# Using half angle formula

• Mar 23rd 2006, 12:24 PM
macca101
Using half angle formula
Using the half angle formula

$

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

How would I go about showing that

$

cos^2(2\theta) = \frac{1}{2} * cos(4\theta)+\frac{1}{2}
$
• Mar 23rd 2006, 03:07 PM
topsquark
Quote:

Originally Posted by macca101
Using the half angle formula

$

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

How would I go about showing that

$

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, 03:35 PM
ThePerfectHacker
Quote:

Originally Posted by macca101
Using the half angle formula

$

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

How would I go about showing that

$

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

You know that,
$\cos^2(x)=\frac{1}{2}\cdot (1+\cos 2x)$
Then in the problem,
$\cos^2(2\theta)$
set $x=2\theta$
Then,
$\frac{1}{2}\cdot(1+\cos [2(2\theta)])=\frac{1}{2}\cdot (1+\cos 4\theta)$
• Mar 26th 2006, 11:48 AM
macca101
Quote:

Originally Posted by ThePerfectHacker
You know that,
$\cos^2(x)=\frac{1}{2}\cdot (1+\cos 2x)$
Then in the problem,
$\cos^2(2\theta)$
set $x=2\theta$
Then,
$\frac{1}{2}\cdot(1+\cos [2(2\theta)])=\frac{1}{2}\cdot (1+\cos 4\theta)$

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

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

To show that

$
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, 04:57 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

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

To show that

$
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.

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

And we know that
$cos^2(2\theta)=(1/2)(1+cos(4\theta))$

So you should just be able to plug that into your sin^4 formula.

-Dan
• Mar 27th 2006, 04:28 AM
macca101
Thanks It was the

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

Thing that was eluding the empty shell I carry on my shoulders
• Mar 27th 2006, 04:50 AM
topsquark
Quote:

Originally Posted by macca101
Thanks It was the

$
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