Question

How do you evaluate \[{\cos ^{ - 1}}\left( {\cos \left( { - \dfrac{\pi }{2}} \right)} \right)\] ?

Answer 1

Hint: Given function is an inverse function along with the normal trigonometric function. We will use the even function theory and inverse function theory both here. We know that \[\cos \left( { - \theta } \right) = \cos \theta \] . Also we know that \[{\cos ^{ - 1}}\left( {\cos \left( \theta \right)} \right) = \theta \] . So we will use these two identities to solve the question above.

Complete step-by-step answer:
Given that,
 \[{\cos ^{ - 1}}\left( {\cos \left( { - \dfrac{\pi }{2}} \right)} \right)\]
We will use the even function property \[\cos \left( { - \theta } \right) = \cos \theta \]
So we can write \[\cos \left( { - \dfrac{\pi }{2}} \right) = \cos \dfrac{\pi }{2}\]
Now we can write,
 \[{\cos ^{ - 1}}\left( {\cos \left( \theta \right)} \right) = \theta \]
So we can write,
 \[{\cos ^{ - 1}}\left( {\cos \left( {\dfrac{\pi }{2}} \right)} \right) = \dfrac{\pi }{2}\]
This is the correct answer.
So, the correct answer is “\[\dfrac{\pi }{2}\] ”.

Note: Note that, this problem is simply based on the even function and the property of inverse functions. We either can assign the value of \[\cos \left( {\dfrac{\pi }{2}} \right)\] and then find the value of the inverse function.
This inverse function property is applicable for other trigonometric functions also. Even functions are those whose negative angle is also equal to positive \[f\left( { - x} \right) = f\left( x \right)\] but odd functions are those whose negative value of the angle will be negative function \[f\left( { - x} \right) = - f\left( x \right)\] .