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Is the function $f(x) = \tan(x)$ odd, even, or neither?

Here is what I have so far:

I know the function is not even because $f(x) ≠ f(-x):$

$$f(-x) = \tan(-x)$$

$$\tan(-x) ≠ \tan(x)$$

Now I want to determine if the function is odd. I know a function is odd if $f(x) = -f(-x)$:

$$-f(-x) = -\tan(-x)$$

How does the negative sign on the outside of the brackets affect $\tan(-x)?$

My instinct is that the function is neither even nor odd, but I would like confirmation.

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Just use the definition of $\tan$ in terms of $\sin$ and $\cos$, and use the fact that $\sin$ is odd, and $\cos$ is even :

$$\tan(-x)=\dfrac{\sin(-x)}{\cos(-x)}=\dfrac{-\sin(x)}{\cos(x)}=-\tan(x).$$

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  • $\begingroup$ If sin is odd and cos is even, would that make tan odd? $\endgroup$ – McB Jan 12 '15 at 21:55
  • $\begingroup$ @McB Yes, by definition. $\endgroup$ – Workaholic Jan 12 '15 at 21:56
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another way to show that is $$\int_{-a}^{a} f(x)\ tan(x) dx =0 ,\ a>0$$ where $f(x)$ is an even function , this fact has a nice geometric explanation, odd function is symmetric about x-axis, so the area between $x=0$ and $x=a$ above is same as the area between $x=0$ and $x=-2$ but in opposite direction.

P.S: $\ f(x)=1$ is sufficient

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