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Question: If $\sin^{-1}x: [-1,1] \rightarrow [\frac{-\pi}{2}, \frac{3\pi}{2}]$ and $\cos^{-1}x: [-1,1] \righta...

If sin⁡−1x:[−1,1]→[−π2,3π2]\sin^{-1}x: [-1,1] \rightarrow [\frac{-\pi}{2}, \frac{3\pi}{2}] and cos⁡−1x:[−1,1]→[π,2π]\cos^{-1}x: [-1,1] \rightarrow [\pi, 2\pi] are defined. Then

sin⁡−1(−x)=\sin^{-1}(-x) =

A

−sin⁡−1x-\sin^{-1}x

B

π−sin⁡−1x\pi - \sin^{-1}x

C

2π−sin⁡−1x2\pi - \sin^{-1}x

D

π+sin⁡−1x\pi + \sin^{-1}x

Answer

A

Explanation

Solution

The problem defines a modified range for sin⁡−1x\sin^{-1}x and cos⁡−1x\cos^{-1}x.

Given:

  1. sin⁡−1x:[−1,1]→[−π2,3π2]\sin^{-1}x: [-1,1] \rightarrow [\frac{-\pi}{2}, \frac{3\pi}{2}]
  2. cos⁡−1x:[−1,1]→[π,2π]\cos^{-1}x: [-1,1] \rightarrow [\pi, 2\pi]

We need to find the expression for sin⁡−1(−x)\sin^{-1}(-x).

Let y=sin⁡−1(−x)y = \sin^{-1}(-x). By definition, this means sin⁡y=−x\sin y = -x. Also, yy must be in the range [−π2,3π2][\frac{-\pi}{2}, \frac{3\pi}{2}].

Let α=sin⁡−1x\alpha = \sin^{-1}x. By definition, this means sin⁡α=x\sin \alpha = x. Also, α\alpha must be in the range [−π2,3π2][\frac{-\pi}{2}, \frac{3\pi}{2}].

Now we have sin⁡y=−x=−sin⁡α\sin y = -x = -\sin \alpha.

The standard definition of sin⁡−1x\sin^{-1}x is such that sin⁡−1(−x)=−sin⁡−1x\sin^{-1}(-x) = -\sin^{-1}x. This property is derived from the fact that sin⁡x\sin x is an odd function and its principal branch is symmetric about the origin.

If the question is from a context where these properties are expected to be maintained unless explicitly stated otherwise, then A is the answer. The issue of the range not being monotonic makes it ambiguous for a strict mathematical definition. However, in the context of JEE/NEET, questions often test standard properties first.