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What is the period of $f(x) = \cos(x) \cos(2x) \cos(3x)$? Please tell me the method plus the logic behind solving these kind of problems .. Plus is there any property for even functions like even functions are always onto ?

Irregular User
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Ujjwal
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    You really need to learn to make titles more specific. If it ends in the number 5 because it's the 5th question about the same general topic you've asked... that's not good. I really don't see why you couldn't put the actual question in the title. – pjs36 Jun 04 '16 at 07:32
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    I highly encourage you to make a more meaningful title. – davidlowryduda Jun 04 '16 at 07:33
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    See: http://math.stackexchange.com/questions/164221/period-of-the-sum-product-of-two-functions – Irregular User Jun 04 '16 at 07:36
  • See http://math.stackexchange.com/questions/873723/how-to-find-the-period-of-the-sum-of-two-trigonometric-functions – lab bhattacharjee Jun 04 '16 at 07:39

2 Answers2

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\begin{align} \cos(3x - x) &= \cos 3x \cos x + \sin 3x \sin x \\ \cos(3x + x) &= \cos 3x \cos x - \sin 3x \sin x \\ \hline \cos x \cos 3x &= \dfrac 12 \cos 2x + \dfrac 12 \cos 4x \\ \cos x \cos 3x &= \dfrac 12 \cos 2x + \dfrac 12 (2 \cos^2 2x - 1) \\ \cos x \cos 3x &= \dfrac 12 \cos 2x + \cos^2 2x - \dfrac 12 \\ \hline \cos x \cos 2x \cos 3x &= \dfrac 12 \cos^2 2x + \cos^3 2x - \dfrac 12 \cos 2x \\ \end{align}

The period is $\pi$.

Steven Alexis Gregory
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That the product has period $\pi$ is easily seen, once we prove $$\cos x\cos2x\cos3x =\frac{1}{4}\left(\cos6x+\cos4x+\cos2x+1\right).$$There are at least two quick ways to do this. One uses $\cos nx =\tfrac{1}{2}\left(z^n+z^{-n}\right)$ with $z:=e^{ix}$; the other uses $\cos A\cos B=\tfrac{1}{2}\left(\cos\left(A+B\right)+\cos\left(A-B\right)\right)$.

Edited for a more detailed explanation: the product is $$\tfrac{1}{8}\left(z+z^{-1}\right)\left(z^2+z^{-2}\right)\left(z^3+z^{-3}\right)=\tfrac{1}{8}\left(z^6+z^{-6}+z^4+z^{-4}+z^2+z^{-2}+2\right).$$The right-hand side is $\tfrac{1}{8}\left(2\cos 6x+2\cos 4x+2\cos 2x+2\right).$

J.G.
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  • Can you please explain to me the first method ? I'm pretty weak with complex numbers ... So if you could please illustrate how to process with the first method .. – Ujjwal Jun 05 '16 at 04:56
  • It might be a bit much for a comment, so I've expanded the answer somewhat. – J.G. Jun 05 '16 at 08:11
  • How did you get 1/8 outside ? Shouldn't it be 1/6 according to the formula ? And is $z^6+z^{-6}$ equal to 2cos6x? If it is , then kindly explain how. My basics of complex numbers are not clear. Plus could you suggest me some posts on here where I can build up my concepts on complex numbers ? – Ujjwal Jun 05 '16 at 11:11
  • @Ujjwal the $\tfrac{1}{8}$ factor is the product of three $\tfrac{1}{2}$ factors from $\cos nx=\tfrac{1}{2}\left(z^n+z^{-n}\right)$ for $n\in\left{1,2,3\right}$. The cases $n\in\left{0,2,4,6\right}$ allow the right-hand side to be rewritten in terms of cosines, including $\cos 0x=1$. I'm not sure what posts here would help you learn more; sorry. – J.G. Jun 05 '16 at 13:42
  • What does the term $e^ix$ really signify ? And what do you mean by writing right hand side for n belonging to 0, 2, 4, 6? And why do we take only even integrals? – Ujjwal Jun 05 '16 at 14:05
  • And I didn't see the explanation properly now I got the reason for the existence of 1/8 outside. – Ujjwal Jun 05 '16 at 14:06
  • @Ujjwal It can be shown that $\cos x + i\sin x = e^{ix}$. (Hint: the functions are equal at $x=0$; the derivative of their ratio can be shown to be $0$.) The relevant values of $n$ on each side of the equation are those for which $z^n+z^{-n}$ appears with some coefficient. – J.G. Jun 05 '16 at 14:19