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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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3 Answers

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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 $$\frac{1}{2} \sin(4\theta) = \sin(2\theta)\cos(2\theta).$$

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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?

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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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