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

1. ## 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

2. 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.

3. Hello, skeske1234!

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

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

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

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

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

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

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

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

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