Soal Identitas Trigonometri dan Pembahasan

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Soal Identitas Trigonometri 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. $\sin x\tan x$

Penyelesaian: Lihat/Tutup

$\frac{\sin x+\tan x}{\cot x+\csc x}$
= $\frac{\sin x+\frac{\sin x}{\cos x}}{\frac{\cos x}{\sin x}+\frac{1}{\sin x}}$
= $\frac{\frac{\sin x.\cos x+\sin x}{\cos x}}{\frac{\cos x+1}{\sin x}}$
= $\frac{\sin x(\cos x+1)}{\cos x}\times \frac{\sin x}{(\cos x+1)}$
= $\sin x.\tan x$
Jawaban: E

Soal Identitas Trigonometri 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 $

Penyelesaian: Lihat/Tutup

$\begin{align}\frac{\tan \alpha .\sin \alpha }{\cos \alpha .\sec \alpha } &= \frac{\frac{\sin \alpha }{\cos \alpha }.\sin \alpha }{\cos \alpha .\frac{1}{\cos \alpha }} \\ &= \frac{{{\sin }^2}\alpha }{\cos \alpha } \end{align}$
Jawaban: A

Soal Identitas Trigonometri 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. $\frac{\sin x}{1+\cos x}$

Penyelesaian: Lihat/Tutup

$\begin{align}\frac{1-\cos x}{\sin x} &= \frac{1-\cos x}{\sin x}\times \frac{1+\cos x}{1+\cos x} \\ &= \frac{1-{{\cos }^2}x}{\sin x(1+\cos x)} \\ &= \frac{{{\sin }^2}x}{\sin x(1+\cos x)} \\ \frac{1-\cos x}{\sin x} &= \frac{\sin x}{1+\cos x} \end{align}$
Jawaban: E

Soal Identitas Trigonometri 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. $\frac{\csc x+\sec x}{\sec x-\csc x}$

Penyelesaian: Lihat/Tutup

$\begin{align}\frac{\tan x+1}{\tan x-1} &= \frac{\left. \frac{\sin x}{\cos x}+1 \right\}\times \frac{1}{\sin x}}{\left. \frac{\sin x}{\cos x}-1 \right\}\times \frac{1}{\sin x}} \\ &= \frac{\frac{1}{\cos x}+\frac{1}{\sin x}}{\frac{1}{\cos x}-\frac{1}{\sin x}} \\ \frac{\tan x+1}{\tan x-1} &= \frac{\sec x+\csc x}{\sec x-\csc x} \end{align}$
Jawaban: E

Soal Identitas Trigonometri 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. ${{\cos }^2}A+2$

Penyelesaian: Lihat/Tutup

$(1-{{\cos }^2}A).{{\cot }^2}A$
= ${{\sin }^2}A.\frac{{{\cos }^2}A}{{{\sin }^2}A}$
= ${{\cos }^2}A$
= $1-{{\sin }^2}A$
Jawaban: D

Soal Identitas Trigonometri 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. $\csc A$

Penyelesaian: Lihat/Tutup

$\tan A+\frac{\cos A}{1+\sin A}$
= $\frac{\sin A}{\cos A}+\frac{\cos A}{1+\sin A}$
= $\frac{\sin A(1+\sin A)+\cos A.\cos A}{\cos A(1+\sin A)}$
= $\frac{\sin A+{{\sin }^2}A+{{\cos }^2}A}{\cos A(1+\sin A}$
= $\frac{(1+\sin A)}{\cos A(1+\sin A)}$
= $\frac{1}{\cos A}$
= $\sec A$
Jawaban: A

Soal Identitas Trigonometri 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 $

Penyelesaian: Lihat/Tutup

$(1-{{\sin }^2}\beta )(1+{{\tan }^2}\beta )$
= $co{{s}^2}\beta .{{\sec }^2}\beta $
= ${{\cos }^2}\beta .\frac{1}{{{\cos }^2}\beta }$
= 1
Jawaban: D

Soal Identitas Trigonometri 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. ${{\cos }^2}x-{{\sin }^2}x$

Penyelesaian: Lihat/Tutup

$\sqrt{2pq}=\sin x\Leftrightarrow 2pq={{\sin }^2}x$
$\begin{align}p-q &= \cos x \\ (p-q)^2 &= {{\cos }^2}x \\ p^2+q^2-2pq &= {{\cos }^2}x \\ p^2+q^2 &= 2pq+{{\cos }^2}x \\ p^2+q^2 &= {{\sin }^2}x+{{\cos }^2}x \end{align}$
Jawaban: B

Soal Identitas Trigonometri 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. $\cos x$

Penyelesaian: Lihat/Tutup

$\begin{align}\frac{1+\sec x}{\tan x+\sin x} &= \frac{1+\frac{1}{\cos x}}{\frac{\sin x}{\cos x}+\sin x} \\ &= \frac{\frac{\cos x+1}{\cos x}}{\frac{\sin x+\sin x.\cos x}{\cos x}} \\ &= \frac{1+\cos x}{\sin x+\sin x.\cos x} \\ &= \frac{(1+\cos x)}{\sin x(1+\cos x)} \\ &= \frac{1}{\sin x} \\ \frac{1+\sec x}{\tan x+\sin x} &= \csc x \end{align}$
Jawaban: D

Soal Identitas Trigonometri 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. $2\cos x$

Penyelesaian: Lihat/Tutup

$\begin{align}\frac{2\tan x}{1+{{\tan }^2}x} &= \frac{2.\frac{\sin x}{\cos x}}{{{\sec }^2}x} \\ &= \frac{\frac{2\sin x}{\cos x}}{\frac{1}{{{\cos }^2}x}} \\ &= \frac{2\sin x}{\cos x}\times \frac{{{\cos }^2}x}{1} \\ \frac{2\tan x}{1+{{\tan }^2}x} &= 2\sin x.\cos x \end{align}$
Jawaban: A

Soal Identitas Trigonometri 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. $\sin x+\cos x$

Penyelesaian: Lihat/Tutup

$\frac{{{\sin }^{4}}x-{{\cos }^{4}}x}{\sin x-\cos x}$
= $\frac{({{\sin }^2}x-{{\cos }^2}x)({{\sin }^2}x+{{\cos }^2}x)}{\sin x-\cos x}$
= $\frac{(\sin x+\cos x)(\sin x-\cos x)(1)}{(\sin x-\cos x)}$
= $\sin x+\cos x$
Jawaban: E

Soal Identitas Trigonometri 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. $\frac{2}{{{\cos }^2}x}+5$

Penyelesaian: Lihat/Tutup

$\begin{align}5{{\tan }^2}x+3 &= 5(se{{c}^2}x-1)+3 \\ &= 5{{\sec }^2}x-5+3 \\ 5{{\tan }^2}x+3 &= \frac{5}{{{\cos }^2}x}-2 \end{align}$
Jawaban: B

Soal Identitas Trigonometri 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. ${{\cos }^2}A+2$

Penyelesaian: Lihat/Tutup

$\begin{align}(1-{{\sin }^2}A).{{\tan }^2}A &= {{\cos }^2}A.\frac{{{\sin }^2}A}{{{\cos }^2}A} \\ &= {{\sin }^2}A \\ (1-{{\sin }^2}A).{{\tan }^2}A &= 1-{{\cos }^2}A \end{align}$
Jawaban: C

Soal Identitas Trigonometri 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

Penyelesaian: Lihat/Tutup

${{\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 Identitas Trigonometri 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. $2\csc x$

Penyelesaian: Lihat/Tutup

$\frac{1+\sin x}{\cos x}+\frac{\cos x}{1+\sin x}$
= $\frac{{{(1+\sin x)}^2}+{{\cos }^2}x}{\cos x(1+\sin x)}$
= $\frac{1+2\sin x+{{\sin }^2}x+{{\cos }^2}x}{\cos x(1+\sin x)}$
= $\frac{1+2\sin x+1}{\cos x(1+\sin x)}$
= $\frac{2+2\sin x}{\cos x(1+\sin x)}$
= $\frac{2(1+\sin x)}{\cos x(1+\sin x)}$
= $\frac{2}{\cos x}$
= $2\sec x$
Jawaban: A

Soal Identitas Trigonometri No. 16

Hasil dari $\sqrt{\frac{1-\sin x}{1+\sin x}}$ = ….
A. $1-\sin x$
B. $\sec x+\tan x$
C. $\sec x-\tan x$
D. $\cos x-\sin x$
E. $\tan x-\sec x$

Penyelesaian: Lihat/Tutup

$\begin{align}\sqrt{\frac{1-\sin x}{1+\sin x}} &= \sqrt{\frac{1-\sin x}{1+\sin x}\times \frac{1-\sin x}{1-\sin x}} \\ &= \sqrt{\frac{{{(1-\sin x)}^2}}{1-{{\sin }^2}x}} \\ &= \sqrt{\frac{{{(1-\sin x)}^2}}{{{\cos }^2}x}} \\ &= \frac{1-\sin x}{\cos x} \\ &= \frac{1}{\cos x}-\frac{\sin x}{\cos x} \\ \sqrt{\frac{1-\sin x}{1+\sin x}} &= \sec x-\tan x \end{align}$
Jawaban: C

Soal Identitas Trigonometri No. 17

${{\sin }^{4}}x-{{\cos }^{4}}x-2{{\sin }^2}x$ = ….
A. $-1$
B. 0
C. 1
D. ${{\sin }^2}x-{{\cos }^2}x$
E. ${{({{\sin }^2}x-{{\cos }^2}x)}^2}$

Penyelesaian: Lihat/Tutup

Ingat: ${{a}^2}-{{b}^2}=(a+b)(a-b)$
${{\sin }^{4}}x-{{\cos }^{4}}x-2{{\sin }^2}x$
= ${{({{\sin }^2}x)}^2}-{{({{\cos }^2}x)}^2}-2{{\sin }^2}x$
= $({{\sin }^2}x+{{\cos }^2}x)({{\sin }^2}x-{{\cos }^2}x)-2{{\sin }^2}x$
= $1.({{\sin }^2}x-{{\cos }^2}x)-2{{\sin }^2}x$
= ${{\sin }^2}x-{{\cos }^2}x-2{{\sin }^2}x$
= $-{{\sin }^2}x-{{\cos }^2}x$
= $-({{\sin }^2}x+{{\cos }^2}x)$
= $-1$
Jawaban: A

Soal Identitas Trigonometri No. 18

Bentuk yang ekuivalen dengan $\frac{{{\sin }^{4}}x+{{\cos }^2}x}{{{\sin }^2}x}+{{\cos }^2}x$ adalah ….
A. ${{\sin }^2}x$
B. ${{\cos }^2}x$
C. ${{\tan }^2}x$
D. ${{\sec }^2}x$
E. ${{\csc }^2}x$

Penyelesaian: Lihat/Tutup

$\frac{{{\sin }^{4}}x+{{\cos }^2}x}{{{\sin }^2}x}+{{\cos }^2}x$
= $\frac{{{\sin }^{4}}x}{{{\sin }^2}x}+\frac{{{\cos }^2}x}{{{\sin }^2}x}+{{\cos }^2}x$
= ${{\sin }^2}x+{{\cot }^2}x+{{\cos }^2}x$
= ${{\sin }^2}x+{{\cos }^2}x+{{\cot }^2}x$
= $1+{{\cot }^2}x$
= ${{\csc }^2}x$
Jawaban: E

Soal Identitas Trigonometri No. 19

$2{{\sec }^2}x-{{\sec }^{4}}x$ senilai dengan ….
A. ${{\sec }^2}x-{{\sec }^{4}}x$
B. $1+{{\tan }^2}x$
C. $1-{{\tan }^2}x$
D. $1+{{\tan }^{4}}x$
E. $1-{{\tan }^{4}}x$

Penyelesaian: Lihat/Tutup

$2{{\sec }^2}x-{{\sec }^{4}}x$
= ${{\sec }^2}x+{{\sec }^2}x-{{\sec }^{4}}x$
= ${{\sec }^2}x+{{\sec }^2}x(1-{{\sec }^2}x)$
= ${{\sec }^2}x+{{\sec }^2}x.(-{{\tan }^2}x)$
= ${{\sec }^2}x(1-{{\tan }^2}x)$
= $(1+{{\tan }^2}x)(1-{{\tan }^2}x)$
= $1-{{\tan }^{4}}x$
Jawaban: E

Soal Identitas Trigonometri No. 20

$\tan x\sin x+\cos x$ senilai dengan ….
A. $\cos x$
B. $\tan x$
C. $\sin x$
D. $\sec x$
E. $\csc x$

Penyelesaian: Lihat/Tutup

$\tan x\sin x+\cos x$
= $\frac{\sin x}{\cos x}.\sin x+\cos x$
= $\frac{{{\sin }^2}x}{\cos x}+\frac{{{\cos }^2}x}{\cos x}$
= $\frac{{{\sin }^2}x+{{\cos }^2}x}{\cos x}$
= $\frac{1}{\cos x}$
= $\sec x$
Jawaban: D

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