Using half angle formula

**macca101** · March 23rd 2006, 11:24 AM

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} $

**topsquark** · March 23rd 2006, 02:07 PM

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

**ThePerfectHacker** · March 23rd 2006, 02:35 PM

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,
$\frac{1}{2}\cdot(1+\cos [2(2\theta)])=\frac{1}{2}\cdot (1+\cos 4\theta)$

**macca101** · March 26th 2006, 10:48 AM

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.

**topsquark** · March 26th 2006, 03:57 PM

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

**macca101** · March 27th 2006, 03:28 AM

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

**topsquark** · March 27th 2006, 03:50 AM

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

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