If, $cos x + cos y = a$ $Cosx+Cosy=a$ and $sin x + sin y = b$ $Sinx+Siny=b$ So Prove That ?? $sin 2 x + sin 2 y = 2 a b {1 - (\frac{2}{a^{2} + b^{2}})}$ $Sin2x+Sin2y = 2ab{1-(2/(a^2+b^2))}$

Question

If, $cos x + cos y = a$ $Cosx+Cosy=a$ and $sin x + sin y = b$ $Sinx+Siny=b$ So Prove That ?? $sin 2 x + sin 2 y = 2 a b {1 - (\frac{2}{a^{2} + b^{2}})}$ $Sin2x+Sin2y = 2ab{1-(2/(a^2+b^2))}$

1 Answer

Impact of this question

15428 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-31T17:45:12

Given that,

$\to cos x + c o y = a$ $rarrcosx+coy=a$ ....[1]

$\to sin x + sin y = b$ $rarrsinx+siny=b$ ....[2]

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

$\to {cos}^{2} x + 2 cos x cos y + {cos}^{2} y + {sin}^{2} x + 2 sin x sin y + {sin}^{2} y = a^{2} + b^{2}$ $rarrcos^2x+2cosxcosy+cos^2y+sin^2x+2sinxsiny+sin^2y=a^2+b^2$

$\to 2 + 2 (cos x cos y + sin x sin y) = a^{2} + b^{2}$ $rarr2+2(cosxcosy+sinxsiny)=a^2+b^2$

$\to 2 (1 + cos (x - y)) = a^{2} + b^{2}$ $rarr2(1+cos(x-y))=a^2+b^2$

$\to cos (x - y) = \frac{a^{2} + b^{2}}{2} - 1$ $rarrcos(x-y)=(a^2+b^2)/2-1$

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

$\to \frac{cos x + cos y}{sin x + sin y} = \frac{a}{b}$ $rarr(cosx+cosy)/(sinx+siny)=a/b$

$\to \frac{2 cos (\frac{x + y}{2}) cos (\frac{x - y}{2})}{2 sin (\frac{x + y}{2}) cos (\frac{x - y}{2})} = \frac{a}{b}$ $rarr(2cos((x+y)/2)cos((x-y)/2))/(2sin((x+y)/2)cos((x-y)/2))=a/b$

$\to cot (\frac{x + y}{2}) = \frac{a}{b}$ $rarrcot((x+y)/2)=a/b$

$\to tan (\frac{x + y}{2}) = \frac{b}{a}$ $rarrtan((x+y)/2)=b/a$

$\to \frac{x + y}{2} = {tan}^{- 1} (\frac{b}{a})$ $rarr(x+y)/2=tan^(-1)(b/a)$

$\to x + y = 2 {tan}^{- 1} (\frac{b}{a})$ $rarrx+y=2tan^(-1)(b/a)$

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

$\to x + y = {sin}^{- 1} ⎛ ⎜ ⎜ ⎝ \frac{2 \cdot (\frac{b}{a})}{1 + {(\frac{b}{a})}^{2}} ⎞ ⎟ ⎟ ⎠ = {sin}^{- 1} (\frac{2 a b}{a^{2} + b^{2}})$ $rarrx+y=sin^(-1)((2*(b/a))/(1+(b/a)^2))=sin^(-1)((2ab)/(a^2+b^2))$

$\to sin (x + y) = \frac{2 a b}{a^{2} + b^{2}}$ $rarrsin(x+y)=(2ab)/(a^2+b^2)$

Now,

$L H S = sin 2 x + sin 2 y$ $LHS=sin2x+sin2y$

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

$= 2 [\frac{2 a b}{a^{2} + b^{2}}] [\frac{a^{2} + b^{2}}{2} - 1]$ $=2[(2ab)/(a^2+b^2)][(a^2+b^2)/2-1]$

$= 2 a b [\frac{2}{a^{2} + b^{2}} \cdot \frac{a^{2} + b^{2}}{2} - \frac{2}{a^{2} + b^{2}}]$ $=2ab[2/(a^2+b^2)*(a^2+b^2)/2-2/(a^2+b^2)]$

$= 2 a b [1 - \frac{2}{a^{2} + b^{2}}] = R H S$ $=2ab[1-2/(a^2+b^2)]=RHS$

Science

Math

Humanities

... and beyond

If, $cos x + cos y = a$ $Cosx+Cosy=a$ and $sin x + sin y = b$ $Sinx+Siny=b$ So Prove That ?? $sin 2 x + sin 2 y = 2 a b {1 - (\frac{2}{a^{2} + b^{2}})}$ $Sin2x+Sin2y = 2ab{1-(2/(a^2+b^2))}$

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

If, cosx+cosy=aCosx+Cosy=a and sinx+siny=bSinx+Siny=b So Prove That ?? sin2x+sin2y=2ab{1−(2a2+b2)}Sin2x+Sin2y = 2ab{1-(2/(a^2+b^2))}

1 Answer

Related questions

Impact of this question

If, $cos x + cos y = a$ $Cosx+Cosy=a$ and $sin x + sin y = b$ $Sinx+Siny=b$ So Prove That ?? $sin 2 x + sin 2 y = 2 a b {1 - (\frac{2}{a^{2} + b^{2}})}$ $Sin2x+Sin2y = 2ab{1-(2/(a^2+b^2))}$