Question

What does `sin(arccos(4))-cot(arccos(1))` equal?

Answer 1

This expression cannot be calculated. See explanation.

Explanation:

The trigonimetric functions `sin` and `cos` have range `<-1;1>`, so there are NO values of `x` for which `cosx=4`, so `arccos(4)` is NOT defined.

Answer 2

This is undefined on two distinct counts:

(1) If dealing with Real valued functions, then `4` is not in the range of `cos(x)`, so `arccos(4)` is undefined.

(2) `cot(arccos(1))` is always undefined.

Explanation:

As a Real valued function of Reals, the range of the function `cos(x)` is `[-1, 1]`, so `arccos(4)` is undefined.

But...

Note that `e^(i theta) = cos theta + i sin theta`

Hence:

`cos theta = (e^(i theta) + e^(-i theta))/2`

`sin theta = (e^(i theta) - e^(-i theta))/(2i)`

This yields definitions for `cos z` and `sin z` for Complex values of `z` using the formulae:

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

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

With these definitions, we find that the Pythagorean identity still holds:

`cos^2 z + sin^2 z = 1" "` for all `z in CC`

Hence:

`sin(arccos(4)) = sqrt(1-4^2) = sqrt(-15) = sqrt(15)i`

How about `cot(arccos(1))` ?

Here we still get an undefined value since:

`arccos(1) = 0` and `cot(0) = cos(0)/sin(0) = 1/0` which is undefined.