# 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?

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

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Micah, Claude Leibovici, Harish Chandra Rajpoot, user26857, zhoraster
If this question can be reworded to fit the rules in the help center, please edit the question.

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 $X$.