Trigonometric Functions Test 26

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Trigonometric Functions Test 26

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Weekly Quiz Competition

Question 1
1 / -0

The principal value of
$$ cos^{-1} \left (cos\dfrac{2\pi}{3} \right ) + sin^{-1} \left (sin\dfrac{2\pi}{3} \right ) $$ is

A
$$ \pi $$

B
$$ \pi/2 $$

C
$$ \pi/3 $$

D
$$ 4\pi/3 $$

Solution
Question 2
1 / -0

Evaluate : $$ tan \left [ 2\, tan^{-1}\dfrac{1}{5} - \dfrac{\pi}{4} \right ] $$

A
$$ \dfrac{5}{4} $$

B
$$ \dfrac{5}{16} $$

C
$$- \dfrac{7}{17} $$

D
$$ \dfrac{7}{17} $$

Solution
Question 3
1 / -0

The value of $$ \dfrac{3 \pi}{4} $$ in sexagesimal system is:

A
$$ 75^{\circ} $$

B
$$ 135^{\circ} $$

C
$$ 120^{\circ} $$

D
$$ 220^{\circ} $$

Solution
Question 4
1 / -0

The value of $$ \dfrac{5}{16} $$ right angles in sexagesimal system is equal to

A
$$28^{\circ} 30' 7'' $$

B
$$ 27^{\circ} 5' 26'' $$

C
$$ 28^{\circ} 7' 30'' $$

D
$$ 29^{\circ} 3' 27'' $$

Solution
Question 5
1 / -0

1 radian is equal to:

A
$$180^{\circ} $$

B
$$200^{\circ}$$

C
$$100^{\circ}$$

D
None of these

Solution
Question 6
1 / -0

How many right angles is equal to $$56^{\circ} 15' $$ ?

A
$$ \dfrac{8}{5} $$ right angles

B
$$ \dfrac{5}{8} $$ right angles

C
$$ \dfrac{3}{5} $$ right angles

D
$$ \dfrac{5}{4} $$ right angles

Solution
Question 7
1 / -0

The positive integer value of $$n >3$$ satisfying the equation
$$ \displaystyle{\dfrac{1}{\sin \left(\dfrac{\pi}{n}\right)}=\dfrac{1}{\sin\left(\dfrac{2\pi}{n}\right)}+\dfrac{1}{\sin\left(\dfrac{3\pi}{n}\right)}
}$$ is

A
$$7$$

B
$$6$$

C
$$4$$

D
$$5$$

Solution

Given equation is
$$\displaystyle\dfrac { 1 }{ \sin { \dfrac { \pi }{ n } } } =\dfrac { 1 }{ \sin { \dfrac { 2\pi }{ n } } } +\dfrac { 1 }{ \sin { \dfrac { 3\pi }{ n } } } $$
$$\displaystyle\dfrac { 1 }{ \sin { \dfrac { \pi }{ n } } } -\dfrac { 1 }{ \sin { \dfrac { 3\pi }{ n } } } =\dfrac { 1 }{ \sin { \dfrac { 2\pi }{ n } } } $$
$$\displaystyle\dfrac { \sin { \dfrac { 3\pi }{ n } } - \sin { \dfrac { \pi }{ n } } }{ \sin { \dfrac { 3\pi }{ n } } \sin { \dfrac { \pi }{ n } } } =\dfrac { 1 }{ \sin { \dfrac { 2\pi }{ n } } } $$
$$\displaystyle\dfrac { 2\cos { \dfrac { 2\pi }{ n } } \sin { \dfrac { \pi }{ n } } }{ \sin { \dfrac { 3\pi }{ n } } \sin { \dfrac { \pi }{ n } } } =\dfrac { 1 }{ \sin { \dfrac { 2\pi }{ n } } } $$
$$\displaystyle\dfrac { 2\cos { \dfrac { 2\pi }{ n } } }{ \sin { \dfrac { 3\pi }{ n } } } =\dfrac { 1 }{ \sin { \dfrac { 2\pi }{ n } } } $$
$$\displaystyle2\cos { \dfrac { 2\pi }{ n } \sin { \dfrac { 2\pi }{ n } } } =\sin { \dfrac { 3\pi }{ n } } $$
$$\displaystyle\sin { \dfrac { 4\pi }{ n } } =\sin { \dfrac { 3\pi }{ n } } $$
The general solution for $$\sin { \theta } =\sin { \alpha } $$ is given by
$$\theta =p\pi +{ (-1) }^{ p }\alpha \quad ,\quad p\in I$$
So, $$ \dfrac { 4\pi }{ n } =p\pi +{ (-1) }^{ p }\dfrac { 3\pi }{ n } \quad ,\quad p\in I$$
If $$p = 2m $$ , then
$$\dfrac { 4\pi }{ n } =2m\pi +\dfrac { 3\pi }{ n } \quad $$
$$\dfrac { \pi }{ n } =2m\pi $$
or $$\dfrac { 1 }{ n } =2m$$ , which is not possible.
So, let $$ p = 2m+1$$ then
$$\dfrac { 4\pi }{ n } =(2m+1)\pi -\dfrac { 3\pi }{ n } \quad $$
$$\dfrac { 7\pi }{ n } =(2m+1)\pi $$
$$\dfrac { 7 }{ n } =2m+1$$
For $$m = 0, n=7 $$
Hence, $$n = 7$$
Question 8
1 / -0

The solution set of the equation
$$\tan(4 {k}+2) {x}- \tan(4 {k}+1) {x} - \tan(4 {k}+2) {x}\cdot\tan(4 {k}+1) {x}=1; {k}\in I$$ is

A
$$\phi$$

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

C
$$\displaystyle \left\{\mathrm{n}\pi+\frac{\pi}{4}, \mathrm{n}\in I\right\}$$

D
$$\displaystyle \left\{2n\pi+\frac{\pi}{4}, n\in I\right\}$$

Solution

Domain of the given equation is
$$x\in R - \left\{ \dfrac{\pi}4,\dfrac{(2k+1)\pi}{(4k+1)2}\right\}$$

Now, the equation can be written as

$$\dfrac{\tan(4k+2)x - \tan(4k+1)x}{1+\tan(4k+2)x\cdot\tan(4k+1)x} = 1$$

$$\Rightarrow \tan(x) = 1$$

$$\left(\text{Using } \tan(A-B) = \dfrac{\tan A - \tan B}{1+\tan A\cdot\tan B}\right)$$

Now, solving for $$\tan x = 1$$, we get
$$ x = n\pi + \dfrac{\pi}{4}$$
But this value is not in domain.
$$\therefore$$ their is no solution

Hence, option A.
Question 9
1 / -0

The general solution of $$\sin x - 3\sin 2x + \sin 3x
= \cos x - 3\cos 2x + \cos 3x$$ is

A
$$n\displaystyle \pi+\dfrac{\pi}{8}$$

B
$$\displaystyle \dfrac{n\pi}{2}+\dfrac{\pi}{8}$$

C
$$(-1)^{n}\dfrac{n\pi }{2}+\dfrac{\pi }{8}$$

D
$$2n\pi+Cos^{-1}\dfrac{3}{2} $$

Solution

using C-D formulae :
we get
$$ 2\sin 2x \cos x - 3\sin 2x = 2\cos 2x \cos x - 3\cos 2x$$
$$ (2\cos x - 3)\sin 2x = (2\cos x - 3)\cos 2x$$
$$ \cos 2x = \sin 2x$$
$$ \tan 2x = 1 $$
$$ x = \dfrac{n\Pi}{2} + \dfrac{\Pi}{8}$$
Question 10
1 / -0

Assertion : lf $$x = \sin (\alpha-\beta)\sin (\gamma-\delta)$$ ,$$y = \sin (\beta-\gamma) \sin (\alpha-\delta)$$, $$z = \sin (\gamma-\alpha) \sin (\beta-\delta)$$, then $$x+y+z = 0$$
Reason : $$2 \sin {A}\sin {B} = \cos {(A-B)}+\cos {(A+B)}$$

A
A is true, R is true and R is correct explanation of A

B
A is true, R is true and R is not correct explanation of A

C
A is true, R is false

D
A is false, R is true.

Solution

Assertion is true as
$$x = \sin (\alpha -\beta )\sin (\gamma -\beta)= (\sin \alpha \cos \beta -\cos \alpha \sin beta ) (\sin \gamma \cos \delta -\cos \gamma sin \delta )$$
$$=\sin \alpha \cos \beta sin \gamma \cos \delta - \sin \alpha \cos \beta \cos \gamma \sin \delta -\cos \alpha \sin \beta \sin \gamma \cos \gamma +\cos \alpha \sin \beta \cos \gamma \sin \delta$$
$$y=\sin (\beta -\gamma)\sin (\alpha -\delta)=(\sin \beta \cos \gamma -\cos \beta \sin \gamma) (\sin \alpha \cos \delta -\cos \alpha sin \gamma)$$
$$=sin \alpha \sin \beta \cos \gamma \cos \delta - \cos \alpha \sin \beta \cos \gamma \sin \delta - \sin \alpha \cos \beta \sin \gamma \cos \delta + cos \alpha \cos \beta \sin \gamma \sin \delta$$
$$z = \sin (\gamma -\alpha) \sin (\beta -\gamma)=(\sin \gamma \cos \alpha - \cos \gamma \sin \alpha ) (\sin \beta \cos \delta -\cos \beta \sin \delta )$$
$$=\cos \alpha \sin \beta \sin \gamma \cos \delta - \cos \alpha \cos \beta \sin \gamma \sin \delta -\sin \alpha \sin \beta \cos \gamma \cos \delta +\sin \alpha \cos \beta \cos \gamma \sin \delta$$
$$x+y+z=0$$
Reason is false as
$$2\sin {A} \sin {B}=\cos (A-B)-\cos (A+B)$$