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So, I have a general question first. What happens to the periodicity when we multiply two periodic trig functions with one another ?

The next one is very specific, what is the period of the function $g(x)=\sin{(ax)}\cos{(bx)}$, where $a$ and $b$ are rational numbers ? I'd be interested in a proof of sorts.

Cheers, Dave

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You might find it helpful to think about the lowest common multiple of the periods. –  in_wolframAlpha_we_trust Apr 2 '12 at 7:53

3 Answers 3

You specify that $a$ and $b$ are rational. Suppose they are respectively $p/q$ and $r/s$ and those are in lowest terms. Let $\ell=\operatorname{lcm}(s,q)$. Then $$ \sin\left( \frac p q x \right)\cos\left( \frac r s x \right) = \sin\left( \frac \bullet \ell x \right) \cos\left( \frac \bullet \ell x\right) = \sin\left( \bullet \frac x \ell \right)\cos\left( \bullet \frac x \ell \right) = \sin\left(c\frac x \ell\right)\cos \left( d \frac x \ell \right) $$ (the first $\bullet$ is the integer $c=\ell p/q$ and the second is the integer $d=\ell r/s$). Now $cx/\ell = 2\pi$ when $x=2\pi\ell/c$, so that is the smallest period of the first factor, and similarly the smallest period of the second factor is $2\pi\ell/d$.

Every multiple of the smallest period is a period, and we seek the smallest period of the product, so we want the smallest common multiple of these two smallest periods that we've found. We want $(\text{some integer}/c)$ and $(\text{some integer}/d)$ to be equal. Multiplying both sides by $cd$, we want $(c\cdot\text{some integer})$ and $(d\cdot\text{some integer})$ to be equal. The period would be $2\pi$ times that common value, thus $2\pi\operatorname{lcm}(c,d)$.

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The product of two periodic trig functions may not be periodic. Try for instance $\sin(\sqrt{2}x)\sin(x)$.

$\sin((p/q)x)$ is periodic of period $(2q/p)\pi$, but make sure you reduce the fraction $2q/p$ to lowest terms.

The period of $\sin{(ax)}\cos{(bx)}$ for $a,b$ rational can be deduced from the previous result. In general, it will be $m\pi$ where $m$ is the lowest common multiple of the denominators of the fractions, except when $a=b$. (There may be a few other special cases.)

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thxs, i think the question was well-defined to exclude anomalies and the exceptions. U failed to follow up on an actual proof underlying the validity of ur statment, don't u agree ? –  Pete Apr 2 '12 at 12:30

When searching the 'prime' period $T_p$ (also called the smallest period) the answer from Michael Hardy is not longer applicable. Example: $$a = 3, b=6 \rightarrow p=3, q=1, r=6, s=1 \rightarrow l = 1, c = 3, d = 6$$ The period calculated from Michael answer is $T = 2\pi\operatorname{lcm}(3,6) = 12\pi$.

The prime period $T_p$ however is $\frac{2\pi}{3}$, which can be seen from a plot.

This paper explains in detail how $T_p$ can be calculated for trigonometric functions.

For $a = 3, b=6$ (following examples 17 and 18 on page 60) $T_p$ can be calculated as follows

For the multiplication part: $\frac{T_1}{T_2} = \frac{3}{6} \rightarrow T_{p,mult} = 2 T_1 = \frac{4\pi}{3}$

For the summation part:$\frac{T_1}{T_2} = \frac{3}{9} \rightarrow T_{p,sum} = 3 T_1 = \frac{2\pi}{3}$

Finally, $T_p = \operatorname{min}(T_{p,mult},T_{p,sum})$ results in $T_p =\frac{2\pi}{3}$.

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