Proof that $\cos^2(\theta) \sin^2(\theta) = \dfrac{1}{4}\sin^2(2\theta)$.

Question

I recently came across the identity $\cos^2(\theta) \sin^2(\theta) = \dfrac{1}{4}\sin^2(2\theta)$.

I was wondering if someone could please take the time to prove that this is true (or link to a proof)?

This is the double-angle formula for sine ($\sin 2x = 2\sin x \cos x$), squared. If you search for that name, you'll find proofs. — StackTD, Aug 25 '17 at 14:24
Do you want to use something other than the square of the double angle formula: $\sin(2\theta)=2\sin(\theta)\cos(\theta)$? — Michael Burr, Aug 25 '17 at 14:25

score 1 · Accepted Answer · answered Aug 25 '17 at 14:25

1

Using the fact that $$\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)\tag{E}$$ you get

$$\sin(2\theta )=2\cos(\theta )\sin(\theta ).$$

You can find a geometrical proof of $(\text{E})$ here.

answered Aug 25 '17 at 14:25

Surb

55,662

1

thanks for the link ! :) – user330587 Aug 25 '17 at 22:26

score 0 · Answer 2 · edited Aug 25 '17 at 14:49

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Many ways how to do this: Ex: $\sin2\alpha=2\sin{\alpha}\cos\alpha$ (standard identity) so squaring gives $\sin^2{2\alpha}=4{\sin^2\alpha}\cos^2\alpha$. Now divide by $4$

edited Aug 25 '17 at 14:49

JohnColtraneisJC

1,919

answered Aug 25 '17 at 14:26

imranfat

10,029

3

MathJax hint: if you put a backslash before common functions you get the proper font and spacing, so \sin 2\alpha gives $\sin 2 \alpha$ instead of $sin2\alpha$ – Ross Millikan Aug 25 '17 at 14:32
@RossMillikan Thanks! – imranfat Aug 25 '17 at 15:51

Proof that $\cos^2(\theta) \sin^2(\theta) = \dfrac{1}{4}\sin^2(2\theta)$.

2 Answers2