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Given $x(t) = 6\cos(t)\cos(3t)$, I would assume period is $2\pi$, because LCM of $(2\pi, 2\frac{\pi}{3}) = 2\pi$. But when I graph it, it comes out to just $\pi$. Does my method only work for sums and differences of 2 trig functions? Is there a more general way to approach this?

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  • $\begingroup$ "LCM" is short for "least common multiple"? $\endgroup$
    – Galen
    Commented Aug 1, 2022 at 16:00

2 Answers 2

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Note that $\cos{t}\cos{3t} = \dfrac{1}{2}(\cos(t+3t)+\cos(t-3t))$. The right hand side has period $\pi$. So yes, the LCM method only works for sums and differences.

I suppose a general way to treat products of trig functions is to convert them to sums and differences, just like above. Then you can use the LCM approach.

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  • $\begingroup$ Just for a sanity check, once we convert to sums and differences we're finding the LCM of (pi/2, pi)? $\endgroup$
    – Jonathan
    Commented Mar 15, 2016 at 23:32
  • $\begingroup$ That's correct. $\endgroup$
    – Aurey
    Commented Mar 15, 2016 at 23:45
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(1). $\;\cos x \cos y=(1/2) [\;\cos (x-y)+\cos (x+y)\;].$ So $6\cos t \cos 3 t=3(\cos 2 t+\cos 4 t)$. And $\pi$ is the least positive period of $\cos 2 t,$ while $\pi$ is a period of $\cos 4 t.$

(2).$\;\cos (x+\pi)=-\cos x. $ And $\cos 3(x+\pi)=-\cos 3 x.$ So $\pi$ is a period of $\cos x \cos 3 x.$

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  • $\begingroup$ Hmm, i don't understand the "pi is a period of cos(t)" part. Can you please elaborate? $\endgroup$
    – Jonathan
    Commented Mar 15, 2016 at 23:33
  • $\begingroup$ typo.... fixed..... meant to say period of cos 4t. $\endgroup$ Commented Mar 15, 2016 at 23:40
  • $\begingroup$ @Aurey. Right. Fixed it. Thank you. $\endgroup$ Commented Mar 16, 2016 at 0:00

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