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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.

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    $\begingroup$ You could always write $\sin^2\alpha \cos^2 \alpha = \sin^2 \alpha - \sin^4 \alpha$, but I'm not sure if that would be OK for your purposes. $\endgroup$ – Minus One-Twelfth May 23 at 10:57
  • $\begingroup$ You can use Fourier transform to simplify it. $\endgroup$ – mathreadler May 23 at 10:57
  • $\begingroup$ @MinusOne-Twelfth I also guessed this and unfortunately no, it would not be enough. $\endgroup$ – BowPark May 23 at 10:58
  • $\begingroup$ "equal to a fixed value?" Certainly not in the sense that it is constant. What exactly are you after? If the answer of @MinusOne-Twelfth is rejected then there must be criteria. $\endgroup$ – drhab May 23 at 10:59
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    $\begingroup$ When you ask for "$\sin (\alpha)$ or a single power of $\sin (\alpha)$ only," do you mean that you will only accept something equivalent to $a + b (\sin(\alpha))^n$ where $a,$ $b,$ and $n$ are constants and $n$ is an integer? If so I think you're out of luck, since your function is sinusoidal with period $\pi/2$ and the powers of $\sin(\alpha)$ all have longer periods (and mostly are not sinusoidal). $\endgroup$ – David K May 23 at 11:54
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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:

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

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

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    $\begingroup$ That's the final answer, but still gives good insight to the following hint. Replace $\sin^2\alpha$ by $1-\cos^2\alpha$ (that's an identity) and use the general doubling formula $\cos(2\theta)=2\cos^2\theta-1$. $\endgroup$ – RMWGNE96 May 23 at 10:57
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    $\begingroup$ Sorry, I can't get your points. This is in terms of $4\alpha$, not $\alpha$. $\endgroup$ – BowPark May 23 at 11:09
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    $\begingroup$ @BowPark Yes, this has $4\alpha$ instead of $\alpha$ inside the trig function, but with this formula you can easily plot the function. It is a sinusoidal shape, but unlike $\sin\alpha$ you have four full cycles instead of just one between $\alpha=0$ and $\alpha=2\pi.$ One cycle goes from a minimum of $0$ at $\alpha=0$ to a maximum of $\frac14$ at $\alpha=\frac\pi4$ and back to $0$ at $\alpha=\frac\pi2.$ $\endgroup$ – David K May 23 at 12:11
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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?

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  • $\begingroup$ Maybe I could try (even if this doesn't belong to the course), but... I still didn't understand what actually is Dr. Sonnhard Graubner's answer. $\endgroup$ – BowPark May 23 at 11:12
  • $\begingroup$ @BowPark A slightly easier way than to use Fourier transforms would be to use Euler formula if you have learned them in course. Like $\sin(x) = \frac{e^{xi} - e^{-ix}}{2i}$ $\endgroup$ – mathreadler May 23 at 11:20

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