# What does #-3sin(arccos(2))-cos(arc cos(3))# equal?

##### 2 Answers

Problem insolvable

#### Explanation:

There are no arcs that their cosine are equal to 2 and 3.

From an analytic point of view, the

For Real

#-3 sin(arccos(2))-cos(arccos(3)) = -3sqrt(3)i-3#

#### Explanation:

As Real valued functions of Real values of

However, it is possible to extend the definition of these functions to Complex functions

Starting with:

#e^(ix) = cos x + i sin x#

#cos(-x) = cos(x)#

#sin(-x) = -sin(x)#

we can deduce:

#cos(x) = (e^(ix)+e^(-ix))/2#

#sin(x) = (e^(ix)-e^(-ix))/(2i)#

Hence we can define:

#cos(z) = (e^(iz)+e^(-iz))/2#

#sin(z) = (e^(iz)-e^(-iz))/(2i)#

for any Complex number

It is possible to find multiple values of

To find suitable candidates, solve

However, note that the identity

#sin(arccos(2)) = +-sqrt(1-2^2) = +-sqrt(-3) = +-sqrt(3) i#

I hope that it's possible to define the principal value in such a way that

In any case,

Putting this all together, we find:

#-3 sin(arccos(2))-cos(arccos(3)) = -3sqrt(3)i-3#