# Why sinxcos2x = (sin3x - sinx)*0.5

1. Apr 12, 2009

### fiziksfun

can someone please show me why sinxcos2x = (sin3x - sinx)*0.5

I've been working on it for thirty minutes

2. Apr 12, 2009

### praharmitra

Re: sinxcos2x

can u expand sin(3x) ?? and cos(2x)

just simplify both sides..and u'll get it..

3. Apr 12, 2009

### qntty

Re: sinxcos2x

You should use the identities

$$\cos{2\theta} = 1-2\sin^2{\theta}$$

$$\sin{3\theta} = 3\sin{\theta} - 4\sin^3{\theta}$$

Perhaps you were putting these identities in terms of cosines. When possible, stick with as few trig functions as possible and replace the $$\cos^2{\theta}$$ with $$1-\sin^2{\theta}$$ or vice versa.

Last edited: Apr 12, 2009