Question

How do you evaluate ${\sin ^{ - 1}}\left( {\cos \left( {\dfrac{{3\pi }}{4}} \right)} \right)$ without a calculator?

Answer 1

Hint: To solve this question, you should know about inverse trigonometric functions.
Trigonometric functions are sin, cos, tan, cosec, cot and sec. Finding the inverse value for these functions are called as Inverse trigonometric functions. In this given case you have to first solve the component inside the bracket and then should find the inverse of those trigonometric functions.

Complete step by step answer:
The given function is ${\sin ^{ - 1}}\left( {\cos \left( {\dfrac{{3\pi }}{4}} \right)} \right)$
Now separate the terms inside the brackets and let us write it as $\theta $ equals.
Therefore we get
$ \Rightarrow \theta = {\sin ^{ - 1}}\left( {\cos \left( {\pi - \dfrac{\pi }{4}} \right)} \right)$
Now apply the reference angle by finding the angle with equivalent trigonometric values in the first quadrant.
Make the expression negative because cosine is negative in the second quadrant.
Therefore we can write it as
$ \Rightarrow \theta = {\sin ^{ - 1}}\left( { - \cos \dfrac{\pi }{4}} \right)$
Now the exact value of $\cos \dfrac{\pi }{4} = \dfrac{{\sqrt 2 }}{2}$
Substitute that value in the above equation, we get
$ \Rightarrow \theta = {\sin ^{ - 1}}\left( { - \dfrac{{\sqrt 2 }}{2}} \right)$
Now if we bring them ${\sin ^{ - 1}}$ to the LHS we will get
$ \Rightarrow \sin \theta = \left( { - \dfrac{{\sqrt 2 }}{2}} \right)$
We know that for $\dfrac{\pi }{4}$ of $\sin $ value we will get $\dfrac{{\sqrt 2 }}{2}$
Therefore the value for ${\sin ^{ - 1}}\left( {\cos \left( {\dfrac{{3\pi }}{4}} \right)} \right)$is $ - \dfrac{\pi }{4}$

Note:
Misconception you will have while solving this problem:
$1)$ The expression ${\sin ^{ - 1}}\left( x \right)$ is not the same as $\dfrac{1}{{\sin x}}$. In other words, the $ - 1$ is not an exponent. Instead, it simply means inverse function.
$2)$ While solving for some angle you should be very careful about the quadrant in which the angle is present. Because the sign of the angle is one main criteria that gives you the correct solution
$3)$ We can also express the inverse sine as $\arcsin $and the inverse cosine as $\arccos $ and the inverse tangent as$\arctan $. They both represent the inverse trigonometric functions only. This notation is common in computer programming languages, and less common in mathematics.