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Question: Find the value of \({{\sin }^{-1}}\left( \sin 10 \right)\) is: A) 10 B) \(10-3\pi \) C) \(3\p...

Find the value of sin⁡−1(sin⁡10){{\sin }^{-1}}\left( \sin 10 \right) is:
A) 10
B) 10−3π10-3\pi
C) 3π−103\pi -10
D) -10

Explanation

Solution

Hint: Take y=sin⁡−1(sin⁡10)y={{\sin }^{-1}}\left( \sin 10 \right) and take sin of both sides of the equation and you will get sin⁡10=sin⁡y\sin 10=\sin y and then we use general solution of sin⁡y=sin⁡x\sin y=\sin x : y=2nπ+xy=2n\pi +x and get a value of y which lie in the range [−π2,π2]\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right] as range of sin⁡−1(sin⁡x){{\sin }^{-1}}\left( \sin x \right) is [−π2,π2]\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right].

Complete step-by-step answer:
Let us assume sin⁡−1(sin⁡10)=x{{\sin }^{-1}}\left( \sin 10 \right)=x
We know that the range of f(x)=sin⁡−1xf\left( x \right)={{\sin }^{-1}}x is [−π2,π2]\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right].
Here sin⁡−1(sin⁡10)=x{{\sin }^{-1}}\left( \sin 10 \right)=x so x∈[−π2,π2]x\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right].
Taking sin of both sides of equations, we will get
⇒sin⁡10=sin⁡x\Rightarrow \sin 10=\sin x.
We know the general solution for the equation. “sin⁡y=sin⁡x\sin y=\sin x “ is y=x+2κπy=x+2\kappa \pi , where κ\kappa is any integer.
So, the general solution for sin⁡x=sin⁡10\sin x=\sin 10 will be x=10+2nπx=10+2n\pi , where n is any integer.
We have to find the value of x such that x=sin⁡−1(sin⁡10)x={{\sin }^{-1}}\left( \sin 10 \right) and x∈[−π2,π2]x\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right] .
For x=sin⁡−1(sin⁡10)x={{\sin }^{-1}}\left( \sin 10 \right) , we have got x=10+2nπx=10+2n\pi , where ‘n’ is any integer and we know that x can belong to [−π2,π2]\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right].
So, we have to find the value of “10+2nπ10+2n\pi “ which lies in [−π2,π2]\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right], where ‘n’ is any integer.
sin⁡θ=1\sin \theta =1 .
As π=3.14,π2=1.57\pi =3.14,\dfrac{\pi }{2}=1.57
So, (2nπ+10)\left( 2n\pi +10 \right) should belong to [−1.57,1.57]\left[ -1.57,1.57 \right]
We know π=3.14\pi =3.14
So, 3π=3(3.14)=9.423\pi =3\left( 3.14 \right)=9.42
And 4π=4(3.14)=12.564\pi =4\left( 3.14 \right)=12.56 .
So, 10 will lie between 3π3\pi and 4π4\pi
i.e. 3π<10<4π3\pi <10<4\pi .
we know sin⁡10=sin⁡(2nπ+10),n∈I\sin 10=\sin \left( 2n\pi +10 \right),n\in I
Let us take n=−1n=-1
2nπ+10=10−2π =10−2(3.14) =10−6.28 =3.72 \begin{aligned} & 2n\pi +10=10-2\pi \\\ & =10-2\left( 3.14 \right) \\\ & =10-6.28 \\\ & =3.72 \\\ \end{aligned}
Hence, sin⁡(10)=sin⁡(3.72)=sin⁡(10−2π)\sin \left( 10 \right)=\sin \left( 3.72 \right)=\sin \left( 10-2\pi \right) ……………….(1)
But 3.72 also don’t lie between [−π2,π2]\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right] i.e. (−1.57,1.57)\left( -1.57,1.57 \right)
Now, we know that sin⁡(π−θ)=sin⁡θ\sin \left( \pi -\theta \right)=\sin \theta .
Taking θ=(10−2π)\theta =\left( 10-2\pi \right) , we will get
sin⁡(π−(10−2π))=sin⁡(10−2π) ⇒sin⁡(π−10+2π)=sin⁡(10−2π) \begin{aligned} & \sin \left( \pi -\left( 10-2\pi \right) \right)=\sin \left( 10-2\pi \right) \\\ & \Rightarrow \sin \left( \pi -10+2\pi \right)=\sin \left( 10-2\pi \right) \\\ \end{aligned}
⇒sin⁡(3π−10)=sin⁡(10−2π)\Rightarrow \sin \left( 3\pi -10 \right)=\sin \left( 10-2\pi \right) ………….(2)
From eq (1) and (2)
sin⁡(10)=sin⁡(3π−10)\sin \left( 10 \right)=\sin \left( 3\pi -10 \right)
And
3π−10=3(3.14)−10 =9.42−10 =0.58 0.58∈[−1.57,1.57] ⇒(3π−10)∈[−π2,π2] \begin{aligned} & 3\pi -10=3\left( 3.14 \right)-10 \\\ & =9.42-10 \\\ & =0.58 \\\ & 0.58\in \left[ -1.57,1.57 \right] \\\ & \Rightarrow \left( 3\pi -10 \right)\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right] \\\ \end{aligned}
So, we have got an angle (3π−10)\left( 3\pi -10 \right) such that sin⁡(3π−10)=sin⁡10\sin \left( 3\pi -10 \right)=\sin 10 and 3π−10∈[−π2,π2]3\pi -10\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right] .
Hence the value of sin⁡−1(sin⁡10)=(3π−10){{\sin }^{-1}}\left( \sin 10 \right)=\left( 3\pi -10 \right) and option (C) is the correct answer.

Note: Students can do mistake by directly taking sin⁡−1(sin⁡10)=10{{\sin }^{-1}}\left( \sin 10 \right)=10 , but be careful that range of y=sin⁡−1(sin⁡x)y={{\sin }^{-1}}\left( \sin x \right) will be [−π2,π2]\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right] and 10 doesn’t belong to this range.