# Derivation of trigonometric identity $\cos\left(\frac{2\pi}{N}\right) = 1 - 2\sin^2\left(\frac{\pi}{N}\right)$ [closed]

How is this trigonometric relation derived in simple terms?

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

You know that $\cos(2\theta) = 1 - 2\sin^2(\theta)$, and.... actually, that's it!
• @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. – Blue Oct 20 '15 at 2:30
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 $X$.