Proving Period of Trigonometric Functions

[Vertciel]

Hello everyone,

I have to prove the periods of all six trigonometric functions and I would like to ask if anyone can help me on how I should do so, without showing the full solution.

For example: Show that the period of $\displaystyle f(\theta) = cos (\theta)$ is $\displaystyle 2\pi$.


Thank you.

 

[tkhunny]

Rephrase and prove:

$\displaystyle \cos(\theta + 2\pi) = \cos(\theta)$

Then think about it. Does that actually PROVE the period is $\displaystyle 2\pi$?

 

[Vertciel]

Thank you for your response.

Here is my proof:

$\displaystyle f(\theta) = cos \theta$ is $\displaystyle 2\pi$.

$\displaystyle f(\theta) = f(\theta + P)$

$\displaystyle cos(\theta) = cos(\theta + 2\pi)$

Since $\displaystyle cos(\theta) = cos(\theta + 2\pi)$ and $\displaystyle cos(\theta) = cos(\theta)$ , $\displaystyle 2\pi$ is the period in which the trigonometric function repeats itself.

Would this be an acceptable proof? Is there anything I can do to reinforce the proof?

 

[tkhunny]

No, sorry. You have proven that the function repeats itself every $\displaystyle 2\pi$. This is insufficient. You could do the same to prove that it repeats every $\displaystyle 4\pi$ or $\displaystyle 8\pi$. You have not proven that it does NOT repeat any faster than $\displaystyle 2\pi$. The "period" is the minimum value. Good try, though. You almost let me talk you into it. Perhaps you could demonstrates that the cosine is positive or negative over various Domains? What do you think?

 

[Vertciel]

Thanks for your reply. Could you please explain what you mean by demonstrating "that the cosine is positive or negative over various domains"?

 

[tkhunny]

$\displaystyle \cos(\theta)\;>\;0$ for $\displaystyle \theta \in (-\frac{\pi}{2},\frac{\pi}{2})$

$\displaystyle \cos(\theta)\;<\;0$ for $\displaystyle \theta \in (\frac{\pi}{2},\frac{3\pi}{2})$

If this is the case, there can't very well be a period less than $\displaystyle \pi$.