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.
$\endgroup$2 Answers
$\begingroup$Just use the definition of $\tan$ in terms of $\sin$ and $\cos$, and use the fact that $\sin$ is odd, and $\cos$ is even :
$$\tan(-x)=\dfrac{\sin(-x)}{\cos(-x)}=\dfrac{-\sin(x)}{\cos(x)}=-\tan(x).$$
$\endgroup$ 2 $\begingroup$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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