Write the following in the simplest form: ^ -1 1+x + 1-x 2 ,0<x<1 - Vidyadip

Question

Write the following in the simplest form:
$\sin^{-1}\Big\{\frac{\sqrt{1+\text{x}}+\sqrt{1-\text{x}}}{2}\Big\},0<\text{x}<1$

✓

Answer

Let, $\text{x}=\cos\theta$
Now,
$\sin^{-1}\Big\{\frac{\sqrt{1+\text{x}}+\sqrt{1-\text{x}}}{2}\Big\}$
$=\sin^{-1}\Big\{\frac{\sqrt{1+\cos\theta}+\sqrt{1-\cos\theta}}{2}\Big\}$
$=\sin^{-1}\Bigg\{\frac{\sqrt{2\cos^2\frac{\theta}{2}}+\sqrt{2\sin^2\frac{\theta}{2}}}{2}\Bigg\}$
$=\sin^{-1}\bigg\{\frac{\cos\frac{\theta}{2}+\sin\frac{\theta}{2}}{\sqrt3}\bigg\}$
$=\sin^{-1}\Big\{\frac{1}{\sqrt2}\sin\frac{\theta}{2}+\frac{1}{\sqrt2}\cos\frac{\theta}{2}\Big\}$
$=\sin^{-1}\Big\{\sin\Big(\frac{\theta}{2}+\frac{\pi}{4}\Big)\Big\}$
$=\frac{\theta}{2}+\frac{\pi}{4}$
$=\frac{\cos^{-1}\text{x}}{2}+\frac{\pi}{4}$
$\therefore\ \sin^{-1}\Big\{\frac{\sqrt{1+\text{x}}+\sqrt{1-\text{x}}}{2}\Big\}=\frac{\cos^{-1}\text{x}}{2}+\frac{\pi}{4}$

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