Trig indentity help

[brandonbot]

I have this trig identity question not sure how to solve it:
sin2x+sin2y=2sin(x+y)cos(x-y)

I got the left side to equal
2(sinxcosx+sinycosy)

And then the right i worked out to
2(sinxcosy+sinycosx)(cosxcosy+sinxsiny)

I'm just not sure what to do from here any help would be appreciated thanks

 

[Deleted member 4993]

I have this trig identity question not sure how to solve it:
sin2x+sin2y=2sin(x+y)cos(x-y)

I got the left side to equal
2(sinxcosx+sinycosy)

And then the right i worked out to
2(sinxcosy+sinycosx)(cosxcosy+sinxsiny)

I'm just not sure what to do from here any help would be appreciated thanks

2(sinxcosy+sinycosx)(cosxcosy+sinxsiny)

Break-it up - using FOIL....

= 2(sin x * cos x * cos²y + sin²x * siny * cos y +.......)

and continue.....

 

[brandonbot]

So i expanded it out and got this

2[(cos2y)(sinx)(cosx)+(sin2x)(siny)(cosy)
+(cos2x)(siny)(cosy)+(sin2y)(cosx)(sinx)]

Couldn't figure out how to raise the two but the ones with two's are squared terms anyway i still don't know how to proceed cause its just a bunch of unlike terms i can't really do anything with them.

 

[soroban]

Hello, brandonbot!

I couldn't follow your work.
I don't think you followed Subhotosh's advice.

$\displaystyle \text{Prove: }\:\sin2x+\sin2y\:=\:2\sin(x+y)\cos(x-y)$

Expand the right side:

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

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

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

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

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

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

. . . $\displaystyle =\;\underbrace{2\sin x\cos x} + \underbrace{2\sin y\cos y}$

. . . $\displaystyle =\qquad \sin2x \;\;+\;\;\sin2y$

 

[brandonbot]

Thanks alot for the help