Trigonometric Functions Test 16

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Trigonometric Functions Test 16

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Weekly Quiz Competition

Question 1
1 / -0

If $$\sin { x } +\cos { x } =a,$$ then $$\left| \sin { x } -\cos { x } \right| $$ equals

A
$$\sqrt { 2-{ a }^{ 2 } } $$

B
$$\sqrt { 2+{ a }^{ 2 } } $$

C
$$\sqrt { { a }^{ 2 }-2 } $$

D
$$\sqrt { { a }^{ 2 }-4 } $$

Solution

$$\left|\sin{x}-\cos{x}\right|=\sqrt{{\left(\sin{x}-\cos{x}\right)}^{2}}$$
$$=\sqrt{{\sin}^{2}{x}+{\cos}^{2}{x}-2{\sin}{x}{\cos}{x}}$$
$$=\sqrt{1-2\sin{x}\cos{x}}$$
Given $$\sin{x}+\cos{x}=a$$
squaring both sides we get
$${\left(\sin{x}+\cos{x}\right)}^{2}={a}^{2}$$
$$\Rightarrow {\sin}^{2}{x}+{\cos}^{2}{x}+2\sin{x}\cos{x}={a}^{2}$$
$$\Rightarrow 1+2\sin{x}\cos{x}={a}^{2}$$
$$\Rightarrow 2\sin{x}\cos{x}={a}^{2}-1$$
$$\therefore \left|\sin{x}-\cos{x}\right|=\sqrt{1-\left({a}^{2}-1\right)}=\sqrt{2-{a}^{2}}$$
Question 2
1 / -0

The number of solution of the equation $$\sin 5x \cos 3x=\sin 6x \cos 2x$$ in the interval $$[0,\pi]$$ is

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

$$\sin 5x\cos 3x=\sin 6x \cos 2x$$
$$\sin \theta\cos \phi =\cfrac{1}{2}[\sin(\theta+\phi)+\sin(\theta-\phi)]$$
$$\cfrac{1}{2}[\sin 8x+\sin 2x]=\cfrac{1}{2}[\sin 8x+\sin 4x]$$
$$\cfrac{1}{2}\sin 2x=\cfrac{1}{2}\sin 4x$$
$$\sin 4x-\sin 2x=0$$
$$2\sin 2x\cos 2x-\sin 2x=0$$
$$\sin 2x[2\cos 2x-1]=0$$
$$\sin 2x=0$$ $$2\cos 2x-1=0$$
$$2x=0$$ $$\cos 2x=\cfrac{1}{2}$$
$$x=0$$ $$2x=\cfrac{\pi}{3}$$
$$x=\cfrac{\pi}{6}$$
$$two$$ solutions.
Question 3
1 / -0

The value of $$\theta$$ satisfying $$\sin{7\theta}=\sin{4\theta}-\sin{\theta}$$ and $$\theta<0<\pi/2$$ are-

A
$$\dfrac{\pi}{9}, \dfrac{\pi}{4}$$

B
$$\dfrac{\pi}{3}, \dfrac{\pi}{9}$$

C
$$\dfrac{\pi}{6}, \dfrac{\pi}{9}$$

D
$$\dfrac{\pi}{3}, \dfrac{\pi}{4}$$

Solution
Question 4
1 / -0

$$2\sin 2\beta + 4\cos (\alpha + \beta)\sin \alpha \sin \beta + \cos 2(\alpha + \beta)$$

A
$$\sin 2\alpha$$

B
$$\cos \, 2 \beta$$

C
$$\cos \, 2 \alpha$$

D
$$\sin \, 2 \beta$$

Solution

$$2 \sin^{2}\beta +4 \cos (\alpha +\beta ) \sin \alpha \sin\beta +\cos 2 (\alpha +\beta )$$

$$= 2 \sin^{2}\beta +4 (\cos \alpha \cos \beta -\sin\alpha \sin\beta ) \sin \alpha \sin\beta +\cos 2 \alpha \cos 2 \beta -\sin 2\alpha \sin 2\beta $$

$$=2\sin^{2}\beta +4\cos\alpha \cos\beta \sin\alpha \sin\beta -1 \sin^{2}\alpha \sin^{2}\beta +\cos^{2}\alpha \cos^{2}\beta -\sin 2\alpha \sin 2\beta $$

$$=2\sin^{2}\beta +\sin^{2}\alpha \sin^{2}\beta -4\sin^{2}\alpha \sin^{2}\beta +\cos 2\alpha \cos^{2}\alpha -\sin^{2}\alpha \sin^{2}\beta $$

$$=(-\cos^{2}\beta )-(2 \sin^{2}\alpha )(2 \sin^{2}\beta )+\cos^{2}\alpha \cos^{2}\beta $$

$$=(1-\cos 2\beta )-(1-\cos 2\alpha )(1-\cos 2\beta )+\cos 2\alpha \cos 2\beta $$

$$=\cos 2\alpha $$
Question 5
1 / -0

Evaluate $$\tan \left[ {\dfrac{\pi }{4}\, + \,\dfrac{1}{2}\,{{\cos }^{ - 1}}\,\dfrac{a}{b}} \right]\, + \,\tan \left[ {\dfrac{\pi }{4}\, - \,\dfrac{1}{2}{{\cos }^{ - 1}}\,\dfrac{a}{b}} \right]\, = $$

A
$$\dfrac{{2a}}{b}$$

B
$$\dfrac{{2b}}{a}$$

C
$$\dfrac{a}{b}$$

D
$$\dfrac{b}{a}$$

Solution

$$\textbf{Given expression}$$

$$ = \tan \left[ {\dfrac{\pi }{4}\, + \,\dfrac{1}{2}\,{{\cos }^{ - 1}}\,\dfrac{a}{b}} \right]\, + \,\tan \left[ {\dfrac{\pi }{4}\, - \,\dfrac{1}{2}{{\cos }^{ - 1}}\,\dfrac{a}{b}} \right]\, $$

$$\textbf{Step : 1 : Do relevant substitution to simplify given equation}$$

Let $$x=\cos ^{ -1 }{ \cfrac { a }{ b } } \Rightarrow \cos{x} = \dfrac{a}{b}$$ $$ ........(1)$$

$$ = \tan\left[\cfrac { \theta }{ 4 } +\cfrac { x }{ 2 } \right]+\tan\left[\cfrac { \theta }{ 4 } -\cfrac { x }{ 2 } \right]$$

$$\textbf{Step : 2 Apply relevant identity }$$:

$$=\cfrac { \tan \cfrac {\theta }{ 4 } +\tan\cfrac { x }{ 2 } }{ 1-\tan\cfrac { \theta }{ 4 } \times \tan \cfrac {x }{ 2 } } +\cfrac { \tan \cfrac {\theta }{ 4 } -\tan\cfrac { x }{ 2 } }{ 1+\tan\cfrac { \theta }{ 4 } \times \tan \cfrac {x }{ 2 } } , $$   $$\left[\tan(A \pm B) = \dfrac{tan A \pm tan B}{1\mp tanA tanB}\right]$$

$$=\cfrac { 1+ \tan\cfrac { x }{ 2 } }{ 1- \tan \cfrac { x }{ 2 } } +\cfrac { 1- \tan\cfrac { x }{ 2 } }{ 1+ \tan \cfrac { x }{ 2 } }$$

$$=\cfrac { { \left(1+ \tan \cfrac{x }{ 2 } \right) }^{ 2 }+{ \left(1- \tan\cfrac{ x }{ 2 } \right) }^{ 2 } }{ 1-{ \tan }^{ 2 }\cfrac { x }{ 2 } } $$

$$=\cfrac { 2\left(1+{ \tan }^{ 2 }\cfrac { x }{ 2 } \right) }{ \left(1- { \tan }^{ 2 }\cfrac {x }{ 2 } \right) } $$ $$[\because(a\pm b)^2 = a^2 + b^2 \pm 2ab]$$

$$=\cfrac { 2 }{ \cos x }$$   $$[\because \cos{x} =\cfrac { \left(1-{ \tan }^{ 2 }\cfrac { x }{ 2 } \right) }{ \left(1+ { \tan }^{ 2 }\cfrac {x }{ 2 } \right) } $$

Put the value  from eq.(1),

$$ =\cfrac { 2 }{ \cfrac { a }{ b } }$$

$$ =\cfrac { 2b }{ a } $$
Question 6
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If $$A+B+C=\pi$$ then $$\sin^3A \cos(B-C)+ \sin^3\cos(C-A) + \sin^3C \cos (A-B) $$ is equal to

A
$$3\sin A \sin B \sin C.$$

B
$$6\sin A \sin B \sin C.$$

C
$$4\sin A \sin B \sin C.$$

D
$$0$$

Solution
Question 7
1 / -0

If $$(\cos\theta +\cos 2\theta)^3=\cos^3\theta +\cos^32\theta$$, then the least positive value of $$\theta$$ is equal to?

A
$$\dfrac{\pi}{6}$$

B
$$\dfrac{\pi}{4}$$

C
$$\dfrac{\pi}{3}$$

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

Solution

We are given
$$(\cos\theta+\cos2\theta)^{3}=\cos^{3}\theta+\cos^{3}2\theta$$
$$\cos^{3}\theta+\cos^{3}2\theta+3\cos^{2}\theta\cos^{2}\theta+3\cos\theta\cos^{2}2\theta$$
$$=\cos^{3}\theta+\cos^{3}2\theta$$
$$(as\ (a+b)^{3}=a^{3}+b^{3}+3a^{2}b+3b^{2}a)$$
So, $$3\cos^{2}\theta\cos 2\theta+3\cos\theta\cos^{2}\theta=0$$
$$3\cos\theta\cos2\theta(\cos\theta+\cos2\theta)=0$$
So, $$\cos\theta.\cos 2\theta.(\cos\theta+2\cos^{2}\theta-1)=0$$
$$\cos\theta.\cos 2\theta.(2\cos^{2}\theta+\cos\theta-1)=0$$
We get $$\cos\theta=0, \cos 2\theta=0$$ & $$2\cos^{2}\theta+\cos\theta-1=0$$
$$\cos\theta=0$$ or $$\cos 2\theta=0$$ or $$2\cos^{2}\theta+\cos\theta-1=0$$
$$\theta=\cos^{-1}0$$ or $$2\theta=\cos^{-1}0$$ or
$$2\cos^{2}\theta+2\cos\theta-\cos\theta-1=0$$
$$2\cos\theta(\cos\theta+1)-1(\cos\theta+1)=0$$
$$(2\cos\theta-1)(\cos\theta+1)=0$$
$$2\cos\theta-1=0$$ or $$\cos\theta+1=0$$
$$\cos\theta=\dfrac{1}{2}$$ or $$\cos\theta+1=0$$
$$\cos\theta=\dfrac{1}{2}$$ or $$\cos\theta=-1$$
$$\theta=\cos^{-1}\dfrac{1}{2}$$ $$\theta=\cos^{-1}(-1)$$
By principal values we get $$\theta$$ as
$$\dfrac{\pi}{2}, \dfrac{\pi}{4}, \dfrac{\pi}{3}, \pi$$
$$\Rightarrow$$ then least positive value of $$\theta$$ is equal to $$\left(\dfrac{\pi}{4}\right)$$
Question 8
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General solution of equation $$\cot \theta + \text{cosec}\theta = \sqrt 3 $$ is

A
$$2n\pi + \frac{\pi }{4}$$

B
$$\left( {2n - 1} \right)\pi $$

C
$$2n\pi + \frac{\pi }{3}$$

D
$$2n\pi + \frac{\pi }{6}$$

Solution

Given expression:
$$\cot\theta+\text{cosec}\ \theta=\sqrt3$$
$$\dfrac{\cos \theta}{\sin \ \theta}+\dfrac{1}{\sin \ \theta}=\sqrt3$$

Taking square both side,
$$\cos ^2\theta+1+2\cos \theta=3-3\cos ^2\theta$$ [\sin ce $$\sin ^2\theta =1-\cos ^2\theta$$]
$$4\cos ^2\theta+2\cos \theta-2=0$$
$$4\cos ^2\theta+4\cos \theta-2\cos \theta-2=0$$
We get
$$(\cos \theta+1)(4\cos \theta -2)=0$$
When,
$$(\cos \theta+1)=0$$
$$\cos \theta =-1$$
$$\cos \pi=\cos (2n+1)\pi=-1$$
Hence $$\theta =(2n+1)\pi.$$

When,
$$(4\cos \theta-2)=0$$
$$\cos \theta =\dfrac{1}{2}$$
$$\cos \dfrac{\pi}{3}=\cos (2n\pi\pm \dfrac{\pi}{3})=\dfrac{1}{2}$$

Hence, $$\theta=(2n\pi\pm \dfrac{\pi}{3})$$
Question 9
1 / -0

If $$sin 2 \theta = cos 3 \theta \, and \, \theta $$ is acute then $$sin 5 \theta$$ = ...

A
0

B
1

C
-1

D
2

Solution
Question 10
1 / -0

The general solution of the equation $${\sin ^{50}}x - {\cos ^{50}}x = 1$$

A
$$2n\pi + \frac{\pi }{2}$$

B
$$2n\pi + \frac{\pi }{3}$$

C
$$n\pi + \frac{\pi }{2}$$

D
$$n\pi + \frac{\pi }{3}$$

Solution

Consider the given equation.

$${{\sin }^{50}}x-{{\cos }^{50}}x=1$$

$${{\sin }^{50}}x=1+{{\cos }^{50}}x$$ …….. (1)

Here, $${{\cos }^{50}}x$$ is always positive. So, the maximum value of $${{\sin }^{50}}x$$ is $$1$$.

So,

$${{\cos }^{50}}x$$ must be zero.

Hence, $${{\cos }^{50}}x=0$$

$$x=\left( 2n+1 \right)\dfrac{\pi }{2},$$ where $$n\in I$$
$$x=n\pi +\dfrac{\pi}{2}$$

Hence, this is the answer.

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Trigonometric Functions Test 16

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Trigonometric Functions Test 16

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SHARING IS CARING

Question 1

$$\sqrt { 2-{ a }^{ 2 } } $$

$$\sqrt { 2+{ a }^{ 2 } } $$

$$\sqrt { { a }^{ 2 }-2 } $$

$$\sqrt { { a }^{ 2 }-4 } $$

Solution

Question 2

$$2$$

$$3$$

$$4$$

$$5$$

Solution

Question 3

$$\dfrac{\pi}{9}, \dfrac{\pi}{4}$$

$$\dfrac{\pi}{3}, \dfrac{\pi}{9}$$

$$\dfrac{\pi}{6}, \dfrac{\pi}{9}$$

$$\dfrac{\pi}{3}, \dfrac{\pi}{4}$$

Solution

Question 4

$$\sin 2\alpha$$

$$\cos \, 2 \beta$$

$$\cos \, 2 \alpha$$

$$\sin \, 2 \beta$$

Solution

Question 5

$$\dfrac{{2a}}{b}$$

$$\dfrac{{2b}}{a}$$

$$\dfrac{a}{b}$$

$$\dfrac{b}{a}$$

Solution

Question 6

$$3\sin A \sin B \sin C.$$

$$6\sin A \sin B \sin C.$$

$$4\sin A \sin B \sin C.$$

$$0$$

Solution

Question 7

$$\dfrac{\pi}{6}$$

$$\dfrac{\pi}{4}$$

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

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

Solution

Question 8

$$2n\pi + \frac{\pi }{4}$$

$$\left( {2n - 1} \right)\pi $$

$$2n\pi + \frac{\pi }{3}$$

$$2n\pi + \frac{\pi }{6}$$

Solution

Question 9

0

1

-1

2

Solution

Question 10

$$2n\pi + \frac{\pi }{2}$$

$$2n\pi + \frac{\pi }{3}$$

$$n\pi + \frac{\pi }{2}$$

$$n\pi + \frac{\pi }{3}$$

Solution

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