How can I check the parity of transcendental functions?

Question

I know how to check it in general ($f(x)=f(-x)$) but I don't understand how I can check it for any transcendental functions, because I cannot check if (for example) $\tan(x)= \tan(-x)$

Answer 1

The following is to clarify some concepts and is too long to put in the comment column above.

Not all transcendental functions are purely odd or even. For example,

\begin{align*} e^{-x} & \ne -e^{x} \\ e^{-x} & \ne e^{x} \end{align*}

However, we can artificially redefine a new function from old.

\begin{align*} g(x) &= \frac{f(x)+f(-x)}{2} \\ \implies g(-x) &= g(x) \\ h(x) &= \frac{f(x)-f(-x)}{2} \\ \implies h(-x) &= -h(x) \\ \end{align*}

• Now $g(x)$ is even and $h(x)$ is odd.

• $g(x)+h(x)=f(x)$ that recovers $f(x)$.

• $g(x)$ and $h(x)$ can be regarded as the even and odd components of $f(x)$ respectively.

In particular, \begin{align*} \cosh x &= \frac{e^{x}+e^{-x}}{2} \\ \sinh x &= \frac{e^{x}-e^{-x}}{2} \\ \end{align*}

Also, \begin{align*} \cos x &= \frac{e^{ix}+e^{-ix}}{2} \\ \sin x &= \frac{e^{ix}-e^{-ix}}{2i} \\ \end{align*}

In your case: \begin{align*} \tan x &= \frac{\sin x}{\cos x} \\ &= \frac{e^{ix}+e^{-ix}}{e^{ix}-e^{-ix}} \\ \tan (-x) &= \frac{e^{-ix}+e^{ix}}{e^{-ix}-e^{ix}} \\ &= -\frac{e^{ix}+e^{-ix}}{e^{ix}-e^{-ix}} \\ &= -\tan x \end{align*} which is odd.