Trigonometric Functions Test 10

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Trigonometric Functions Test 10

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Weekly Quiz Competition

Question 1
1 / -0

If $$\cos ^{ -1 }{ \left( \cfrac { p }{ a } \right) } +\cos ^{ -1 }{ \left( \cfrac { p }{ b } \right) } =\alpha $$, then $$\cfrac { { p }^{ 2 } }{ { a }^{ 2 } } +k\cos { \alpha } +\cfrac { { p }^{ 2 } }{ { b }^{ 2 } } =\sin ^{ 2 }{ \alpha } $$, where $$k$$ is equal to

A
$$\cfrac { 2pq }{ ab } $$

B
$$-\cfrac { 2pq }{ ab } $$

C
$$-\cfrac { pq }{ ab} $$

D
$$\cfrac { pq }{ ab } $$

Solution

Given, $$\cos ^{ -1 }{ \left( \cfrac { p }{ a } \right) } +\cos ^{ -1 }{ \left( \cfrac { q }{ b } \right) } =\alpha \quad $$

$$\Rightarrow \cos ^{ -1 }{ \left[ \cfrac { p }{ a } .\cfrac { q }{ b } -\sqrt { \left( 1-\cfrac { { p }^{ 2 } }{ { a }^{ 2 } } \right) } \sqrt { \left( 1-\cfrac { { q }^{ 2 } }{ { b }^{ 2 } } \right) } \right] } =\alpha \quad \quad $$

$$\Rightarrow \cfrac { pq }{ ab } -\sqrt { \left( 1-\cfrac { { p }^{ 2 } }{ { a }^{ 2 } } \right) \left( 1-\cfrac { { q }^{ 2 } }{ { b }^{ 2 } } \right) } =\cos { \alpha } $$

$$\Rightarrow { \left( \cfrac { pq }{ ab } -\cos { \alpha } \right) }^{ 2 }=1-\cfrac { { p }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { q }^{ 2 } }{ { b }^{ 2 } } +\cfrac { { p }^{ 2 }{ q }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 } } =\cfrac { { p }^{ 2 }{ q }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 } } +\cos ^{ 2 }{ \alpha } -\cfrac { 2pq\cos { \alpha } }{ ab } $$

$$\Rightarrow \cfrac { { p }^{ 2 } }{ { a }^{ 2 } } -\cfrac { 2pq }{ ab } \cos { \alpha } +\cfrac { { q }^{ 2 } }{ { b }^{ 2 } } =1-\cos ^{ 2 }{ \alpha } =\sin ^{ 2 }{ \alpha } $$
But $$\cfrac { { p }^{ 2 } }{ { a }^{ 2 } } +k\cos { \alpha } +\cfrac { { q }^{ 2 } }{ { b }^{ 2 } } =\sin ^{ 2 }{ \alpha } $$

$$\Rightarrow k=-\cfrac { 2pq }{ ab } $$
Question 2
1 / -0

If $$\cos { \alpha } +\cos { \beta } +\cos { \gamma } =\sin { \alpha } +\sin { \beta } +\sin { \gamma } =0$$, then the value of $$\cos { 3\alpha } +\cos { 3\beta } +\cos { 3\gamma } $$ is

A
$$0$$

B
$$\cos { \left( \alpha +\beta +\gamma \right) } $$

C
$$3\cos { \left( \alpha +\beta +\gamma \right) } $$

D
$$3\sin { \left( \alpha +\beta +\gamma \right) } $$

Solution

Let $$a=\cos { \alpha } +i\sin { \alpha } ,b=\cos { \beta } +i\sin { \beta } ,c=\cos { \gamma } +i\sin { \gamma } $$

The, $$a+b+c=\left( \cos { \alpha } +\cos { \beta } +\cos { \gamma } \right) +i\left( \sin { \alpha } +\sin { \beta } +\sin { \gamma } \right) =0+i0=0$$

$$\Rightarrow { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }=3abc$$

$$\Rightarrow \left( \cos { 3\alpha } +i\sin { 3\alpha } \right) +\left( \cos { 3\beta } +i\sin { \beta } \right) +\left( \cos { 3\gamma } +i\sin { 3\gamma } \right) $$

$$=3\left[ \cos { \left( \alpha +\beta +\gamma \right) } +i\sin { (\alpha +\beta +\gamma ) } \right] $$

by equating the real components

$$\Rightarrow \cos { 3\alpha } +\cos { 3\beta } +\cos { 3\gamma } =3\cos { \left( \alpha +\beta +\gamma \right) } $$
Question 3
1 / -0

If $$2\sin^{2}\theta + \sqrt {3}\cos \theta + 1 = 0$$, then the value of $$\theta$$ is

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

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

C
$$\dfrac {5\pi}{6}$$

D
$$\pi$$

Solution

Given, $$2\sin^{2}\theta + \sqrt {3}\cos \theta + 1 = 0$$
$$\Rightarrow 2(1 - \cos^{2}\theta) + \sqrt {3}\cos \theta + 1 = 0$$
$$\Rightarrow 2\cos^{2}\theta - \sqrt {3}\cos \theta - 3 = 0$$
$$\therefore \cos \theta = \dfrac {\sqrt {3} \pm \sqrt {3 + 4 \times 3\times 2}}{2\times 2}$$
$$= \dfrac {\sqrt {3} \pm 3\sqrt {3}}{4}$$
$$\Rightarrow \cos \theta = - \dfrac {\sqrt {3}}{2}$$
and $$\cos \theta = \sqrt {3}$$
$$\therefore \cos \theta = -\dfrac {\sqrt {3}}{2}$$
$$(\because \cos \theta$$ cannot be greater than $$1$$)
$$\Rightarrow \theta = \dfrac {5\pi}{6}$$.
Question 4
1 / -0

The set of values of a for which the equation $$\sin x(\sin x+\cos x)=a$$ has real solutions is

A
$$[1-\sqrt{2}, 1+\sqrt{2}]$$

B
$$[2-\sqrt{3}, 2+\sqrt{3}]$$

C
$$[0, 2+\sqrt{3}]$$

D
$$\left[\displaystyle\frac{1-\sqrt{2}}{2}, \frac{1+\sqrt{2}}{2}\right]$$

Solution

We have,
$$\sin x(\sin x+\cos x)=a$$
$$\Rightarrow 2\sin^2x+2\sin x\cos x=2a$$
$$\Rightarrow 1-\cos 2x+\sin 2x=2a$$
$$\Rightarrow \sin 2x-\cos 2x=2a-1$$
This equation will have real solutions, if
$$|2a-1|\leq \sqrt{1+1}$$
$$\Rightarrow 1-\sqrt{2}\leq 2a\leq 1+\sqrt{2}$$
$$\Rightarrow a\epsilon\left[\displaystyle\frac{1-\sqrt{2}}{2}, \frac{1+\sqrt{2}}{2}\right]$$
Question 5
1 / -0

The value of $$\cos ^{ 4 }{ \left( \cfrac { \pi }{ 8 } \right) } +\cos ^{ 4 }{ \left( \cfrac { 3\pi }{ 8 } \right) } +\cos ^{ 4 }{ \left( \cfrac { 5\pi }{ 8 } \right) } +\cos ^{ 4 }{ \left( \cfrac { 7\pi }{ 8 } \right) } $$ is

A
$$0$$

B
$$1/2$$

C
$$3/2$$

D
$$1$$

Solution

Since $$\boxed{\cfrac { 1+\cos 2{ \theta } }{ 2 } =\cos ^{ 2 }{ \theta }} $$

$$\Rightarrow \cos^4(\dfrac{\pi}{8})={ \left[ \cfrac { 1+\cos { (2\pi /8) } }{ 2 } \right] }^{ 2 }={ \left[ \cfrac { 1+\cos { (\pi /4) } }{ 2 } \right] }^{ 2 }$$
Similarly
$$\cos ^{ 4 }{ \cfrac { 3\pi }{ 8 } } ={ \left[ \cfrac { 1+\cos { (3\pi /4) } }{ 2 } \right] }^{ 2 }$$
$$\cos ^{ 4 }{ \cfrac { 5\pi }{ 8 } } ={ \left[ \cfrac { 1+\cos { (5\pi /4) } }{ 2 } \right] }^{ 2 }$$
$$\cos ^{ 4 }{ \cfrac { 7\pi }{ 8 } } ={ \left[ \cfrac { 1+\cos { (7\pi /4) } }{ 2 } \right] }^{ 2 }$$
$$\therefore$$ Given expression will become
$$\cfrac { 1 }{ 4 } \left[ 2{ \left( 1+\cfrac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }+2{ \left( 1+\cfrac { 1 }{ \sqrt { 2 } } \right) }^{ 2 } \right] =\cfrac { 1 }{ 4 } \times 2(2+1)=\cfrac { 1 }{ 4 } \times 2\times 3=\cfrac { 3 }{ 2 } $$
Question 6
1 / -0

Which one of the following equations has no solution?

A
$$\csc { \theta } -\sec { \theta } =\csc { \theta } \cdot \sec { \theta } $$

B
$$\csc { \theta } \cdot \sec { \theta } =1$$

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

D
$$\sqrt { 3 } \sin { \theta } -\cos { \theta } =2$$

Solution

(a) $$\csc { \theta } -\sec { \theta } =\csc { \theta } \cdot \sec { \theta }$$
$$\Rightarrow \dfrac { \cos { \theta } -\sin { \theta } }{ \cos { \theta } \sin { \theta } } =\dfrac { 1 }{ \cos { \theta } \sin { \theta } } $$
$$\Rightarrow \cos { \theta } =1+\sin { \theta } $$
At $$\theta =0$$ above equation satisfies.
(b) $$\csc { \theta } \cdot \sec { \theta } =1$$
$$\Rightarrow \sin { \theta } \cos { \theta } =1$$
$$\Rightarrow 2\sin { \theta } \cos { \theta } =2$$
$$\Rightarrow \sin { 2\theta } =2$$
As we know $$\sin { \theta } $$ is not greater than $$1$$.
Therefore, the above equation has no solution exist.
Question 7
1 / -0

One of the principal solutions of $$ \sqrt3 \sec x = - 2 $$ is equal to :

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

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

C
$$ \dfrac {5 \pi} {6} $$

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

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

Solution

Given, $$ \sqrt3 \sec x = - 2 $$
$$ \Rightarrow \sec x = - \dfrac {2}{\sqrt3} $$
$$ \Rightarrow \cos x = - \dfrac {\sqrt3}{2} = - \cos \dfrac {\pi}{6} $$
$$ \Rightarrow x = \pi - \dfrac {\pi}{6} $$
$$ \Rightarrow x = \dfrac {5\pi}{6} $$
Question 8
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The number of solutions of the equation $$\sin\theta +\cos\theta =\sin 2\theta$$ in the interval $$[-\pi, \pi]$$ is?

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

$$\sin\theta +\cos \theta=\sin 2\theta=(\sin\theta+\cos \theta)^2-1$$

$$ \\ \Rightarrow (sin\theta+\cos\theta)^2-(\sin\theta+\cos\theta)-1=0$$

$$\therefore sin\theta+\cos\theta=\dfrac{1-\sqrt5}{2}$$

$$\Rightarrow\cos(\pi/4-\theta)=\dfrac{1-\sqrt5}{2}=-\dfrac{(\sqrt5-1)}{2}$$

Thus, there are two possible solutions for $$\theta$$ between $$-\pi $$ to $$\pi$$.
Question 9
1 / -0

Let $$X = \left \{x\epsilon \mathbb {R} : \cos (\sin x) = \sin (\cos x)\right \}$$. The number of solutions of $$X$$ is?

A
$$0$$

B
$$2$$

C
$$4$$

D
Not finite

Solution

$$\cos (\sin x)=\sin (\cos x)\implies \cos (\sin x)=\cos (\dfrac{\pi}{2}-\cos x)$$
$$\sin x=\dfrac{\pi}{2}-\cos x\implies \sin x+\cos x=\dfrac{\pi}{2}=1.72$$
But $$\sin x+\cos x$$ maximum value is $$\sqrt{2}=1.414$$
So $$\sin x+\cos x\neq \dfrac{\pi}{2}$$
Number of solutions of $$X$$ is $$0$$
Question 10
1 / -0

The number of real solutions of the equation $$2\sin 3x+\sin 7x-3=0$$ which lie in the interval $$[-2\pi, 2\pi]$$ is?

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

$$2\text{ sin}3x+\text{ sin}7x=3$$

For this to be true, both $$\text{sin}3x$$ and $$\text{sin}7x$$ must be equal to $$1$$

Now,
For $$\text{ sin}3x=1$$

$$\Rightarrow x=\dfrac{\pi}{6},\dfrac{5\pi}{6},\dfrac{9\pi}{6},\dfrac{-\pi}{2},\dfrac{-7\pi}{6},\dfrac{-11\pi}{6}$$

For $$\text{ sin}7x=1$$
$$\Rightarrow x=\dfrac{(4n+1)\pi}{14}$$ for $$n=0,1,2,3,4,5,6$$

$$=-\dfrac{(4n-1)\pi}{14}$$ for $$n=1,2,3,4,5,6,7$$

So, only possible values of $$x$$ are $$\dfrac{3\pi}{2}$$ and $$\dfrac{-\pi}{2}$$

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Trigonometric Functions Test 10

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Trigonometric Functions Test 10

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SHARING IS CARING

Question 1

$$\cfrac { 2pq }{ ab } $$

$$-\cfrac { 2pq }{ ab } $$

$$-\cfrac { pq }{ ab} $$

$$\cfrac { pq }{ ab } $$

Solution

Question 2

$$0$$

$$\cos { \left( \alpha +\beta +\gamma \right) } $$

$$3\cos { \left( \alpha +\beta +\gamma \right) } $$

$$3\sin { \left( \alpha +\beta +\gamma \right) } $$

Solution

Question 3

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

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

$$\dfrac {5\pi}{6}$$

$$\pi$$

Solution

Question 4

$$[1-\sqrt{2}, 1+\sqrt{2}]$$

$$[2-\sqrt{3}, 2+\sqrt{3}]$$

$$[0, 2+\sqrt{3}]$$

$$\left[\displaystyle\frac{1-\sqrt{2}}{2}, \frac{1+\sqrt{2}}{2}\right]$$

Solution

Question 5

$$0$$

$$1/2$$

$$3/2$$

$$1$$

Solution

Question 6

$$\csc { \theta } -\sec { \theta } =\csc { \theta } \cdot \sec { \theta } $$

$$\csc { \theta } \cdot \sec { \theta } =1$$

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

$$\sqrt { 3 } \sin { \theta } -\cos { \theta } =2$$

Solution

Question 7

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

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

$$ \dfrac {5 \pi} {6} $$

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

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

Solution

Question 8

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 9

$$0$$

$$2$$

$$4$$

Not finite

Solution

Question 10

$$1$$

$$2$$

$$3$$

$$4$$

Solution

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