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

Find the Principal value of sin⁡−1(cos⁡3π4){\sin ^{ - 1}}\left( {\cos \dfrac{{3\pi }}{4}} \right)

Explanation

Solution

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,cosec x\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⁡3π4\cos \dfrac{{3\pi }}{4} and then for that value find the correspondingsin⁡θ\sin \theta and after it apply the formula sin⁡−1(sin⁡x)=x{\sin ^{ - 1}}\left( {\sin x} \right) = x where x∈[−π2, π2]x \in \left[ { - \dfrac{\pi }{2},\,\dfrac{\pi }{2}} \right]

Complete step by step solution:
In the expression sin⁡−1(cos⁡3π4){\sin ^{ - 1}}\left( {\cos \dfrac{{3\pi }}{4}} \right) determine the value of cos⁡3π4\cos \dfrac{{3\pi }}{4}:

cos⁡3π4=cos⁡1350 =cos⁡(90+45) =−12  \cos \dfrac{{3\pi }}{4} = \cos {135^0} \\\ = \cos (90 + 45) \\\ = - \dfrac{1}{{\sqrt 2 }} \\\

So, we are going to substitute cos⁡3π4=−12\cos \dfrac{{3\pi }}{4} = - \dfrac{1}{{\sqrt 2 }} in the given expression.
Now, we need to solve sin⁡−1(−12){\sin ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right)
As we know that−12 - \dfrac{1}{{\sqrt 2 }}can be written as sin⁡(−π4)\sin \left( { - \dfrac{\pi }{4}} \right) or, sin⁡(−π4)=−12\sin \left( { - \dfrac{\pi }{4}} \right) = - \dfrac{1}{{\sqrt 2 }}
So, substitute −12 - \dfrac{1}{{\sqrt 2 }}equal to sin⁡(−π4)\sin \left( { - \dfrac{\pi }{4}} \right) as:
Now our expression is sin⁡−1(sin⁡(−π4)){\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(sin⁡x)=x{\sin ^{ - 1}}\left( {\sin x} \right) = x for x∈[−π2, π2]x \in \left[ { - \dfrac{\pi }{2},\,\dfrac{\pi }{2}} \right] , here xx is called principal value.
So sin⁡−1(sin⁡(−π4)){\sin ^{ - 1}}\left( {\sin \left( { - \dfrac{\pi }{4}} \right)} \right) will be equal to −π4 - \dfrac{\pi }{4}
If we compare with the above formula then we get x=−π4x = - \dfrac{\pi }{4}
Here we can clearly see that −π4∈[−π2, π2] - \dfrac{\pi }{4} \in \left[ { - \dfrac{\pi }{2},\,\dfrac{\pi }{2}} \right]
Hence, Principal value of: sin⁡−1(cos⁡3π4){\sin ^{ - 1}}\left( {\cos \dfrac{{3\pi }}{4}} \right) is −π4 - \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(sin⁡x)=x{\sin ^{ - 1}}\left( {\sin x} \right) = x only when x∈[−π2, π2]x \in \left[ { - \dfrac{\pi }{2},\,\dfrac{\pi }{2}} \right] it means sin⁡−1(sin⁡(3π4)){\sin ^{ - 1}}\left( {\sin \left( {\dfrac{{3\pi }}{4}} \right)} \right)cannot be written equal to 3π4\dfrac{{3\pi }}{4} because −π4∈[−π2, π2] - \dfrac{\pi }{4} \in \left[ { - \dfrac{\pi }{2},\,\dfrac{\pi }{2}} \right]
Here we need to put the correct value of cos⁡3π4\cos \dfrac{{3\pi }}{4}that is −π4 - \dfrac{\pi }{4} as cos⁡3π4\cos \dfrac{{3\pi }}{4}can be found by
cos⁡3π4\cos \dfrac{{3\pi }}{4}can be written as cos⁡(π−π4)=−cos⁡π4\cos \left( {\pi - \dfrac{\pi }{4}} \right) = - \cos \dfrac{\pi }{4}that is equal to −12 - \dfrac{1}{{\sqrt 2 }}.