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Question

Mathematics Question on Inverse Trigonometric Functions

Find the value of if sin-1x=yx=y, then

A

0≤y ≤π0\leq y \,\leq\pi

B

−π2≤y≤π2-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}

C

0<y<π0< y<\pi

D

−π2<y<π2-\frac{\pi}{2}< y<\frac{\pi}{2}

Answer

−π2≤y≤π2-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}

Explanation

Solution

It is given that sin-1x=y.x=y.
We know that the range of the principal value branch of sin-1 is [−π2,π2].\bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg].

Therefore, −π2≤y≤π2-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}