Trigonometric Functions Test 22

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Trigonometric Functions Test 22

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Weekly Quiz Competition

Question 1
1 / -0

if $$\sqrt{2} cos \theta$$ = $$\sqrt { \dfrac { 2+\sqrt { 2+\sqrt { 2 } } }{ 2 } }$$ then the value of $$\theta$$ is

A
$$\dfrac{\pi}{16}$$

B
$$\dfrac{\pi}{32}$$

C
$$\dfrac{\pi}{64}$$

D
$$\dfrac{\pi}{128}$$

Solution

Given $$\sqrt{2}\cos \theta=\sqrt{\dfrac{2+\sqrt{2+\sqrt{2}}}{2}}$$

Squaring on both sides

$$2\cos ^2 \theta=1+\dfrac{\sqrt{2+\sqrt{2}}}{2}$$

$$\implies (2\cos^2 \theta-1)=\dfrac{\sqrt{2+\sqrt{2}}}{2}$$

$$\implies \cos 2\theta=\dfrac{\sqrt{2+\sqrt{2}}}{2}$$

Squaring on both sides

$$\implies \cos^2 2\theta=\dfrac{2+\sqrt{2}}{4}$$

$$\implies 2\cos^2 2\theta=1+\dfrac{1}{\sqrt{2}}$$

$$\implies \cos 4\theta=\dfrac{1}{\sqrt{2}}$$

$$\implies 4\theta=\dfrac{\pi}{4}$$

$$\implies \theta=\dfrac{\pi}{16}$$
Question 2
1 / -0

If $$tan\theta +sin\theta =m$$ and $$tan\theta -sin\theta =n$$ then $$m^2-n^2=$$

A
$$\sqrt {mn}$$

B
mn

C
2mn

D
$$4\sqrt {mn}$$

Solution

$$m^2-n^2$$
$$=(\tan \theta +\sin \theta)^2-(\tan \theta-\sin \theta)^2$$
$$=(\tan^2 \theta +\sin^2 \theta +2\tan \theta \sin \theta)-(\tan^2\theta +\sin^2 \theta -2\tan \theta \sin \theta)$$
$$=4\tan \theta \sin \theta$$

Now, $$4\sqrt{mn}=4\sqrt{\tan^2 \theta-\sin^2 \theta}$$
$$=4\sqrt{\dfrac{\sin^2 \theta}{\cos^2 \theta}-\sin^2 \theta}$$

$$=4\sin \theta \sqrt{\dfrac{1-\cos^2 \theta}{\cos ^2 \theta}}$$

$$=4\sin \theta \sqrt{\tan^2 \theta}=4\sin \theta \tan \theta$$

Thus, $$m^2-n^2=4\sqrt{mn}$$
Question 3
1 / -0

The most general solution of $$\tan\theta =-1, \cos\theta =\dfrac { 1 }{ \sqrt { 2 } } $$ is -

A
$$n\pi +\dfrac { 7\pi }{ 4 } ,n\epsilon I$$

B
$$n\pi +{ (-1) }^{ n }\dfrac { 7\pi }{ 4 } ,n\epsilon I$$

C
$$2n\pi +\dfrac { 7\pi }{ 4 } ,n\epsilon I$$

D
None of these

Solution
Question 4
1 / -0

$$cos (4 \pi + \theta)$$ = ...

A
$$sin \theta$$

B
$$cos \theta$$

C
$$- sin \theta$$

D
$$-cos \theta$$

Solution

As we know that
$$\cos(2 n\pi+\theta)=\cos \theta$$
So $$\cos(4\pi+\theta)=\cos \theta$$
Question 5
1 / -0

Consider the operator a = X + $$\dfrac{d}{dx}$$ acting on smooth function of x. The commutator [a, cos x] is

A
(A) - sin x

B
(B) cos x

C
(C) -cos x

D
0
Question 6
1 / -0

If $${ P }_{ n }=1+{ cos2 }^{ n-1 }\theta -{ sin2 }^{ n-1 }\theta $$ and $${ Q }_{ n }=1+{ tan2 }^{ n-1 }\theta $$ such that $$\overset { r=n }{ \underset { r=1 }{ \Pi } } { P }_{ r }{ Q }_{ r-1 }=1$$ (here $$\Pi $$ denotes product of terms) is true for

A
$$\theta =\frac { \pi }{ 33 } $$

B
$$\theta =\frac { \pi }{ 11 } $$

C
$$\theta =\frac { 2\pi }{ 15 } $$

D
$$\theta =\frac { \pi }{ 37 } $$
Question 7
1 / -0

The value of $$\dfrac{\sin^{2}2\theta}{\cos^{2}\theta}+\dfrac{\sin^{2}4\theta}{\cos^{2}2\theta}+\dfrac{\sin^{2}\theta}{\cos^{2}4\theta}$$ is $$\left(where \theta=\dfrac{\pi}{7}\right)$$

A
$$5$$

B
$$\dfrac{11}{2}$$

C
$$\dfrac{13}{2}$$

D
$$7$$
Question 8
1 / -0

If $$4\cos{\theta}-3\sec{\theta}=2\tan{\theta}$$, then $${\theta}$$ is equal to

A
$$n\pi+(-1)^{n}\dfrac{\pi}{10}$$

B
$$n\pi+(-1)^{n}\dfrac{\pi}{6}$$

C
$$n\pi+(-1)^{n}\dfrac{3\pi}{10}$$

D
$$n\pi$$

Solution

Given,

$$4\cos \theta -3\sec \theta =2\tan \theta $$

$$4\cos \theta -3\times \dfrac{1}{\cos \theta} =2\times \dfrac{\sin \theta}{\cos \theta} $$

multiply by $$\cos \theta $$

$$4\cos ^2\theta -3=2\sin \theta $$...........$$(i)$$

$$\because \left (\cos ^2\theta =1-\sin ^2\theta\right ) $$

$$4-4\sin ^2\theta -3-2\sin \theta =0$$............$$from (i)$$

$$4\sin ^2\theta +2\sin \theta -1=0$$

$$\therefore \sin \theta =\dfrac{-1\pm \sqrt{5}}{4}$$

$$\Rightarrow \theta =n\pi +(-1)^n\dfrac{\pi}{10}$$
Question 9
1 / -0

If $$cosx + cos y - cos(x+y) =$$ $$\dfrac{3}{2}$$, then

A
$$x + y = 0$$

B
$$x = 2y$$

C
$$x = y$$

D
$$2x = y$$

Solution
Question 10
1 / -0

In $$\Delta ABC, \dfrac{1}{2} a^2 \dfrac{sin B \, sin C}{sin A} = $$

A
$$\Delta $$

B
$$2 \Delta$$

C
$$3 \Delta $$

D
$$4 \Delta $$

Solution

In $$\triangle ABC$$
$$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\implies \dfrac{\sin B}{\sin A}=\dfrac{b}{a}$$

$$\therefore \dfrac{1}{2}a^2\dfrac{\sin B\sin C}{\sin A}=\dfrac{1}{2} a^2\times \dfrac{\sin B}{\sin A}\times \sin C=\dfrac{1}{2}a^2\times \dfrac{b}{a}\times \sin C=\dfrac{1}{2} a b\sin C=\Delta$$

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Re-Attempt Weekly Quiz Competition

Trigonometric Functions Test 22

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Trigonometric Functions Test 22

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Question 1

$$\dfrac{\pi}{16}$$

$$\dfrac{\pi}{32}$$

$$\dfrac{\pi}{64}$$

$$\dfrac{\pi}{128}$$

Solution

Question 2

$$\sqrt {mn}$$

mn

2mn

$$4\sqrt {mn}$$

Solution

Question 3

$$n\pi +\dfrac { 7\pi }{ 4 } ,n\epsilon I$$

$$n\pi +{ (-1) }^{ n }\dfrac { 7\pi }{ 4 } ,n\epsilon I$$

$$2n\pi +\dfrac { 7\pi }{ 4 } ,n\epsilon I$$

None of these

Solution

Question 4

$$sin \theta$$

$$cos \theta$$

$$- sin \theta$$

$$-cos \theta$$

Solution

Question 5

(A) - sin x

(B) cos x

(C) -cos x

0

Question 6

$$\theta =\frac { \pi }{ 33 } $$

$$\theta =\frac { \pi }{ 11 } $$

$$\theta =\frac { 2\pi }{ 15 } $$

$$\theta =\frac { \pi }{ 37 } $$

Question 7

$$5$$

$$\dfrac{11}{2}$$

$$\dfrac{13}{2}$$

$$7$$

Question 8

$$n\pi+(-1)^{n}\dfrac{\pi}{10}$$

$$n\pi+(-1)^{n}\dfrac{\pi}{6}$$

$$n\pi+(-1)^{n}\dfrac{3\pi}{10}$$

$$n\pi$$

Solution

Question 9

$$x + y = 0$$

$$x = 2y$$

$$x = y$$

$$2x = y$$

Solution

Question 10

$$\Delta $$

$$2 \Delta$$

$$3 \Delta $$

$$4 \Delta $$

Solution

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