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I searched extensively for an answer, but couldn't find one that specifically explained what I was looking for. In working through a problem in my textbook, part of it involves simplifying an expression using power reduction. This is the step:

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

I don't get this step though. The power reduction equations are easy enough to understand, but the book does not explain how they apply to multiples of angles. The double angle identities don't really help either since they turn the trig functions into squares again.

Why does applying power reduction on a squared double angle function quadruple the angle?

Thanks!

T. Rettig
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  • I don't quite understand where you have having difficulty. Do you agree that $\cos^2\theta = \frac{1+\cos (2\theta)}{2}$? – Argon Jun 29 '17 at 03:55
  • Hi Argon. Thanks for replying. It's not whether I agree or disagree. I just want to understand the step(s) to get from one to the other. The power reduction equations in the textbook only deal with single powers of sine and cosine not squares. How squaring cosine for a double angle leads to the angle becoming a quadruple angle is not apparent to me. – T. Rettig Jun 30 '17 at 04:42
  • Do the proofs from here help? The cosine sum-of-angles formula when $\alpha=\beta=2\theta$ gives the relationship connecting $\cos(4\theta)$ with $\cos^2(2\theta)$ (NB $\sin^2(2\theta)=1-\cos^2(2\theta)$). – Argon Jun 30 '17 at 14:05
  • Yes those help a great deal. Thanks Argon! Those triangle relationships diagrams were especially helpful. Those trigonometric diagrams are excellent. – T. Rettig Jul 02 '17 at 20:48

2 Answers2

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It is because $$\cos (2\theta)=\cos(\theta+\theta)=\cos^2(\theta)-\sin^2(\theta)=2\cos^2(\theta)-1$$

Jay Zha
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  • Thanks Yujie! However, it's not clear to me how 2cos^2-1 results in theta being doubled. I recognize it as another of the double angle identities for cos(2\theta), along with 1-2sin^2(\theta), but it's still not clear how this results in the doubling of theta with power reduction. – T. Rettig Jun 30 '17 at 04:53
  • @T.Rettig Hey, that is because $\cos(a+b)=\cos(a)\cos(b) - \sin(a)\sin(b)$, http://home.windstream.net/okrebs/page101.html – Jay Zha Jun 30 '17 at 05:38
  • @T.Rettig And here is a proof the the formula I mentioned in my above comment: http://pages.pacificcoast.net/~cazelais/173/idns-proof.pdf – Jay Zha Jun 30 '17 at 05:51
  • Thanks again Yujie! That proof was very helpful for understanding the fundamental sum and difference identities. The diagram in particular was great for understanding the relationship of the difference between angles. – T. Rettig Jul 02 '17 at 20:45
  • You are welcome :) – Jay Zha Jul 02 '17 at 21:15
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$\cos^2(\text{thing}) = \dfrac{1 + \cos(2 \cdot \text{thing})}{2},$ no matter what that thing happens to be. If it's $2\theta$, then you'll wind up with $2 \cdot 2\theta = 4\theta$ on the right.

The angle inside $\cos^2()$ always twice the angle inside $\cos()$; the power reduction formula always doubles the angle inside $\cos^2()$. If that angle inside $\cos^2()$ is already something doubled, that's where the quadrupling comes from.

pjs36
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  • Thanks pjs36! This is helpful. I guess I thought it was odd that with what the power reduction formula turns into that it would only "reach inside" the cosine angle and double it, but if that is the case then that is the case. Is there a proof or way to explain how the power reduction changes cos^{2}(\theta) to (1+cos(2*\theta)\2)? – T. Rettig Jun 30 '17 at 04:50
  • That's really what the other answer is good for; the double-angle formula for cosine comes in a few varieties \begin{align} \cos(2 \theta) &= \cos^2(\theta) - \sin^2(\theta) \ &= 2\cos^2(\theta) - 1 \ &= 1 - 2\sin^2 (\theta),\end{align} but they provide the link between squared trig functions and $\cos(2 \theta)$, that form the basis of the power-reduction formulas. Any trig/precalculus book should be able to provide the surrounding context. – pjs36 Jun 30 '17 at 14:43