Question

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

Answer 1

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

`-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 = 1 / cos theta` there's no ambiguity in the first term:

`- sec arccos 7 = - 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 (y/x)`. The way to think of these forms is as right triangles. `arctan(y/x)` is a right triangle whose opposite is `y` and adjacent is `x`. So its hypotenuse is `\sqrt{x^2+y^2}` and its cosecant is hypotenuse over opposite, so

`csc arctan(y/x) = pm 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( pm sqrt{1^2+2^2}/2 ) = pm sqrt{5}`

Over the complex numbers,

`- sec arccos 7 + 2 csc arctan 2 = -1/7 pm sqrt{5}`