Using the half angle formula
$\displaystyle
cos^2(\theta) = \frac{1}{2} * (1+cos(2\theta))
$
How would I go about showing that
$\displaystyle
cos^2(2\theta) = \frac{1}{2} * cos(4\theta)+\frac{1}{2}
$
You know that,Originally Posted by macca101
$\displaystyle \cos^2(x)=\frac{1}{2}\cdot (1+\cos 2x)$
Then in the problem,
$\displaystyle \cos^2(2\theta)$
set $\displaystyle x=2\theta$
Then,
$\displaystyle \frac{1}{2}\cdot(1+\cos [2(2\theta)])=\frac{1}{2}\cdot (1+\cos 4\theta)$
Ok Thanks for that.Originally Posted by ThePerfectHacker
I thought I was stuck then boy I'm I struggling now. The question goes on ...
Use the answer above and another half angle formula
I Think
$\displaystyle
sin^2(\theta) = \frac{1}{2} * (1-cos(2\theta))
$
To show that
$\displaystyle
sin^4(\theta) = \frac{3}{8}- \frac{1}{2} * cos(2\theta)+\frac{1}{8}* cos(4\theta)
$
Apologies if I should have started a new thread for this. Thought it would be OK to continue as it relates to the original post.
$\displaystyle sin^4(\theta) = sin^2(\theta) \, sin^2(\theta)=(1/4)(1-cos(2\theta))^2$Originally Posted by macca101
$\displaystyle =(1/4)(1-2cos(2\theta)+cos^2(2\theta))$
And we know that
$\displaystyle cos^2(2\theta)=(1/2)(1+cos(4\theta))$
So you should just be able to plug that into your sin^4 formula.
-Dan