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#