$\sin^2(2x)=\dfrac{1-\cos(4x)}{2}=\dfrac{1}{2}(1-\cos(4x))$
I know these two identities should be used here, but I'm not sure what's exactly legal when they are used together at the same time.
$\sin^2(x)+\cos^2(x)=1$
$\sin(2x)=2\sin(x)\cos(x)$
$\sin^2(2x)=\dfrac{1-\cos(4x)}{2}=\dfrac{1}{2}(1-\cos(4x))$
I know these two identities should be used here, but I'm not sure what's exactly legal when they are used together at the same time.
$\sin^2(x)+\cos^2(x)=1$
$\sin(2x)=2\sin(x)\cos(x)$
Hello, maxpancho!
$\displaystyle \text{How is this derived? }\:\sin^2(2x)\;=\;\dfrac{1-\cos(4x)}{2}$
We have: .$\cos(A+B) \:=\:\cos A\cos B - \sin A\sin B$
Let $\displaystyle A=B\!:\;\cos(2A) \:=\:\cos^2\!A - \sin^2\!A \;=\;(1-\sin^2\!A) - \sin^2\!A$
. . . . . . . . .$\displaystyle \cos(2A) \:=\:1-2\sin^2\!A \quad\Rightarrow\quad 2\sin^2\!A \;=\;1 - \cos(2A)$
. . . . . . . . . . $\displaystyle \sin^2\!A \;=\;\frac{1-\cos(2A)}{2}$
Let $\displaystyle A =2x\!:\;\;\sin^2(2x) \;=\;\frac{1-\cos(4x)}{2}$