Question

Find the value of $\arcsin \left( \sin \left( 160{}^\circ \right) \right)$ .
[a] $160{}^\circ$
[b] $70{}^\circ$
[c] $-20{}^\circ$
[d] $20{}^\circ$

Answer 1

Hint: Use the fact that if $y=\arcsin x$ , then $x=\sin y$ . Assume that $\arcsin \left( \sin \left( 160{}^\circ \right) \right)=y$
Hence form an equation in y. Use the fact that if $\sin x=\sin y$ then $x=n\pi +{{\left( -1 \right)}^{n}}y$ . Use the fact that $\arcsin x\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$ . Hence find the value of $\arcsin \left( \sin \left( 160{}^\circ \right) \right)$ .

Complete step-by-step answer:
Complete step-by-step answer:
Before dwelling into the solution of the above question, we must understand how ${{\sin }^{-1}}x$ is defined even when $\sin x$ is not one-one.
We know that sinx is a periodic function.
Let us draw the graph of sinx

As is evident from the graph sinx is a repeated chunk of the graph of sinx within the interval $\left[ A,B \right]$ , and it attains all its possible values in the interval $\left[ A,C \right]$ . Here $A=\dfrac{-\pi }{2},B=\dfrac{3\pi }{2}$ and $C=\dfrac{\pi }{2}$
Hence if we consider sinx in the interval [A, C], we will lose no value attained by sinx, and at the same time, sinx will be one-one and onto.
Hence $\arcsin x$ is defined over the domain $\left[ -1,1 \right]$ , with codomain $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$ as in the domain $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$ , sinx is one-one and ${{R}_{\sin x}}=\left[ -1,1 \right]$ .
Now since $\arcsin x$ is the inverse of sinx it satisfies the fact that if $y=\arcsin x$ , then $\sin y=x$ .
So let $y=\arcsin \left( \sin \left( 160{}^\circ \right) \right)$
Hence, we have
$\sin \left( y \right)=\sin \left( 160{}^\circ \right)$
We know that if $\sin x=\sin y$ then $x=n\pi +{{\left( -1 \right)}^{n}}y,n\in \mathbb{Z}$ .
Hence, we have
$y=n\pi +{{\left( -1 \right)}^{n}}160\times \dfrac{\pi }{180}=n\pi +{{\left( -1 \right)}^{n}}\dfrac{8}{9}\pi$
Now since $\arcsin \left( x \right)\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right],$ we have $y\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$
Taking n =1, we get
$y=\pi -\dfrac{8\pi }{9}=\dfrac{\pi }{9}$
Hence $\arcsin \left( \sin \left( 160{}^\circ \right) \right)=\dfrac{\pi }{9}=\dfrac{180{}^\circ }{9}=20{}^\circ$
Hence option [d] is correct.

Note: [1] The above-specified codomain for arcsinx is called principal branch for arcsinx. We can select any branch as long as $\sin x$ is one-one and onto and Range $=\left[ -1,1 \right]$ . Whenever not mentioned, we take the principal branch for the codomain of arcsin(x).
[2] Alternative solution:
We know that \arcsin \left( \sin x \right)=\left\\{ \begin{matrix} \vdots & \vdots \\\ x & ,x\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right] \\\ \pi -x & x\in \left[ \dfrac{\pi }{2},\dfrac{3\pi }{2} \right] \\\ \vdots & \vdots \\\ \end{matrix} \right.
Since $160{}^\circ \in \left[ \dfrac{\pi }{3},\dfrac{3\pi }{2} \right]$ , we have $\arcsin \left( \sin \left( 160{}^\circ \right) \right)=180{}^\circ -160{}^\circ =20{}^\circ$
Hence option [a] is correct.
[3] Graph of arcsin(sinx):