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Question: Solve \({\sin ^{ - 1}}\left( {\cos {\text{ x}}} \right)\)....

Solve sin⁡−1(cos⁡ x){\sin ^{ - 1}}\left( {\cos {\text{ x}}} \right).

Explanation

Solution

Hint- Here we will proceed by using the one of the property of inverse trigonometric function i.e. (cos⁡ θ) = sin(π2−θ) \left( {\cos {\text{ }}\theta } \right){\text{ = sin}}\left( {\dfrac{\pi }{2} - \theta } \right){\text{ }}. Then we will multiply it with sin⁡−1{\sin ^{ - 1}} to get the required result.

Complete step-by-step answer:

As we know that,
⇒(cos⁡ θ) = sin(π2−θ) \Rightarrow \left( {\cos {\text{ }}\theta } \right){\text{ = sin}}\left( {\dfrac{\pi }{2} - \theta } \right){\text{ }}
Therefore,
⇒sin⁡−1(cos⁡ x)\Rightarrow {\sin ^{ - 1}}\left( {\cos {\text{ x}}} \right)
⇒sin⁡−1(sin⁡(π2+θ))\Rightarrow {\sin ^{ - 1}}\left( {\sin \left( {\dfrac{\pi }{2} + \theta } \right)} \right)
Also we know that,
f−1(f(x))=x{f^{ - 1}}\left( {f\left( x \right)} \right) = x
Which implies that-
=(π2+θ)= \left( {\dfrac{\pi }{2} + \theta } \right)
Hence the answer is (π2+θ)\left( {\dfrac{\pi }{2} + \theta } \right)

Note- In order to solve this type of questions, we must know all the inverse trigonometric functions which are sin⁡−1θ,cos⁡−1θ,tan⁡−1θ,cos⁡ec−1θ,sec⁡−1θ,cot⁡−1θ{\sin ^{ - 1}}\theta ,{\cos ^{ - 1}}\theta ,{\tan ^{ - 1}}\theta ,\cos e{c^{ - 1}}\theta ,{\sec ^{ - 1}}\theta ,{\cot ^{ - 1}}\theta as here also we used one of its function i.e. (cos⁡ θ) = sin(π2−θ) \left( {\cos {\text{ }}\theta } \right){\text{ = sin}}\left( {\dfrac{\pi }{2} - \theta } \right){\text{ }}so that we can also tackle similar type of questions.