Question

Prove (Cosx+cosy)^2 + (sinx-siny)^2=4cos^2(x+y/2)

Answer 1

LHS=( cosx + cosy) = 2cos(x+y/2). cos(x-y/2)

( sinx - siny ). = 2cos(x+y/2).sin(x-y/2)

Answer 2

$\textbf{To prove:}$

$(cosx+cosy)^2+(sinx-siny)^2=4\,cos^2(\frac{x+y}{2})$

$\textbf{Solution:}$

$\text{Consider,}$

$(cosx+cosy)^2+(sinx-siny)^2$

$\text{Using}$

$\boxed{(a+b)^2=a^2+b^2+2ab}$

$\boxed{(a-b)^2=a^2+b^2-2ab}$

$=(cos^2x+cos^2y+2\,cosx\,cosy)+(sin^2x+sin^2y-2\,sinx\,siny)$

$=(cos^2x+sin^2x)+(cos^2y+sin^2y)+2(cosx\,cosy-sinx\,siny)$

$\text{Using}\;\boxed{cos^2\theta+sin^2\theta=1}$

$=1+1+2(cosx\,cosy-sinx\,siny)$

$=2+2\,cos(x+y)$

$=2(1+cos(x+y))$

$\text{We know that}$

$\boxed{cosA=2\,cos^2\frac{A}{2}-1\implies\,1+cosA=2\,cos^\frac{A}{2}}$

$=2(2\,cos^2(\frac{x+y}{2}))$

$=4\,cos^2(\frac{x+y}{2})$

$\implies\bf(cosx+cosy)^2+(sinx-siny)^2=4\,cos^2(\frac{x+y}{2})$

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