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