Solveeit Logo
Login

Question

Question: \(\cos ^{-1}(\sqrt{1-x^{2}})\) in terms of sin^-1x...

cos⁡−1(1−x2)\cos ^{-1}(\sqrt{1-x^{2}}) in terms of sin^-1x

A

sin⁡−1x\sin^{-1}x

B

π2−sin⁡−1x\frac{\pi}{2} - \sin^{-1}x

C

π2−∣sin⁡−1x∣\frac{\pi}{2} - |\sin^{-1}x|

D

∣sin⁡−1x∣|\sin^{-1}x|

Answer

∣sin⁡−1x∣|\sin^{-1}x|

Explanation

Solution

Let y=cos⁡−1(1−x2)y = \cos^{-1}(\sqrt{1-x^2}). We know that for A≥0A \ge 0, cos⁡−1A=sin⁡−11−A2\cos^{-1}A = \sin^{-1}\sqrt{1-A^2}. Here, A=1−x2A = \sqrt{1-x^2}, which is always non-negative when defined. So, y=sin⁡−11−(1−x2)2=sin⁡−11−(1−x2)=sin⁡−1x2=sin⁡−1∣x∣y = \sin^{-1}\sqrt{1 - (\sqrt{1-x^2})^2} = \sin^{-1}\sqrt{1 - (1-x^2)} = \sin^{-1}\sqrt{x^2} = \sin^{-1}|x|.

Now, consider the properties of sin⁡−1∣x∣\sin^{-1}|x|: If x≥0x \ge 0, then ∣x∣=x|x|=x, so sin⁡−1∣x∣=sin⁡−1x\sin^{-1}|x| = \sin^{-1}x. If x<0x < 0, then ∣x∣=−x|x|=-x, so sin⁡−1∣x∣=sin⁡−1(−x)\sin^{-1}|x| = \sin^{-1}(-x). Since sin⁡−1\sin^{-1} is an odd function, sin⁡−1(−x)=−sin⁡−1x\sin^{-1}(-x) = -\sin^{-1}x. Thus, sin⁡−1∣x∣\sin^{-1}|x| can be written as sin⁡−1x\sin^{-1}x for x≥0x \ge 0 and −sin⁡−1x-\sin^{-1}x for x<0x < 0. This is exactly the definition of ∣sin⁡−1x∣|\sin^{-1}x|. Therefore, cos⁡−1(1−x2)=∣sin⁡−1x∣\cos^{-1}(\sqrt{1-x^2}) = |\sin^{-1}x|.