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

1 Answer
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

#-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} #