# Why $\cos^2 (2x) = \frac{1}{2}(1+\cos (4x))$?

Why: $$\cos ^2(2x) = \frac{1}{2}(1+\cos (4x))$$

I don't understand this, how I must to multiply two trigonometric functions?

Thanks a lot.

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I've changed algebra tag to algebra-precalculus, since we don't use algebra tag anymore, see meta for details. –  Martin Sleziak Oct 29 '12 at 13:34
This Wikipedia article is a useful resource for various formulas about trigonometric functions: List of trigonometric identities –  Martin Sleziak Oct 29 '12 at 13:36

Recall the formula $$\cos(2 \theta) = 2 \cos^2(\theta) - 1$$ This gives us $$\cos^2(\theta) = \dfrac{1+\cos(2 \theta)}{2}$$ Plug in $\theta = 2x$, to get what you want.

EDIT The identity $$\cos(2 \theta) = 2 \cos^2(\theta) - 1$$ can be derived from $$\cos(A+B) = \cos(A) \cos(B) - \sin(A) \sin(B)$$ Setting $A = B = \theta$, we get that $$\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = \cos^2(\theta) - (1-\cos^2(\theta)) = 2 \cos^2(\theta) - 1$$

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Thanks, this is a complete answer, now understand. –  calbertts Oct 28 '12 at 2:23

It’s just the double-angle formula for the cosine: for any angle $\alpha$, $\cos 2\alpha=\cos^2\alpha-\sin^2\alpha\;,$ and since $\sin^2\alpha=1-\cos^\alpha$, this can also be written $\cos2\alpha=2\cos^2\alpha-1$. Now let $\alpha=2x$: you get $\cos4x=2\cos^22x-1$, so $\cos^22x=\frac12(\cos4x+1)$.

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Thanks by your time, this helped me. –  calbertts Oct 28 '12 at 2:23

$$\cos(4x) = \cos^2 (2x) - \sin^2 (2x) = 2\cos^2 (2x) - 1$$

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Thanks by your time. –  calbertts Oct 28 '12 at 2:25