Simplifying $\sin^2 (\alpha) \cos^2 (\alpha)$

Question

In a more complex expression, I have the term (the only one depending on $\alpha$)

$$\sin^2 (\alpha) \cos^2 (\alpha)$$

and I would like to further simplify it, if possible in either of the following two ways.

Question 1: is it possible that the product of the squares of sine and cosine of an angle $\alpha$ is equal to a fixed value? I have no clue about this and the equality $\sin^2(\alpha) + \cos^2(\alpha) = 1$ seems not to help.

The initial expression can always be rewritten as:

$$\sin^2 (\alpha) \cos^2 (\alpha) = \frac{1}{4} \sin^2 (2\alpha)$$

Question 2: is it possible to rewrite this in terms of $\sin (\alpha)$ or a single power of $\sin (\alpha)$ only? Again, I don't know how to proceed: Sum and difference formulae would get the expression back to the beginning.

Answer 1

If you follow your link to the sum and difference of angles formulae, you'll see just below the double angle formulae, which are direct consequences. One of those is

$$\cos 2x = 1-2\sin^2 x.$$

Let $x = 2\alpha$ and solve for $\sin^2 2\alpha$ to get

$$\sin^2 2\alpha = \frac{1-\cos 4\alpha}{2}.$$

Plug this into the expression you already have:

$$\sin^2\alpha\cos^2\alpha = \frac{1}{4} \sin^2 2\alpha = \frac{1-\cos 4\alpha}{8}.$$

Answer 2

Hint: It is $$\frac{1}{8}(1-\cos(4x))$$

Answer 3

You can for example use Fourier transforms together with the convolution theorem to solve it and get Dr. Sonnhard Graubner's answer.

It is kind of a generalization of all the trigonometric laws.

But maybe Fourier transforms are too hard for your course?