# prove that sin2x+sin2y=2sin(x+y)cos(x-y)

• Mar 29th 2009, 01:54 PM
skeske1234
prove that sin2x+sin2y=2sin(x+y)cos(x-y)
so i am stuck on this question and have been for a while

this is my work that i have:
can someone show me what to do after this?

2sinxcosx+2sinycosy=2(sinxcosy+cosxsiny)(cosxcosy+ sinxsiny)
2sinxcosx+2sinycosy=(2sinxcosy+2cosxsiny)(cosxcosy +sinxsiny)
2sinxcosx+2sinycosy=2sinxcosxcos^2y+2sin^2xsinycos y+2cos^2xcosysiny+2cosxsin^2ysinx
• Mar 29th 2009, 02:07 PM
stapel
Quote:

Originally Posted by skeske1234
prove that sin2x+sin2y=2sin(x+y)cos(x-y)

Try working from the left-hand side:

. . .sin(2x) + sin(2y) = 2sin(x)cos(x) + 2sin(y)cos(y)

. . . . .= 2[sin(x)cos(x) + sin(y)cos(y)]

Now work from the right-hand side, ignoring the "2" for the time being:

. . . . .sin(x + y) = sin(x)cos(y) + cos(x)sin(y)

. . . . .cos(x - y) = cos(x)cos(y) + sin(x)sin(y)

Multiplying:

. . . . .sin(x)cos(x)cos^2(y) + cos^2(x)sin(y)cos(y)

. . . . . . . .+ sin^2(x)sin(y)cos(y) + sin(x)sin^2(y)cos(x)

Rearranging a bit:

. . . . .sin(x)cos(x)[cos^2(y) + sin^(y)]

. . . . . . . .+ sin(y)cos(y)[cos^(x) + sin^2(x)]

This simplifies as:

. . . . .sin(x)cos(x) + sin(y)cos(y)

Throw that "2" back in, and you've got equality.

Of course, the above is not a proper "proof". For that, you'll need to start at one side, work down the where "they meet in the middle", and then work back up to the other side. (Wink)
• Mar 29th 2009, 02:15 PM
Soroban
Hello, skeske1234!

Quote:

Prove that: .$\displaystyle \sin2x + \sin2y \;=\;2\sin(x+y)\cos(x-y)$
I started with the left side . . .

$\displaystyle 2\sin(x+y)\cos(x-y)$

. . $\displaystyle = \;2\bigg[\sin x\cos y + \cos x\sin y\bigg]\bigg[\cos x\cos y + \sin x\sin y\bigg]$

. . $\displaystyle = \;2\bigg[\sin x\cos x\cos^2\!y + \sin^2\!x\sin y\cos y + \cos^2\!x\sin y\cos y + \sin x\cos x\sin^2\!y\bigg]$

. . $\displaystyle =\;2\bigg[(\sin x\cos x\cos^2\!y \;+\; \sin x\cos x\sin^2\!y) \;+\; (\sin^2\!x\sin y\cos y \;+\; \cos^2\!x\sin y\cos y)\bigg]$

. . $\displaystyle = \;2\bigg[\sin x\cos x\underbrace{(\cos^2\!y + \sin^2\!y)}_{\text{This is 1}} + \underbrace{(\sin^2\!x + \cos^2\!x)}_{\text{This is 1}}\sin y\cos y\bigg]$

. . $\displaystyle = \;2\bigg[\sin x\cos x + \sin y\cos t\bigg]$

. . $\displaystyle = \;\underbrace{2\sin x\cos x}_{\searrow} + \underbrace{2\sin y\cos y}_{\swarrow}$
. . $\displaystyle =\;\qquad\;\; \overbrace{\sin2x} + \overbrace{\sin 2y}$