# What does -sec(arccos(7))+2csc(arctan(2)) equal?

May 31, 2018

$\arccos 7$ is undefined over the reals. Over the complex numbers, using a multivalued interpretation of the inverse trig functions, this expression equals

$- \frac{1}{7} \pm \sqrt{5}$

#### Explanation:

I prefer the interpretation where small letter $\arccos$ etc. are multivalued, representing all the angles whose cosine is a given value, etc.

Since $\sec \theta = \frac{1}{\cos} \theta$ there's no ambiguity in the first term:

$- \sec \arccos 7 = - \frac{1}{7}$

Of course this glosses over the detail that there are no real angles whose cosine is seven. Over the reals, $\arccos 7$ is undefined.

The second term ask for $\sec \arctan \left(\frac{y}{x}\right)$. The way to think of these forms is as right triangles. $\arctan \left(\frac{y}{x}\right)$ is a right triangle whose opposite is $y$ and adjacent is $x$. So its hypotenuse is $\setminus \sqrt{{x}^{2} + {y}^{2}}$ and its cosecant is hypotenuse over opposite, so

$\csc \arctan \left(\frac{y}{x}\right) = \pm \frac{\sqrt{{x}^{2} + {y}^{2}}}{y}$

Whenever we get one of these with a square root we'll have an ambiguity about the sine if we consider the inverse functions multivalued.

$2 \csc \arctan 2 = 2 \left(\pm \frac{\sqrt{{1}^{2} + {2}^{2}}}{2}\right) = \pm \sqrt{5}$

Over the complex numbers,

$- \sec \arccos 7 + 2 \csc \arctan 2 = - \frac{1}{7} \pm \sqrt{5}$