# Proving an identity.

• Apr 12th 2011, 01:29 AM
dagbayani481
Proving an identity.
Hello! In need of help yet again. I've seemed have gotten stumped :|

$\displaystyle sin(3x)+sin(x) = 2sin(2x)cos(x)$

Took the left side and changed it to:
$\displaystyle sin(2x+x)+sin(x)$

From there I got it equaled to:

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

from then on I get confused. I tried using the double angle formulas to expand the sin(2x) and the cos(2x) but that gets me no where. Any ideas on what I could do next?
• Apr 12th 2011, 01:56 AM
Chris L T521
Quote:

Originally Posted by dagbayani481
Hello! In need of help yet again. I've seemed have gotten stumped :|

$\displaystyle sin(3x)+sin(x) = 2sin(2x)cos(x)$

Took the left side and changed it to:
$\displaystyle sin(2x+x)+sin(x)$

From there I got it equaled to:

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

from then on I get confused. I tried using the double angle formulas to expand the sin(2x) and the cos(2x) but that gets me no where. Any ideas on what I could do next?

You're on the right track...here's an interesting observation:

\displaystyle \begin{aligned}\sin(2x)\cos(x) + \cos(2x)\sin(x)+\sin(x) &= \sin(2x)\cos(x)+2\sin x\left(\dfrac{\cos(2x)+1}{2}\right)\\ &=\sin(2x)\cos(x)+2\sin x\cos^2x\\ &=\sin(2x)\cos(x)+(2\sin(x)\cos(x))\cos(x)\\ &= \sin(2x)\cos(x)+\sin(2x)\cos(x)\\ &=2\sin(2x)\cos(x)\end{aligned}

(Whew)

I hope this makes sense!
• Apr 12th 2011, 05:22 PM
Krizalid
But it's a direct conclusion by transforming the left side into a product by using sum-product identity.