Question

Prove that:
$sin ^{ - 1} \frac{ \sqrt{1 + x} + \sqrt{1 - x} }{2} = \frac{\pi}{4} + \frac{cos ^{ - 1} x}{2}$
Where
$0 < x < 1$
​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\sf{sin^{-1}\left(\dfrac{\sqrt{1+x}+\sqrt{1-x}}{2}\right)}$

$\bf{\bigstar\,\,Put\,\,\,x=cos(2\,\theta)\,\,\,\,\,\,,0<cos(2\theta)<1}$

So,

$\sf{sin^{-1}\left(\dfrac{\sqrt{1+cos(2\,\theta)}+\sqrt{1-cos(2\,\theta)}}{2}\right)}$

$\sf{=sin^{-1}\left(\dfrac{\sqrt{2\,cos^2(\theta)}+\sqrt{sin^2(\theta)}}{2}\right)}$

$\sf{=sin^{-1}\left\{\dfrac{\sqrt{2}\left(cos(\theta)+sin(\theta)\right)}{2}\right\}}$

$\sf{=sin^{-1}\left\{\dfrac{cos(\theta)+sin(\theta)}{\sqrt{2}}\right\}}$

$\sf{=sin^{-1}\left\{\dfrac{1}{\sqrt{2}}\,cos(\theta)+\dfrac{1}{\sqrt{2}}\,sin(\theta)\right\}}$

$\sf{=sin^{-1}\left\{sin\left(\dfrac{\pi}{4}+\theta\right)\right\}}$

$\sf{=\dfrac{\pi}{4}+\theta}$

$\sf{=\dfrac{\pi}{4}+\dfrac{1}{2}\cdot(2\theta)}$

$\sf{=\dfrac{\pi}{4}+\dfrac{1}{2}\cdot\,cos^{-1}(x)}$

$\sf{=\dfrac{\pi}{4}+\dfrac{cos^{-1}(x)}{2}}$