Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What's the prime period of the following function?

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

share|improve this question

1 Answer 1

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?

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.