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.
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.
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 $\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = \cos^2(\theta) - (1-\cos^2(\theta)) = 2 \cos^2(\theta) - 1$
Itβs 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 $\cos^22x=\frac12(\cos4x+1)$.
$\cos(4x) = \cos^2 (2x) - \sin^2 (2x) = 2\cos^2 (2x) - 1$