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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

The range of sin⁡−1x{\sin ^{ - 1}}x is between (−π2,π2)\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right). Equate the expression to θ\theta . Now apply the basic trigonometric identity of cos⁡θ=sin⁡(π2−θ)\cos \theta = \sin \left( {\dfrac{\pi }{2} - \theta } \right) and Simplify the expression. Substitute back the value of cos⁡3π4\cos \dfrac{{3\pi }}{4} and find the principal value.

Complete step-by-step answer:
A principal value of a function is the value selected at a point in the domain of a multiple-valued function, chosen so that the function has a single value at the point. he principal value of sin⁡−1x{\sin ^{ - 1}}x for x>0x > 0 , is the length of the arc of a unit circle centred at the origin which subtends an angle at the centre whose sine is x. For this reason sin⁡−1x{\sin ^{ - 1}}x is also denoted by arcsinx⁡\operatorname{arcsinx} .
The principal value of sin⁡−1x{\sin ^{ - 1}}x branches to,
sin⁡−1x∈(−π2,π2){\sin ^{ - 1}}x \in \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right).
Hence the principal value of the given function will be between the range(−π2,π2)\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right) .
Now we have been given the function, sin⁡−1(cos⁡3π4){\sin ^{ - 1}}\left( {\cos \dfrac{{3\pi }}{4}} \right).
Let us take the principal value of sin⁡−1(cos⁡3π4){\sin ^{ - 1}}\left( {\cos \dfrac{{3\pi }}{4}} \right) as θ\theta . Thus we get,
θ=sin⁡−1(cos⁡3π4)\theta = {\sin ^{ - 1}}\left( {\cos \dfrac{{3\pi }}{4}} \right)
We know the basic representation of cosine function as,
cos⁡θ=sin⁡(π2−θ),θ∈(−∞,∞)\cos \theta = \sin \left( {\dfrac{\pi }{2} - \theta } \right),\theta \in ( - \infty ,\infty )
Now put θ=3π4\theta = \dfrac{{3\pi }}{4}in the above expression. Hence we get,

cos⁡3π4=sin⁡(π2−3π4)=sin⁡(4π−6π8)=sin⁡(−2π8)  sin⁡(−2π8)=sin⁡(−π4)  cos⁡3π4=sin⁡(−π4)  \cos \dfrac{{3\pi }}{4} = \sin \left( {\dfrac{\pi }{2} - \dfrac{{3\pi }}{4}} \right) = \sin \left( {\dfrac{{4\pi - 6\pi }}{8}} \right) = \sin \left( { - \dfrac{{2\pi }}{8}} \right) \ \\\ \sin \left( { - \dfrac{{2\pi }}{8}} \right) = \sin \left( { - \dfrac{\pi }{4}} \right) \ \\\ \cos \dfrac{{3\pi }}{4} = \sin \left( { - \dfrac{\pi }{4}} \right) \ \\\

Now let us replace the value of cos⁡3π4\cos \dfrac{{3\pi }}{4}in the equation represented by θ\theta .

θ=sin⁡−1(cos⁡3π4)=sin⁡−1sin⁡(−π4)  θ=−π4  \theta = {\sin ^{ - 1}}\left( {\cos \dfrac{{3\pi }}{4}} \right) = {\sin ^{ - 1}}\sin \left( { - \dfrac{\pi }{4}} \right) \ \\\ \theta = - \dfrac{\pi }{4} \ \\\

Thus we got the principal value of the expression as (−π4)\left( { - \dfrac{\pi }{4}} \right).
∴sin⁡−1(cos⁡3π4)=(−π4)\therefore {\sin ^{ - 1}}\left( {\cos \dfrac{{3\pi }}{4}} \right) = \left( { - \dfrac{\pi }{4}} \right).

Note: To solve a question like these you should be familiar with the domain and range of the sine functions as well as the domain and range of the inverse sine functions. For us the range of inverse sine function is (−π2,π2)\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right) and the domain of inverse function of sine is [−1,1][ - 1,1].