Question

Find the Principal value of \[{\sin ^{ - 1}}\left( {\cos \dfrac{{3\pi }}{4}} \right)\]

Answer 1

Hint- The relation between the angle and two length’s sides of a right-angled triangle is known as the trigonometric functions. The basic and widely used trigonometric functions are $\sin x,\cos x,\tan x,\sec x,\cot x,{\text{cosec }}x$. In this question two functions are involved, first is an inverse-trigonometric function which gives the angle and the second is the trigonometric function which gives the length of the side. First of all, we need to determine the value of \[\cos \dfrac{{3\pi }}{4}\] and then for that value find the corresponding\[\sin \theta \]and after it apply the formula \[{\sin ^{ - 1}}\left( {\sin x} \right) = x\] where \[x \in \left[ { - \dfrac{\pi }{2},\,\dfrac{\pi }{2}} \right]\]

Complete step by step solution:
In the expression \[{\sin ^{ - 1}}\left( {\cos \dfrac{{3\pi }}{4}} \right)\] determine the value of \[\cos \dfrac{{3\pi }}{4}\]:
\[
  \cos \dfrac{{3\pi }}{4} = \cos {135^0} \\
   = \cos (90 + 45) \\
   = - \dfrac{1}{{\sqrt 2 }} \\
 \]
So, we are going to substitute \[\cos \dfrac{{3\pi }}{4} = - \dfrac{1}{{\sqrt 2 }}\] in the given expression.
Now, we need to solve \[{\sin ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right)\]
As we know that\[ - \dfrac{1}{{\sqrt 2 }}\]can be written as \[\sin \left( { - \dfrac{\pi }{4}} \right)\] or, \[\sin \left( { - \dfrac{\pi }{4}} \right) = - \dfrac{1}{{\sqrt 2 }}\]
So, substitute \[ - \dfrac{1}{{\sqrt 2 }}\]equal to \[\sin \left( { - \dfrac{\pi }{4}} \right)\] as:
Now our expression is \[{\sin ^{ - 1}}\left( {\sin \left( { - \dfrac{\pi }{4}} \right)} \right)\]
Now we are going to apply the below formula as we discussed in the above hint part
\[{\sin ^{ - 1}}\left( {\sin x} \right) = x\] for \[x \in \left[ { - \dfrac{\pi }{2},\,\dfrac{\pi }{2}} \right]\] , here $x$ is called principal value.
So \[{\sin ^{ - 1}}\left( {\sin \left( { - \dfrac{\pi }{4}} \right)} \right)\] will be equal to \[ - \dfrac{\pi }{4}\]
If we compare with the above formula then we get \[x = - \dfrac{\pi }{4}\]
Here we can clearly see that \[ - \dfrac{\pi }{4} \in \left[ { - \dfrac{\pi }{2},\,\dfrac{\pi }{2}} \right]\]
Hence, Principal value of: \[{\sin ^{ - 1}}\left( {\cos \dfrac{{3\pi }}{4}} \right)\] is \[ - \dfrac{\pi }{4}\]
Hence after following each and every step given in the hint part, we obtained our final answer.

Note: It should be noted here that the principal value of the inverse-trigonometric function should be in the predefined range. We can write \[{\sin ^{ - 1}}\left( {\sin x} \right) = x\] only when \[x \in \left[ { - \dfrac{\pi }{2},\,\dfrac{\pi }{2}} \right]\] it means \[{\sin ^{ - 1}}\left( {\sin \left( {\dfrac{{3\pi }}{4}} \right)} \right)\]cannot be written equal to \[\dfrac{{3\pi }}{4}\] because \[ - \dfrac{\pi }{4} \in \left[ { - \dfrac{\pi }{2},\,\dfrac{\pi }{2}} \right]\]
Here we need to put the correct value of \[\cos \dfrac{{3\pi }}{4}\]that is \[ - \dfrac{\pi }{4}\] as \[\cos \dfrac{{3\pi }}{4}\]can be found by
\[\cos \dfrac{{3\pi }}{4}\]can be written as \[\cos \left( {\pi - \dfrac{\pi }{4}} \right) = - \cos \dfrac{\pi }{4}\]that is equal to \[ - \dfrac{1}{{\sqrt 2 }}\].