How to find the period of $f(x)= (\sin x)(\cos x)$?

Question

I thought could mess with the method that you find the period of the sum of two functions (ex. $F(x)=\sin x+\cos x$), but I am not getting anywhere. How could I go about finding it?

Thank you in advance.

Answer 1

Given that

$$\sin(2x)=2\sin(x)\cos(x)\iff \frac 12 \sin(\color{red}{2}x)=\sin(x)\cos(x)$$

the period is of the function $g(x)=\sin(x)\cos(x)$ is $$T=\frac {2\pi}{\color{red}{2}}=\pi$$

Answer 2

The simplest and most standard way to answer this is to use the double-angle formula:

$$\sin(2x) = 2\sin x \cos x$$

to get:

$$ \sin x \cos x = \frac{1}{2} \sin(2x).$$

The $\frac{1}{2}\ $ has no effect on the period as it is a stretch in the vertical direction.

$\ f(x)=\sin(2x)\ $ is a stretch, scale factor $\ \frac{1}{2}\ $ in the horizontal direction of $\ g(x) = \sin(x).$

Therefore the period of $\ f(x)=\sin(2x)\ $ is half the period of $\ g(x) = \sin(x).$

$$$$

Alternative method:

From the identities: $\ \sin \left(x+\pi\right) = -\sin(x)\quad \forall\ x\in\mathbb{R}\quad $ and $\quad \cos \left(x+\pi\right) = -\cos(x)\quad \forall\ x\in\mathbb{R},\quad $

we see that,

$$ \sin(x) \cos(x) = \sin \left(x+\pi\right) \cos \left(x+\pi\right) \quad \forall\ x\in\mathbb{R}.$$

This implies that the period of $\ \sin(x) \cos(x)\ $ is $\ \frac{\pi}{k}\ $ for some $\ k\in\mathbb{N}.$

But by inspecting the sign (positive or negative) of $\ \sin\ $ and $\ \cos\ $ in each of the two intervals $\ \left(0,\frac{\pi}{2}\right),\ \left(\frac{\pi}{2},\frac{2\pi}{2}\right),\ $ we see that

$$ \cos(x)\sin(x) \begin{cases} > 0\quad \text{ if }\ x\in \left(0,\frac{\pi}{2}\right)\\ \\ < 0\quad \text{ if }\ x\in \left(\frac{\pi}{2},\frac{2\pi}{2}\right)\\ \end{cases} $$

Suppose the period of $\ f(x) = \cos(x)\sin(x)\ $ is equal to $\ \alpha < \pi.\ $ Then, $\ f\left(\frac{\pi}{2}-\frac{\alpha}{2} \right) = f\left(\frac{\pi}{2}-\frac{\alpha}{2} + \alpha \right) = f\left(\frac{\pi}{2} + \frac{\alpha}{2} \right)\quad (*).\ $ But, noting that $\ \frac{\alpha}{2} < \frac{\pi}{2},\ $ we see that $\ (*)\ $ is impossible, since the sign of $\ f\left(\frac{\pi}{2}-\frac{\alpha}{2} \right)\ $ must be different to the sign of $\ f\left(\frac{\pi}{2} + \frac{\alpha}{2} \right). $