Question

How do I find the exact value of tan⁻¹(tan ula1" title=" \frac{4 \pi }{5} " alt=" \frac{4 \pi }{5} " align="absmiddle" class="latex-formula"> ) ? I know the answer is negative, but I don't know why the inverse of tan and tan don't cancel out for the answer to just be 4pi/5 ?

Answer 1

This is because $\tan x$ is a multivalued function and is not invertible over its entire domain. We restrict its domain to the interval $-\dfrac\pi2$, which gives one complete branch of values (or one period). $\dfrac{4\pi}5>\dfrac\pi2$ and thus $\dfrac{4\pi}5$ is outside the domain.

A different way to go about this is to find the value of $\tan\dfrac{4\pi}5$ first, then compute the inverse tangent of that result. But finding the trigonometric values of multiples of $\dfrac\pi5$ is somewhat tricky and perhaps more work than is needed.

Instead, we can use a trigonometric identity to find the value of $\tan$ whenever its argument falls outside the "standard" branch. 

We know that $\tan(x\pm\pi)=\tan x$ (because $\tan$ is $\pi$-periodic), so $\tan\dfrac{4\pi}5=\tan\left(\dfrac{4\pi}5-\pi\right)=\tan\left(-\dfrac\pi5\right)$. And now the function can be inverted, so that

$\tan^{-1}\left(\tan\dfrac{4\pi}5\right)=\tan^{-1}\left(\tan\left(-\dfrac\pi5\right)\right)=-\dfrac\pi5$