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Question: How do you evaluate \({\sin ^{ - 1}}\left( {\cos \left( {\dfrac{{3\pi }}{4}} \right)} \right)\) with...

How do you evaluate sin⁡−1(cos⁡(3π4)){\sin ^{ - 1}}\left( {\cos \left( {\dfrac{{3\pi }}{4}} \right)} \right) without a calculator?

Explanation

Solution

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(cos⁡(3π4)){\sin ^{ - 1}}\left( {\cos \left( {\dfrac{{3\pi }}{4}} \right)} \right)
Now separate the terms inside the brackets and let us write it as θ\theta equals.
Therefore we get
⇒θ=sin⁡−1(cos⁡(π−π4))\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
⇒θ=sin⁡−1(−cos⁡π4)\Rightarrow \theta = {\sin ^{ - 1}}\left( { - \cos \dfrac{\pi }{4}} \right)
Now the exact value of cos⁡π4=22\cos \dfrac{\pi }{4} = \dfrac{{\sqrt 2 }}{2}
Substitute that value in the above equation, we get
⇒θ=sin⁡−1(−22)\Rightarrow \theta = {\sin ^{ - 1}}\left( { - \dfrac{{\sqrt 2 }}{2}} \right)
Now if we bring them sin⁡−1{\sin ^{ - 1}} to the LHS we will get
⇒sin⁡θ=(−22)\Rightarrow \sin \theta = \left( { - \dfrac{{\sqrt 2 }}{2}} \right)
We know that for π4\dfrac{\pi }{4} of sin⁡\sin value we will get 22\dfrac{{\sqrt 2 }}{2}
Therefore the value for sin⁡−1(cos⁡(3π4)){\sin ^{ - 1}}\left( {\cos \left( {\dfrac{{3\pi }}{4}} \right)} \right)is −π4 - \dfrac{\pi }{4}

Note:
Misconception you will have while solving this problem:
1)1) The expression sin⁡−1(x){\sin ^{ - 1}}\left( x \right) is not the same as 1sin⁡x\dfrac{1}{{\sin x}}. In other words, the −1 - 1 is not an exponent. Instead, it simply means inverse function.
2)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)3) We can also express the inverse sine as arcsin⁡\arcsin and the inverse cosine as arccos⁡\arccos and the inverse tangent asarctan⁡\arctan . They both represent the inverse trigonometric functions only. This notation is common in computer programming languages, and less common in mathematics.