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How is this trigonometric relation derived in simple terms?

$$\cos\left(\frac{2\pi}{N}\right) = 1 - 2\sin^2\left(\frac{\pi}{N}\right)$$

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closed as off-topic by Micah, Claude Leibovici, Harish Chandra Rajpoot, user26857, zhoraster Oct 20 '15 at 8:31

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You know that $\cos(2\theta) = 1 - 2\sin^2(\theta)$, and.... actually, that's it!

(These are the double-angle identities.)

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  • $\begingroup$ thanks, Is there a way to visualise this? I would like to understand it better rather than just know that it is true. $\endgroup$ – Duncan Gravill Oct 20 '15 at 2:21
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    $\begingroup$ Sure. Imagine the graph of cos x. The graph of cos 2x is twice as compressed horizontally. Now the graph of sin^2 x is like the graph of sin x, except it's (roughly) twice as compressed horizontally and it's always above the x-axis, since x^2 > 0 for all x. When you stretch it by a factor of 2 vertically and then move it down by one, it becomes the graph of cos x. Try plotting everything out on a graphing calculator and hopefully it'll make sense! $\endgroup$ – eyqs Oct 20 '15 at 2:24
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    $\begingroup$ @FunkyFresh84: The double-angle identities arise from the Angle-Addition formulas, which this diagram helps to visualize. For the double-angle cosine formula, take $\alpha=\beta$ to get $\cos 2\alpha = \cos^2\alpha - \sin^2\alpha$. Then apply a Pythagorean relation to convert to $(1-\sin^2\alpha)-\sin^2\alpha=1-2\sin^2\alpha$.) There are other diagrams you can make that show the relation "in one step", without having to invoke Pythagoras, but I think it's important to see the general case as well. $\endgroup$ – Blue Oct 20 '15 at 2:30
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    $\begingroup$ Another good post with a different geometrical proof is here. $\endgroup$ – eyqs Oct 20 '15 at 2:34
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In isosceles triangle $ABC$ with $AB=AC=1,$ let $D$ be the midpoint of$ BC , $ let $\angle BAD=\angle CAD=X. $ We have $\angle BDA=\pi/2$ , so $$CB= 2DB=2AB \sin X =2\sin X. $$ $$\text {Hence } CB^2=4\sin^2 X.$$The Theorem of Pythagoras implies the Cosine Formula : $$CB^2=AB^2+AC^2-2.AC.AB.\cos \angle BAC.$$ With $AB=AC=1$ and $\angle BAC=2X$ we have $$CB^2=2-2\cos 2X.$$ Therefore $4\sin^2X=2-2\cos^2 2X.$ And $X$ can be any angle between $0$ and $\pi/2$.It is easy now to show this equation holds for all $X$.

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