# How do you calculate  [tan^-1(-1) + cos^-1(-4/5)]?

May 15, 2016

${278.13}^{o} , {351.87}^{o} , {458.13}^{o} \mathmr{and} {531.87}^{o}$, choosing the range $\left[{0}^{o} , {360}^{0}\right]$, for the inverse functions.. .

#### Explanation:

It is anticlockwise rotation for angles. So, negative angles do not appear.

Let $a = {\tan}^{- 1} \left(- 1\right)$.

Then, $\tan a = - 1 < 0$.

So, a is in the 2nd or 4th quadrant.

The solutions in $\left[{0}^{o} , {360}^{0}\right]$ are $a = {135}^{o} \mathmr{and} {315}^{o}$.

Let $b = {\cos}^{- 1} \left(- \frac{3}{4}\right)$.

Then, $\cos a = - \frac{3}{4} < 0$.

So, a is in the 2nd or 3rd quadrant.

The solutions in $\left[{0}^{o} , {360}^{0}\right]$ are $b = {143.13}^{o} \mathmr{and} {216.87}^{o}$

The given expression is

$a + b = {278.13}^{o} , {351.87}^{o} , {458.13}^{o} \mathmr{and} {531.87}^{o}$ .

If negative angles are used like #[-180^o, 180^0] for the range of

separate solutions, instead, the results would change, accordingly.