Trigonometry Identity Rearrangement

**Simplicity** · June 15th 2008, 01:51 AM

Can someone show me how $\sin ^22\theta$ equal to $1-\cos 4\theta$ ?

Thanks in advance.

**Moo** · June 15th 2008, 01:54 AM

Hello,

Originally Posted by Air

Can someone show me how $\sin ^22\theta$ equal to $1-\cos 4\theta$ ?

Thanks in advance.

$\cos 4 \theta=\cos (2 \cdot 2 \theta)$

Using the identity $\cos 2a=1-2 \sin^2a$ , we get :

$\cos 4 \theta=1-2 \sin^2 2\theta$

$\sin^2 2 \theta=\frac{1-\cos 4 \theta}{\color{red}2}$

**Simplicity** · June 15th 2008, 01:58 AM

...So if we had something like:

$\sin^2 6\theta$ , would that be $\frac{1 - \cos 12 \theta}{2}$ ?

**Moo** · June 15th 2008, 01:59 AM

Originally Posted by Air

...So if we had something like:

$\sin^2 6\theta$ , would that be $\frac{1 - \cos 12 \theta}{2}$ ?

Yes =)

Always refer to double angle properties ^^
It's just a matter of substitution.

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