# trig prove

• September 12th 2012, 07:41 AM
srirahulan
trig prove
If $sin^-1x+sin^-1y=\pi/3$ Prove That $x^2+xy+y^2=\frac{4}{3}$
• September 12th 2012, 08:26 AM
kalyanram
Re: trig prove
Quote:

Originally Posted by srirahulan
If $sin^-1x+sin^-1y=\pi/3$ Prove That $x^2+xy+y^2=\frac{4}{3}$

The problem is incorrect it has to be $x^2+xy+y^2=\frac{3}{4}$. Can you solve it now.
Spoiler:

$\sin^{-1}x+\sin^{-1}y = \frac{\pi}{3}$

$\implies \sin^{-1} x = \frac{\pi}{3} - \sin^{-1} y$

$\implies x = sin \left( \frac{\pi}{3} - sin^{-1} y \right)$

$\implies x = \frac{\sqrt{3}}{2}\cos \left( sin^{-1} y \right) - \frac{1}{2} y$

$\implies x + \frac{1}{2} y = \frac{\sqrt{3}}{2} \sqrt{1- y^2}$

$\implies \left( x + \frac{1}{2} y \right)^2 = \frac{3}{4} \left(1- y^2 \right)$

$\implies x^2 + y^2 + xy = \frac{3}{4}$