# If, Cosx+Cosy=a and Sinx+Siny=b So Prove That ?? Sin2x+Sin2y = 2ab{1-(2/(a^2+b^2))}

May 31, 2018

Given that,

$\rightarrow \cos x + c o y = a$....[1]

$\rightarrow \sin x + \sin y = b$....[2]

Squaring and adding [1] and [2], we get,

$\rightarrow {\cos}^{2} x + 2 \cos x \cos y + {\cos}^{2} y + {\sin}^{2} x + 2 \sin x \sin y + {\sin}^{2} y = {a}^{2} + {b}^{2}$

$\rightarrow 2 + 2 \left(\cos x \cos y + \sin x \sin y\right) = {a}^{2} + {b}^{2}$

$\rightarrow 2 \left(1 + \cos \left(x - y\right)\right) = {a}^{2} + {b}^{2}$

$\rightarrow \cos \left(x - y\right) = \frac{{a}^{2} + {b}^{2}}{2} - 1$

Dividing equation [1] by [2], we get,

$\rightarrow \frac{\cos x + \cos y}{\sin x + \sin y} = \frac{a}{b}$

$\rightarrow \frac{2 \cos \left(\frac{x + y}{2}\right) \cos \left(\frac{x - y}{2}\right)}{2 \sin \left(\frac{x + y}{2}\right) \cos \left(\frac{x - y}{2}\right)} = \frac{a}{b}$

$\rightarrow \cot \left(\frac{x + y}{2}\right) = \frac{a}{b}$

$\rightarrow \tan \left(\frac{x + y}{2}\right) = \frac{b}{a}$

$\rightarrow \frac{x + y}{2} = {\tan}^{- 1} \left(\frac{b}{a}\right)$

$\rightarrow x + y = 2 {\tan}^{- 1} \left(\frac{b}{a}\right)$

As, $2 {\tan}^{- 1} x = {\sin}^{- 1} \left(\frac{2 x}{1 + {x}^{2}}\right)$,we have,

$\rightarrow x + y = {\sin}^{- 1} \left(\frac{2 \cdot \left(\frac{b}{a}\right)}{1 + {\left(\frac{b}{a}\right)}^{2}}\right) = {\sin}^{- 1} \left(\frac{2 a b}{{a}^{2} + {b}^{2}}\right)$

$\rightarrow \sin \left(x + y\right) = \frac{2 a b}{{a}^{2} + {b}^{2}}$

Now,

$L H S = \sin 2 x + \sin 2 y$

$= 2 \sin \left(x + y\right) \cdot \cos \left(x - y\right)$

$= 2 \left[\frac{2 a b}{{a}^{2} + {b}^{2}}\right] \left[\frac{{a}^{2} + {b}^{2}}{2} - 1\right]$

$= 2 a b \left[\frac{2}{{a}^{2} + {b}^{2}} \cdot \frac{{a}^{2} + {b}^{2}}{2} - \frac{2}{{a}^{2} + {b}^{2}}\right]$

$= 2 a b \left[1 - \frac{2}{{a}^{2} + {b}^{2}}\right] = R H S$