2\cot x$
(D) $2\tan x$
(E) $2\csc x$
$\begin{align}\frac{1+\sin x}{\cos x}+\frac{\cos x}{1+\sin x} &=\frac{{{(1+\sin x)}^{2}}+{{\cos }^{2}}x}{\cos x(1+\sin x)} \\ &=\frac{1+2\sin x+{{\sin }^{2}}x+{{\cos }^{2}}x}{\cos x(1+\sin x)} \\ &=\frac{1+2\sin x+1}{\cos x(1+\sin x)} \\ &=\frac{2+2\sin x}{\cos x(1+\sin x)} \\ &=\frac{2(1+\sin x)}{\cos x(1+\sin x)} \\ &=\frac{2}{\cos x} \\ \frac{1+\sin x}{\cos x}+\frac{\cos x}{1+\sin x} &=2\sec x \end{align}$
Jawaban: A
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