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What's the prime period of the following function?

$$\frac{\sin 2x + \cos 2x}{\sin 2x - \cos 2x}$$

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up vote 3 down vote accepted

$$\begin{align*} \frac{\sin 2x+\cos 2x}{\sin 2x-\cos 2x}&=\frac{\sin 2x+\cos 2x}{\sin 2x-\cos 2x}\cdot\frac{\sin 2x+\cos 2x}{\sin 2x+\cos 2x}\\\\ &=\frac{(\sin 2x+\cos 2x)^2}{\sin^2 2x-\cos^2 2x}\\\\ &=-\frac{\sin^2 2x+2\sin 2x\cos 2x+\cos^2 2x}{\cos 4x}\\\\ &=-\frac{1+\sin 4x}{\cos 4x}\\\\ &=-\sec 4x-\tan 4x\;. \end{align*}$$

The first term has primitive period $\dfrac{2\pi}4=\dfrac{\pi}2$, and the second has primitive period $\dfrac{\pi}4$. Can you finish it from there?

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It would be $\dfrac{\pi}4$, right? could you please give me a link about how to find period of functions and what does happen to period by adding or subtracting or multiplying two functions? – Gigili Jan 23 '12 at 10:33
@zizi: No, it would be $\pi/2$. Take a look at the graph: what you see between $-3\pi/8$ and $\pi/8$ is one full period. What happens to the period when you add or multiply two period functions can get complicated, depending on the functions and the periods; I don’t offhand know of a good online reference. – Brian M. Scott Jan 23 '12 at 10:37
got it, thanks. – Gigili Jan 23 '12 at 10:48

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