How do I find the exact value of tan⁻¹(tan (4 \pi )/(5) ) ? I know the answer is negative, but I don't know why the inverse of tan and tan don't cancel out for the ans…

Question

How do I find the exact value of tan⁻¹(tan
`(4 \pi )/(5)` ) ? 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
`-\frac\pi2<x<\frac\pi2`, which gives one complete branch of values (or one period).
`\frac{4\pi}5>\frac\pi2` and thus
`\frac{4\pi}5` is outside the domain.

A different way to go about this is to find the value of
`\tan\frac{4\pi}5` first, then compute the inverse tangent of that result. But finding the trigonometric values of multiples of
`\frac\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\frac{4\pi}5=\tan\left(\frac{4\pi}5-\pi\right)=\tan\left(-\frac\pi5\right)`. And now the function can be inverted, so that


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