Period of cosine curve

**Mike9182** · August 13th 2009, 05:13 PM

Correct Answer Explanation:

How do you come up with the first step of the correct answer? I understnd the simplification from there.

**pickslides** · August 13th 2009, 05:20 PM

easier way: for $\cos(nx)$ the period is $\frac{2\pi}{n}$

therefore for $\cos\left(\frac{\pi}{4}x\right)$ the period is $\frac{2\pi}{\frac{\pi}{4}} = 8$

**Mike9182** · August 13th 2009, 05:26 PM

thanks, although how do you get the first step of the correct answer explanation I posted?

**pankaj** · August 13th 2009, 05:31 PM

Originally Posted by Mike9182

Correct Answer Explanation:

How do you come up with the first step of the correct answer? I understnd the simplification from there.

In the first step you are using the fact that $f(x)=\cos x$ is periodic with period $2\pi$

In general if fundamental period of $f(x)$ is $T$ ,then fundamental period of of $f(ax+b)$ is $\frac{T}{|a|}$ , where, $a\neq 0$

The fundamental period of $f(x)=\cos x$ is $2\pi$ .

Therefore,the fundamental period of $f\left(\frac{\pi}{4}x\right)$ is $\frac{2\pi}{\frac{\pi}{4}}=8$

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