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I need to find the period of the following trigonometric function: $$f(x) = \tan2x + \cos2x$$

Any suggestions?

Blue
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Kevin
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  • What did you try so far? Please use MathJax for writing mathematical expressions. – flawr Aug 01 '16 at 15:01
  • Solve $f(x+T)=f(x)$ for the smallest nonzero $T$. In particular, $\tan 2T+\cos 2T=1$, but that's not sufficient. –  Aug 01 '16 at 15:03
  • $$\tan2(x+h)-\tan2x+\cos2(x+h)-\cos2x=\dfrac{\sin2h}{\cos2(x+h)\cos2x}-2\sin h\sin(2x+h)$$

    $$=2\sin h\dfrac{\cos h-\cos2(x+h)\cos2x\sin(2x+h)}{\cos2(x+h)\cos2x}$$

     – lab bhattacharjee Aug 01 '16 at 15:04
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    $\tan n\theta$ has period of $\frac{\pi}{n}$ while $\cos n\theta$ has period of $\frac{2\pi}{n}$. The required period are the $\operatorname{lcm}$ of periods each term. Can you proceed? – Ng Chung Tak Aug 01 '16 at 15:43

1 Answers1

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I know that this question is old, but still, I liked it and I think I can answer it.

First, note that $$\frac{\text{d}}{\text{d}x} \text{tan}(2x)+\text{cos}(2x)=\frac{2}{\text{cos}^2(2x)}-2\text{sin}(2x).$$

Since the following are all equivalent: $$\frac{2}{\text{cos}^2(2x)}-2\text{sin}(2x)>0$$ $$\frac{2}{\text{cos}(2x)}>2\text{sin}(2x)\text{cos}(2x)$$ $$2>\text{cos}(2x)\text{sin}(4x),$$ the last being clearly true since $\text{cos}(2x)\leq 1$ and $\text{sin}(4x)\leq 1$, we then have the original function is strictly increasing when the derivative is not undefined, that is, when $x$ is not a half-integer multiple of $\frac{\pi}{4}$. So, the period has to be at least $\frac{\pi}{2}$, and the problem is solved when we verify such period works.

Useful link.

Anonymous Pi
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