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}

$

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- Mar 23rd 2006, 11:24 AMmacca101Using 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 PMtopsquarkQuote:

Originally Posted by**macca101**

-Dan - Mar 23rd 2006, 02:35 PMThePerfectHackerQuote:

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,

$\displaystyle \frac{1}{2}\cdot(1+\cos [2(2\theta)])=\frac{1}{2}\cdot (1+\cos 4\theta)$ - Mar 26th 2006, 10:48 AMmacca101Quote:

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. - Mar 26th 2006, 03:57 PMtopsquarkQuote:

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 - Mar 27th 2006, 03:28 AMmacca101
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 AMtopsquarkQuote:

Originally Posted by**macca101**

-Dan