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

2 Answers
Dec 16, 2015

Problem insolvable

Explanation:

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

From an analytic point of view, the #arccos# function is only defined on #[-1,1]# so #arccos(2)# & #arccos(3)# don't exist.

Dec 20, 2015

For Real #cos# and #sin# this has no solutions, but as functions of Complex numbers we find:

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

Explanation:

As Real valued functions of Real values of #x#, the functions #cos(x)# and #sin(x)# only take values in the range #[-1, 1]#, so #arccos(2)# and #arccos(3)# are undefined.

However, it is possible to extend the definition of these functions to Complex functions #cos(z)# and #sin(z)# as follows:

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 #z#.

It is possible to find multiple values of #z# that satisfy #cos(z) = 2# or #cos(z) = 3#, so there could be some choices to be made to define the principal value #arccos(2)# or #arccos(3)#.

To find suitable candidates, solve #(e^(iz)+e^(-iz))/2 = 2#, etc.

However, note that the identity #cos^2 z + sin^2 z = 1# holds for any Complex number #z#, so we can deduce:

#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 #sin(arccos(2)) = sqrt(3) i# rather than #-sqrt(3) i#.

In any case, #cos(arccos(3)) = 3# by definition.

Putting this all together, we find:

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