Question

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

Answer 1

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.

Answer 2

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`