Prove that, sin2x+sin2y=2sin(x+y)cos(x-y).

help please :)

Thanks in advance.

Printable View

- May 2nd 2010, 12:27 PMHelloooIdentity
Prove that, sin2x+sin2y=2sin(x+y)cos(x-y).

help please :)

Thanks in advance. - May 2nd 2010, 05:53 PMsa-ri-ga-ma
sin(x+y) = sinx*cosy + cosx*siny

cos(x-y) = cosx*cosy + sinx*siny

sin(x+y)*cos(x-y) = sinxcosy*cosxcosy + sinxcosy*sinxsiny + cosxsiny*cosxcosy + cosxsiny*sinxsiny

= sinxcosx*cos^2y + sinycosy*sin^2x + sinycosy*cos^2x + sinxcosx*sin^2y

So

2sin(x+y)cos(x-y) = 2[sinxcosx(sin^2y + cos^2y) + sinycosy(sin^2x + sin^2y)]

= 2sinxcosx + 2sinycosy

= sin2x + sin2y - May 2nd 2010, 06:09 PMKrizalid
we can short the solution by using the fact that $\displaystyle \sin (a)\cos (b)=\frac{\sin (a+b)+\sin (a-b)}{2},$ in our case put $\displaystyle (a,b)=(x+y,x-y)$ and the rest follows trivially.