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

Last updated date: 15th Mar 2023

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Answer

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Hint: We need to know the range and basic values of inverse sine function to solve this problem.

Given ${\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \frac{{3\pi }}{2}$

Splitting R.H.S.

$ \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 ${\sin ^{ - 1}}x$ is $\frac{\pi }{2}$

And here sum of three inverse of sine is $3 \times \frac{\pi }{2}$

i.e., every sine inverse function is equal to $\frac{\pi }{2}$ here

$ \Rightarrow {\sin ^{ - 1}}x = \frac{\pi }{2},{\sin ^{ - 1}}y = \frac{\pi }{2},{\sin ^{ - 1}}z = \frac{\pi }{2}$

$ \Rightarrow x = \sin \frac{\pi }{2},y = \sin \frac{\pi }{2},z = \sin \frac{\pi }{2}$

$ \Rightarrow x = 1,y = 1,z = 1$

$\therefore x + y + z = 1 + 1 + 1 = 3$

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

Given ${\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \frac{{3\pi }}{2}$

Splitting R.H.S.

$ \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 ${\sin ^{ - 1}}x$ is $\frac{\pi }{2}$

And here sum of three inverse of sine is $3 \times \frac{\pi }{2}$

i.e., every sine inverse function is equal to $\frac{\pi }{2}$ here

$ \Rightarrow {\sin ^{ - 1}}x = \frac{\pi }{2},{\sin ^{ - 1}}y = \frac{\pi }{2},{\sin ^{ - 1}}z = \frac{\pi }{2}$

$ \Rightarrow x = \sin \frac{\pi }{2},y = \sin \frac{\pi }{2},z = \sin \frac{\pi }{2}$

$ \Rightarrow x = 1,y = 1,z = 1$

$\therefore x + y + z = 1 + 1 + 1 = 3$

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

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