# Thread: Double-Angle Formula For Sin Question Problem

1. ## Double-Angle Formula For Sin Question Problem

I've been asked the prove that
$\displaystyle sin^{2}xcos^{2}x=\frac{1}{8}(1-cos(4x))$
by using the double-angle for $\displaystyle sin(2\theta)$ and $\displaystyle cos(2\theta)$ in turn.
I'm stuck at the first section of the solution given for the question, I think if I can understand this first part I will be OK after that. The first part of the solution is:

From the double-angle formula for $\displaystyle sin(2\theta)$, with $\displaystyle \theta=x$, we have
$\displaystyle sin^{2}xcos^{2}x=(sinxcosx)^{2}=\frac{1}{4}sin^{2} (2x)$.

Where has all this come from?

The only formula I've been given for the double-angle formula for$\displaystyle sin(2\theta)$ is
$\displaystyle sin(2\theta)=2sin(\theta)cos(\theta)$
Any help clearing this up would be greatly appreciated.

2. Try going in reverse, starting with $\displaystyle Sin(2x)$:

$\displaystyle Sin(2x)=2Sin(x)Cos(x)$

Therefore $\displaystyle Sin^2(2x)=(2Sin(x)Cos(x))^2$

$\displaystyle =4Sin^2(x)Cos^2(x)$

You have $\displaystyle Sin^2(x)Cos^2(x)$

$\displaystyle =\frac{1}{4}Sin^2(2x)$

3. \displaystyle \displaystyle \begin{align*}\frac{1}{8}\left[1 - \cos{(4x)}\right] &= \frac{1}{8}\left\{1- \left[\cos^2{(2x)} - \sin^2{(2x)}\right]\right\} \\ &= \frac{1}{8}\left[1 - \cos^2{(2x)} + \sin^2{(2x)} \right] \\ &= \frac{1}{8}\left[\sin^2{(2x)} + \sin^2{(2x)}\right] \\ &= \frac{1}{8}\left[2\sin^2{(2x)}\right] \\ &= \frac{1}{4}\sin^2{(2x)} \\ &= \frac{1}{4}\left[\sin{(2x)}\right]^2 \\ &= \frac{1}{4}\left[2\sin{(x)}\cos{(x)}\right]^2 \\ &= \frac{1}{4}\left[4\sin^2{(x)}\cos^2{(x)}\right] \\ &= \sin^2{(x)}\cos^2{(x)}\end{align*}

4. Many thanks for that.

I understand both solutions now.

I wish text books were more thorugh with their solutions, at least for the first year of studying or so anyway.

Thanks again, I really appreciate it.