Question

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

Answer 1

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

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

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