Question

How do you evaluate $\arcsin (\dfrac{\sqrt{2}}{2})$ ?

Answer 1

In the given question, we have been given an inverse trigonometric function, and knowledge of trigonometric function is very important to solve these types of questions as they are simply reverse of each other.

Complete step by step solution:
First of all as we know $\dfrac{\sqrt{2}}{2}$ can be written as $\dfrac{1}{\sqrt{2}}$.
Hence, $\arcsin (\dfrac{\sqrt{2}}{2})$ can be written as $\arcsin (\dfrac{1}{\sqrt{2}})$.
First, recognize that the domain of the function $\arcsin (x)$ is $[-1,1]$.
Now as we know, $\arcsin (x)$and $\sin (x)$ are inverse functions. Which simply means that $\arcsin (\sin (x))=x$ and $\sin (\arcsin (x))=x$. We can see tangent and arctangent as "undoing" one another.
For example, we know that $\tan (\dfrac{\pi }{4})=1$ which means that $\arctan (1)=\dfrac{\pi }{4}$.
So, for the value of $\arcsin (\dfrac{1}{\sqrt{2}})$ is essentially asking, the angle whose tan gives $\dfrac{1}{\sqrt{2}}$.
Since, $\sin (\dfrac{\pi }{4})=\dfrac{1}{\sqrt{2}}$,
We can reverse this with the arcsin function to see that

$\arcsin (\dfrac{\sqrt{2}}{2})=\dfrac{\pi }{4}$.

Note:

1. In $\sin (x)$, x is an angle.
2. In $\arcsin (x)$, x is the value of the tangent function.
3. We can say that $\sin (x)$ and $\arcsin (x)$ are opposite to each other and this knowledge can be utilized in solving such questions.