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I'm sure there is some very basic Algebra I'm missing out on but. . .

how does $\sin^2(2\theta)$ end up equaling $4\sin^2(\theta)\cos^2(\theta)$?

I assume this is derived from the $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$

This is in a Khan academy Calculus example I'm working through. I'm about 15 years out of my most recent algebra class, so please be kind :)

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Your assumption is correct.

Then $$ (\sin(2\theta))^2=(2\sin(\theta)\cos(\theta))^2=4\sin^2\theta\cos^2\theta. $$

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  • $\begingroup$ Ohh, is that because you go through and square each of the three elements (2, cos, and sin)? I understand this is a very basic question, but I just want to confirm I'm correct. $\endgroup$ – Matt Oct 19 '17 at 0:40
  • $\begingroup$ @Matt Yes. $(abc)^2 = a^2b^2c^2$, so $$(2\sin\theta\cos\theta)^2 = 2^2\sin^2\theta\cos^2\theta = 4\sin^2\theta\cos^2\theta$$ $\endgroup$ – N. F. Taussig Oct 19 '17 at 0:42
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$\sin^2(2θ)$ is the same as $(\sin(2θ))^2$. Replace $\sin(2θ)$ with $2\sin(θ)\cos(θ)$ to get $(2\sin(θ)\cos(θ))^2=4\sin^2(θ)\cos^2(θ)$.

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    $\begingroup$ surely this didn't need more than one answer.... $\endgroup$ – qbert Oct 19 '17 at 0:43
  • $\begingroup$ Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Oct 19 '17 at 0:44

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