Question

How do you evaluate ${{\sin }^{-1}}\left( \dfrac{\sqrt{2}}{2} \right)$ ?

Answer 1

Hint : We first find the principal value of x for which ${{\sin }^{-1}}\left( \dfrac{\sqrt{2}}{2} \right)$ . In that domain, equal value of the same ratio gives equal angles. We find the angle value for x. At the end we also find the general solution for the equation ${{\sin }^{-1}}\left( \dfrac{\sqrt{2}}{2} \right)$ .

Complete step-by-step answer :
It’s given that ${{\sin }^{-1}}\left( \dfrac{\sqrt{2}}{2} \right)$ . The value in fraction is $\dfrac{\sqrt{2}}{2}$ . This is equal to $\dfrac{1}{\sqrt{2}}$ .
We need to find x for which ${{\sin }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)$ .
We know that in the principal domain or the periodic value of
$-\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2}$ for $\sin x$ , if we get $\sin a=\sin b$ where $-\dfrac{\pi }{2}\le a,b\le \dfrac{\pi }{2}$ then $a=b$ .
We have the value of $\sin \left( \dfrac{\pi }{4} \right)$ as $\dfrac{1}{\sqrt{2}}$ . $-\dfrac{\pi }{2}<\dfrac{\pi }{4}<\dfrac{\pi }{2}$ .
Therefore, $\sin \left( x \right)=\dfrac{1}{\sqrt{2}}=\sin \left( \dfrac{\pi }{4} \right)$ which gives $x=\dfrac{\pi }{4}$ .
For ${{\sin }^{-1}}\left( \dfrac{\sqrt{2}}{2} \right)$ , the value of x is $x=\dfrac{\pi }{4}$ .
We also can show the solutions (primary and general) of the equation $\sin \left( x \right)=\dfrac{1}{\sqrt{2}}$ through the graph. We take $y=\sin \left( x \right)=\dfrac{1}{\sqrt{2}}$ . We got two equations $y=\sin \left( x \right)$ and $y=\dfrac{1}{\sqrt{2}}$ . We place them on the graph and find the solutions as their intersecting points.

We can see the primary solution in the interval $-\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2}$ is the point A as $x=\dfrac{\pi }{4}$ .
All the other intersecting points of the curve and the line are general solutions.
So, the correct answer is “ $x=\dfrac{\pi }{4}$ ”.

Note : Although for elementary knowledge the principal domain is enough to solve the problem. But if mentioned to find the general solution then the domain changes to $-\infty \le x\le \infty$ . In that case we have to use the formula $x=n\pi +{{\left( -1 \right)}^{n}}a$ for $\sin \left( x \right)=\sin a$ where $-\dfrac{\pi }{2}\le a\le \dfrac{\pi }{2}$ . For our given problem ${{\sin }^{-1}}\left( \dfrac{\sqrt{2}}{2} \right)$ , the general solution will be $x=n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{4}$ . Here $n\in \mathbb{Z}$ .