# Thread: More integration solved by use of double angle formulae

1. ## More integration solved by use of double angle formulae

Within the same question as I previously posted about,
there is a jump not explaining conclusions that I can't quite see how the derivation was made.

cos cubed 2theta = (1 - sin squared 2theta)cos 2theta

Can anyone help?

Thanks

2. Think about rewriting $\displaystyle \cos^{3}(2\theta)=\cos^{2}(2\theta)\,\cos(2\theta)$, while nobody's looking.

3. Oh, and by the way, I would highly recommend using parentheses around all function arguments. It makes your writing clearer.

4. Originally Posted by Ackbeet
Think about rewriting $\displaystyle \cos^{3}(2\theta)=\cos^{2}(2\theta)\,\cos(2\theta)$, while nobody's looking.

I have but I still can't figure out where (1 - (sin squared 2theta)) actually comes from (yes (cos squared 2theta), but how?)

Thanks

5. So, you know that $\displaystyle \cos^{2}(\theta)+\sin^{2}(\theta)=1$, I'm sure. What is $\displaystyle \cos^{2}(x)+\sin^{2}(x)$? How about $\displaystyle \cos^{2}(y)+\sin^{2}(y)$? And finally, my favorite, how about
$\displaystyle \cos^{2}(\text{stickfigure})+\sin^{2}(\text{stickf igure})$?

If all those are equal to the same thing, then you can see that no matter what you put inside the parentheses, you'll get one. So you could put in, say, $\displaystyle 2\theta$.

6. omg thank you so much, that is sooo easy! I've been working all day, clearly my head's given up on me. Thanks