I can't figure out how to simplify $\sin2\theta \cos2\theta$ to be $\frac12 \sin4\theta $. I assume double angle identities are involved, but I can't get it and would like some help.
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$\begingroup$We know the double angle formula for sine is $\sin(2x) = 2\sin(x)\cos(x)$.
For convenience, let $x = 2\theta$. Then $4\theta$ can be written as$$4\theta = 2(2\theta) = 2x.$$It then follows that$$\frac{1}{2}\sin(4\theta) = \frac{1}{2}\sin(2x) = \frac{1}{2} \cdot 2\sin(x)\cos(x) = \sin(x)\cos(x).$$But wait--we said $x = 2\theta$ ! Plugging this back in allows us to conclude that$$\frac{1}{2} \sin(4\theta) = \sin(2\theta)\cos(2\theta).$$
$\endgroup$ $\begingroup$Note:\begin{equation} \sin(2x) = 2\sin(x)\cos(x) \\ \end{equation}
Therefore we have $\frac{\sin(4x)}{2} = \frac{\sin(2(2x))}{2}= \frac{2\sin(2x)\cos(2x)}{2}=\sin(2x)\cos(2x)$
Does this clarify?
$\endgroup$ 0 $\begingroup$If you are familiar with the double angle identity:
$\sin(2x) = 2\sin(x)\cos(x)$ or written another way
${1 \over 2} \sin(2x) = \sin(x)\cos(x)$
Let $x = 2\theta$ and you'll get:
${1 \over 2} \sin(4\theta) = \sin(2 \theta)\cos(2 \theta)$
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