1. ## Prooving and Identity

Hi, I'm struggling to understand how to prove this:

(sin^4) T = ( 3 - 4 cos 2 T + cos 4 T ) / 8

(where T is theta)

I start on the left side and "break" this up into a square:

( sin^2 T) ^ 2

From here, my book does the following:

( 1 - cos 2 T / 2 ) ^ 2

Why?

I know sin^2 is also = 1 - cos ^ 2
How do I know to use the double angle identity?? It doesn't seem to apply if I'm just working with sin^2 T?

Thanks!

2. One of the ways to write the Double Angle Identiy for Cosine is:

$\cos(2\theta)=1-2sin^{2}\theta\Rightarrow$
$2sin^{2}\theta=1-\cos(2\theta)\Rightarrow$
$sin^{2}\theta=\frac{1-\cos(2\theta)}{2}$

You can rearrange this to come to the step your book has taken.

3. Hello, Snowcrash!

Prove: . $\sin^4\!\theta\:=\: \frac{
3 - 4\cos2\theta + \cos4\theta}{8}$

I start on the left side and "break" this up into a square: . $(\sin^2\!\theta)^ 2$

From here, my book does the following: . $\left(\frac{1 - \cos2\theta}{2}\right)^ 2$
Why?

I know: . $\sin^2\!\theta \:=\:1 - \cos^2 \!\theta$

How do I know to use the double angle identity?

How else are you going to change your $(1-\cos^2\!\theta)^2$
. . into an expression with . $\cos2\theta\,\text{ and }\,\cos4\theta$ ?