# Using half angle formula

• Mar 23rd 2006, 11:24 AM
macca101
Using half angle formula
Using the half angle formula

$\displaystyle cos^2(\theta) = \frac{1}{2} * (1+cos(2\theta))$

How would I go about showing that

$\displaystyle cos^2(2\theta) = \frac{1}{2} * cos(4\theta)+\frac{1}{2}$
• Mar 23rd 2006, 02:07 PM
topsquark
Quote:

Originally Posted by macca101
Using the half angle formula

$\displaystyle cos^2(\theta) = \frac{1}{2} * (1+cos(2\theta))$

How would I go about showing that

$\displaystyle cos^2(2\theta) = \frac{1}{2} * cos(4\theta)+\frac{1}{2}$

Just double the angle in your formula. I suspect you are trying to "over think" the problem here...

-Dan
• Mar 23rd 2006, 02:35 PM
ThePerfectHacker
Quote:

Originally Posted by macca101
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,
$\displaystyle \cos^2(x)=\frac{1}{2}\cdot (1+\cos 2x)$
Then in the problem,
$\displaystyle \cos^2(2\theta)$
set $\displaystyle x=2\theta$
Then,
$\displaystyle \frac{1}{2}\cdot(1+\cos [2(2\theta)])=\frac{1}{2}\cdot (1+\cos 4\theta)$
• Mar 26th 2006, 10:48 AM
macca101
Quote:

Originally Posted by ThePerfectHacker
You know that,
$\displaystyle \cos^2(x)=\frac{1}{2}\cdot (1+\cos 2x)$
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.

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.
• Mar 26th 2006, 03:57 PM
topsquark
Quote:

Originally Posted by macca101
Ok Thanks for that.

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.

$\displaystyle sin^4(\theta) = sin^2(\theta) \, sin^2(\theta)=(1/4)(1-cos(2\theta))^2$
$\displaystyle =(1/4)(1-2cos(2\theta)+cos^2(2\theta))$

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
• Mar 27th 2006, 03:28 AM
macca101
Thanks It was the

$\displaystyle sin^4(\theta) = sin^2(\theta) \, sin^2(\theta)$

Thing that was eluding the empty shell I carry on my shoulders
• Mar 27th 2006, 03:50 AM
topsquark
Quote:

Originally Posted by macca101
Thanks It was the

$\displaystyle sin^4(\theta) = sin^2(\theta) \, sin^2(\theta)$

Thing that was eluding the empty shell I carry on my shoulders

Don't feel so bad. Soon you'll be randomly adding and subtracting 1 from equations just like the rest of us. ;) (The tricks we use are seem so inane sometimes!)

-Dan