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Question: If \({\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \frac{{3\pi }}{2},\)then write the value ...

If sin⁡−1x+sin⁡−1y+sin⁡−1z=3π2,{\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \frac{{3\pi }}{2},then write the value of x+y+z.x + y + z.

Explanation

Solution

Hint: We need to know the range and basic values of inverse sine function to solve this problem.
Given sin⁡−1x+sin⁡−1y+sin⁡−1z=3π2{\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \frac{{3\pi }}{2}
Splitting R.H.S.
⇒sin⁡−1x+sin⁡−1y+sin⁡−1z=π2+π2+π2\Rightarrow {\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \frac{\pi }{2} + \frac{\pi }{2} + \frac{\pi }{2}
As the maximum value in the range of sin⁡−1x{\sin ^{ - 1}}x is π2\frac{\pi }{2}
And here sum of three inverse of sine is 3×π23 \times \frac{\pi }{2}
i.e., every sine inverse function is equal to π2\frac{\pi }{2} here
⇒sin⁡−1x=π2,sin⁡−1y=π2,sin⁡−1z=π2\Rightarrow {\sin ^{ - 1}}x = \frac{\pi }{2},{\sin ^{ - 1}}y = \frac{\pi }{2},{\sin ^{ - 1}}z = \frac{\pi }{2}
⇒x=sin⁡π2,y=sin⁡π2,z=sin⁡π2\Rightarrow x = \sin \frac{\pi }{2},y = \sin \frac{\pi }{2},z = \sin \frac{\pi }{2}
⇒x=1,y=1,z=1\Rightarrow x = 1,y = 1,z = 1
∴x+y+z=1+1+1=3\therefore x + y + z = 1 + 1 + 1 = 3

Note: The domain of sin inverse function is [-1, 1] and range is[−π2,π2]\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]. That means the maximum value that inverse sine function can take is π2\frac{\pi }{2}. If we observe that the given problem on the RHS values is 3π2\frac{{3\pi }}{2} and on LHS we have a sum of three inverse sine functions. So we are splitting the RHS into three π2\frac{\pi }{2}s. The sum can achieve a value of 3π2\frac{{3\pi }}{2}, if and only if each inverse sine function takes their maximum value π2\frac{\pi }{2}. This is the logic we need to keep in mind while solving these kinds of problems.