Trigonometric Functions Test 27

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Trigonometric Functions Test 27

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Weekly Quiz Competition

Question 1
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If $$\mathrm{s}in(\pi \mathrm{c}\mathrm{o}s\theta)= \mathrm{c}os$$$$(\pi\sin\theta)$$ , then which of the following is correct

A
$$\cos \displaystyle \theta=\frac{3}{2\sqrt{2}}$$

B
$$\cos (\theta-\dfrac{\pi}{2})=\dfrac{1}{2\sqrt{2}}$$

C
$$\cos (\theta-\dfrac{\pi}{4})=\dfrac{1}{2\sqrt{2}}$$

D
$$\cos \left(\dfrac{\pi\theta}{4}\right )=\dfrac{1}{2\sqrt{2}}$$

Solution

$$\sin(\pi \cos\theta)= \cos (\pi\sin\theta)$$

$$\Rightarrow \sin(\pi \cos\theta)= \sin (\dfrac{\pi}{2}-\pi\sin\theta)$$

$$\Rightarrow (\pi \cos\theta)= (\dfrac{\pi}{2}-\pi\sin\theta)$$ or $$ \pi \cos\theta +\pi \sin \theta =\dfrac{\pi}{2} $$

$$\displaystyle \Rightarrow \cos\theta + \sin\theta = \frac{1}{2}$$ or $$ \cos\theta -\sin\theta = \dfrac{1}{2} $$

Let us consider the first case,

$$\displaystyle \Rightarrow \frac{1}{\sqrt{2}}\cos\theta + \frac{1}{\sqrt{2}}\sin\theta = \frac{1}{2\sqrt{2}}$$

$$ \displaystyle \Rightarrow cos\left(\frac{\pi}{4}-\theta \right)=\frac{1}{2\sqrt{2}}$$

Similarly, we can show that,

$$\cos \left(\theta +\dfrac{\pi}{4} \right) = \dfrac{1}{2\sqrt2}$$
Question 2
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If $$\tan\theta=\displaystyle \dfrac{x\sin\phi}{1-x\cos\phi}$$, $$\displaystyle \tan\phi=\dfrac{y\sin\theta}{1-y\cos\theta}$$ , then $$\displaystyle \dfrac{x}{y}=$$

A
$$\displaystyle \dfrac{\sin\phi}{\sin\theta}$$

B
$$\displaystyle \dfrac{\sin\theta}{\sin\phi}$$

C
$$\displaystyle \dfrac{1- cos\phi}{1-\cos\theta}$$

D
$$\displaystyle \dfrac{1- cos\theta}{1-cos\phi}$$

Solution

$$\displaystyle tan \theta =\frac{x sin \phi }{1-x cos \phi }$$
$$sin \theta -x sin \theta cos \phi =x cos \theta sin \phi $$
$$\displaystyle x=\frac{sin \theta }{sin (\theta + \phi )}$$
Similarly,$$\displaystyle y=\frac{sin \phi }{sin (\theta +\phi )}$$
$$\displaystyle \therefore \frac{x}{y}=\frac{sin \theta }{sin \phi }$$
Question 3
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If $$ 32\tan^{8}\theta=2\cos^{2}\alpha-3\cos\alpha$$ and $$3\cos2\theta=1$$ then the general value of $$\alpha$$ is

A
$$n\displaystyle \pi\pm\frac{\pi}{3}$$

B
$$2\displaystyle {n}\pi\pm\frac{2\pi}{3}$$

C
$$n\pi\pm\pi$$

D
$$2n\displaystyle \pi\pm\frac{\pi}{2}$$

Solution

Given,
$$32 \tan^{8}\theta=2\cos^2 \alpha-3\cos \alpha$$...............(i)

$$3\cos2\theta = 1$$

This gives $$\dfrac{1 - \tan^{2}\theta}{1 + \tan^{2}\theta} = \dfrac13$$

So, $$ \tan^{2}\theta = \dfrac{1}{2}$$

Given equation (i) becomes

$$32 (\tan^{2} \alpha)^4 = 2\cos^{2}\alpha - 3\cos\alpha$$

$$\Rightarrow 32\times \dfrac1{16}=2 \cos^2{\alpha}-3\cos \alpha$$

$$\Rightarrow 2=2 \cos^2{\alpha}-3\cos \alpha$$

$$\Rightarrow 2 \cos^2{\alpha}-3\cos \alpha-2=0$$

$$\Rightarrow 2\cos^2 \alpha-4\cos \alpha+\cos \alpha-2=0$$

$$\Rightarrow 2\cos \alpha(\cos \alpha -2)+1(\cos \alpha-2)=0$$

$$\Rightarrow (2\cos \alpha+1)(\cos \alpha-2)=0$$

So, $$ \cos\alpha = -\dfrac{1}{2}$$ or $$\cos\alpha = 2$$

$$\because \cos \alpha \neq 2$$

$$\therefore \cos\alpha = -\dfrac12 \Rightarrow \alpha=\dfrac{2\pi}{3}$$

So. general solution is $$\alpha=2n\pi \pm \dfrac{2\pi}{3}$$

Hence, the correct answer is B
Question 4
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The values of $$x$$ between $$0$$ and $$2p$$ which satisfy the equation $$\sin x\sqrt{8\cos^{2}x}=1$$ are in A.P. The common difference of the A.P. is

A
$$\dfrac p8$$

B
$$\dfrac p4$$

C
$$\dfrac {3p}8$$

D
$$\dfrac {5p}8$$

Solution

$$\sin x \sqrt{8 \cos ^{2}x}=1 \sin x > 0$$
$$8 \sin ^{2}x \cos ^{2}x =1$$
$$2(\sin 2x)^{2}=1$$
$$\displaystyle \sin 2x=+\frac{1}{\sqrt{2}} , -\frac{1}{\sqrt{2}}$$
$$\displaystyle 2x=\frac{\pi }{4} , \frac{3\pi }{4}$$
$$\displaystyle x=\frac{\pi }{8}, \frac{3\pi }{8}$$ are in A.P.
$$\therefore $$ common diff $$\displaystyle =\frac{3\pi }{8} - \frac{\pi }{8} =\frac{2\pi }{8}=\frac{\pi }{4}$$
Question 5
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$$|\tan x + \sec x| = |\tan x| - |\sec x|, x \epsilon [0,2\pi]$$if and only if x belongs to the interval

A
$$[0,\pi]$$

B
$$[0,\dfrac{\pi}{2})\cup (\dfrac{\pi}{2},\pi]$$

C
$$[\pi,\dfrac{3\pi}{2})\cup (\dfrac{3\pi}{2},2\pi]$$

D
none of these

Solution

$$|\tan x + \sec x| = |\tan x| - |\sec x|$$
Then $$\tan x . \sec x \leq 0$$ and $$ |\tan x| \geq |\sec x|$$
$$Now, |\tan x| \geq |\sec x|$$

$$\Rightarrow \tan ^2x \geq \sec ^2x$$

$$\Rightarrow \tan ^2x \geq 1+ \tan ^2x \Rightarrow 0 \geq 1 $$ This is not possible.
Question 6
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The total number of solutions of $$\cos x=\sqrt{1-\sin 2x}$$ in $$\left [ 0,2\pi \right ]$$ is equal to

A
$$2$$

B
$$3$$

C
$$5$$

D
None of these

Solution

$$\cos x=\sqrt{1-\sin 2x}=\left | \sin x -\cos x\right |$$

(a) $$\sin x< \cos x\Rightarrow x\in \left [ 0,\frac{\pi }{4})\cup (\frac{5\pi }{4},2\pi \right ]$$
Then the given equation is $$\cos x=\cos x-\sin x$$
$$ \sin x=0\Rightarrow x=0,\pi ,2\pi \left [ from equ.(i) \right ]$$
$$x=\pi $$ is not possible

(b) $$\sin x\geq \cos x\Rightarrow x\in \left [ \frac{\pi }{4},\frac{5\pi }{4} \right ]$$

$$ \cos x=\sin x -\cos x$$

$$ \tan x=2$$

$$x=\tan ^{-1}2$$

Hence, there are three solutions.
Question 7
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Find the value of $$\cot^{-1}\left(\dfrac{\sqrt {1-\sin x}+\sqrt {1+\sin x}}{\sqrt {1-\sin x}-\sqrt {1+\sin x}}\right)$$

A
$$\pi -\dfrac{x}{2}$$

B
$$\pi$$

C
$$\dfrac{x}{2}$$

D
$$\dfrac{\pi}{2}$$

Solution
Question 8
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Directions For Questions

The trigonometric equation is-
$$\sin x+3 \sin 2x+\sin 3x=\cos x+3 \cos 2x+\cos 3x$$
when $$x$$ lies in first four quadrants. It means $$x\epsilon [0, 2\pi]$$, then-

...view full instructions

The difference between greatest and least solution of $$x$$ is-

A
$$\dfrac {3\pi}{2}$$

B
$$\dfrac {\pi}{2}$$

C
$$\pi$$

D
$$10\pi$$

Solution

Given equation is
$$\sin\,x+3\,\sin\,2x+\sin\,3x=\cos\,x+3\,\cos\,2x+\cos\,3x$$
$$(\sin x+\sin\,3x)+3\sin\,2x=(\cos x+\cos\,3x)+3\cos\,2x$$
$$(2\sin\,2x\,\cos x+3\sin\,2x)-(2\cos\,2x\,\cos x+3\cos\,2x)=0$$
$$\sin\,2x(2\cos\,x+3)-\cos\,2x(2\cos\,x+3)=0$$
$$\Rightarrow (\sin 2x-\cos 2x)(2\cos x+3)=0$$
$$cos\,x=-\displaystyle\frac{3}{2}$$
which is not possible
$$\sin\,2x=\cos\,2x$$
$$\tan\,2x=1$$
$$\tan 2x=tan\displaystyle\frac{\pi}{4}, \tan\frac{5\pi}{4}$$
$$x=\displaystyle\frac{\pi}{8},\dfrac{5\pi}{8}$$
Required difference $$\dfrac{5\pi}{8}-\dfrac{\pi}{8}=\dfrac{\pi}{2}$$
Question 9
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The value of $$\dfrac{\tan \alpha}{1-\cot \alpha}+\dfrac{\cot \alpha}{1-\tan \alpha}$$ is identically equal to

A
$$\sec \alpha.\csc \alpha$$

B
$$\sin \alpha. \cos \alpha$$

C
$$\sec \alpha. \csc \alpha +1$$

D
$$\sin \alpha . \cos \alpha +1$$

Solution

$$\dfrac{\tan\alpha}{1-\cot\alpha}+\dfrac{\cot\alpha}{1-\tan\alpha}$$
$$=\dfrac{\tan\alpha}{1-\dfrac{1}{\tan\alpha}}+\dfrac{\cot\alpha}{1-\tan\alpha}$$
$$=\dfrac{\tan^2\alpha}{\tan\alpha-1}-\dfrac{\cot\alpha}{\tan\alpha-1}$$
$$=\dfrac{\tan^2-\dfrac{1}{\tan\alpha}}{\tan\alpha-1}$$
$$=\dfrac{\tan^3\alpha-1}{(\tan\alpha)(\tan\alpha-1)}$$
$$=\dfrac{(\tan\alpha-1)(\tan^2\alpha+\tan\alpha+1)}{(\tan\alpha)(\tan\alpha-1)}$$
$$=\dfrac{\tan^2\alpha+\tan\alpha+1}{\tan\alpha}$$
$$={\tan\alpha+1+\cot\alpha}$$
$$=\dfrac{\sin\alpha}{\cos\alpha}+\dfrac{\cos\alpha}{\sin\alpha}+1$$
$$=\dfrac{\sin^2\alpha+\cos^2\alpha}{\cos\alpha \sin\alpha}+1$$
$$=\dfrac{1}{\cos\alpha \sin\alpha}+1$$
$$=\sec\alpha\:\csc\alpha+1$$

Hence answer is C
Question 10
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If $$ \cos \theta - \sin \theta =\sqrt{2} \sin \theta$$, then $$ \cos \theta + \sin \theta$$ is

A
$$\sqrt{2}\cos \theta$$

B
$$\sqrt{2} \sin \theta$$

C
$$0$$

D
$$1$$

Solution

Given,
$$\cos \theta - \sin \theta =\sqrt{2} \sin \theta$$

Squaring, we get
$$\cos^2\theta+\sin^2\theta-2 \sin \theta \cos \theta=2 \sin^2 \theta$$
or $$ 1-2 \sin \theta \cos \theta=2(1-\cos^2 \theta)$$
or $$1+2 \sin \theta \cos \theta =2 \cos^2 \theta$$
or $$ \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 2 \cos^2 \theta$$
or $$ (\sin \theta + \cos \theta)^2=2 \cos^2 \theta$$

$$\therefore \sin \theta + \cos \theta =\sqrt{2} \cos \theta$$

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Trigonometric Functions Test 27

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Trigonometric Functions Test 27

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Question 1

$$\cos \displaystyle \theta=\frac{3}{2\sqrt{2}}$$

$$\cos (\theta-\dfrac{\pi}{2})=\dfrac{1}{2\sqrt{2}}$$

$$\cos (\theta-\dfrac{\pi}{4})=\dfrac{1}{2\sqrt{2}}$$

$$\cos \left(\dfrac{\pi\theta}{4}\right )=\dfrac{1}{2\sqrt{2}}$$

Solution

Question 2

$$\displaystyle \dfrac{\sin\phi}{\sin\theta}$$

$$\displaystyle \dfrac{\sin\theta}{\sin\phi}$$

$$\displaystyle \dfrac{1- cos\phi}{1-\cos\theta}$$

$$\displaystyle \dfrac{1- cos\theta}{1-cos\phi}$$

Solution

Question 3

$$n\displaystyle \pi\pm\frac{\pi}{3}$$

$$2\displaystyle {n}\pi\pm\frac{2\pi}{3}$$

$$n\pi\pm\pi$$

$$2n\displaystyle \pi\pm\frac{\pi}{2}$$

Solution

Question 4

$$\dfrac p8$$

$$\dfrac p4$$

$$\dfrac {3p}8$$

$$\dfrac {5p}8$$

Solution

Question 5

$$[0,\pi]$$

$$[0,\dfrac{\pi}{2})\cup (\dfrac{\pi}{2},\pi]$$

$$[\pi,\dfrac{3\pi}{2})\cup (\dfrac{3\pi}{2},2\pi]$$

none of these

Solution

Question 6

$$2$$

$$3$$

$$5$$

None of these

Solution

Question 7

$$\pi -\dfrac{x}{2}$$

$$\pi$$

$$\dfrac{x}{2}$$

$$\dfrac{\pi}{2}$$

Solution

Question 8

$$\dfrac {3\pi}{2}$$

$$\dfrac {\pi}{2}$$

$$\pi$$

$$10\pi$$

Solution

Question 9

$$\sec \alpha.\csc \alpha$$

$$\sin \alpha. \cos \alpha$$

$$\sec \alpha. \csc \alpha +1$$

$$\sin \alpha . \cos \alpha +1$$

Solution

Question 10

$$\sqrt{2}\cos \theta$$

$$\sqrt{2} \sin \theta$$

$$0$$

$$1$$

Solution

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