# Half angle formula,

1. Dec 20, 2009

### tweety1234

1. The problem statement, all variables and given/known data

Can someone please explain how my book got $sin^{2} \theta = \frac{1-cos2\theta}{2}$ and $cos^{2}\theta = \frac{1+cos2\theta}{2}$

As I thought the half angle formula's were $sin \frac{\theta}{2} = \sqrt{\frac{1-cos\theta}{2}}$

$cos\frac{\theta}{2} = \sqrt\frac{1+cos\theta}{2}}$

So how can it also be the top one's aswell?

Thanks.

2. Dec 20, 2009

### Staff: Mentor

Replace θ with 2θ.

3. Dec 20, 2009

### tweety1234

Oh I see, can't believe I missed that.

Thanks.