Question

What does `sin(arccos(3))+csc(arc cot(4))` equal?

Answer 1

`sin(arccos(3))+csc(arc cot(4)) = 2sqrt(2) i + sqrt(17)`

Explanation:

If `sin`, `cos`, etc are being treated as Real valued functions of Real values, then `arccos(3)` is not defined, since `3` lies outside the range of `cos(x)` which is `[-1, 1]`.

However, inspired by Euler's formula `e^(ix) = cos x + i sin x`, it is possible to define `sin`, `cos` etc for Complex values as follows:

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

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

With these definitions, note that `sin^2 z + cos^2 z = 1` for any `z in CC`.

Suppose `cos z = a` for some `a in CC`. What is `z`?

Let `t = e^(iz)`. Then `e^(-iz) = 1/t` and we have:

`(t+1/t)/2 = cos z = a`

Hence:

`t+1/t = 2a`

Hence:

`t^2-2at+1 = 0`

Solving using the quadratic formula, we get:

`e^(iz) = t = (2a+-sqrt(4a^2-4))/2 = a+-sqrt(a^2-1)`

So:

`iz = ln(a+-sqrt(a^2-1))`

Hence:

`z = -i ln(a+-sqrt(a^2-1)) = +-i ln(a+sqrt(a^2-1))`

It is reasonable that the definition of `arccos(x)` for `x > 1` should choose the positive sign for this expression.

Hence:

`arccos(3) = i ln(3+sqrt(8)) = i ln(3+2sqrt(2))`

Then:

`sin(arccos(3))`

`= (e^(-ln(3+2sqrt(2)))-e^(ln(3+2sqrt(2))))/(2i)`

`=1/(2i)(1/(3+2sqrt(2)) - (3+2sqrt(2)))`

`=i/2((3+2sqrt(2))-(3-2sqrt(2)))`

`=color(blue)(2sqrt(2)i)`

`color(white)()`
On the other hand, `csc(arc cot(4))` is all about Real values and somewhat easier to construct.

Consider a right angled triangle with sides `1`, `4` and `sqrt(17)`

Let `theta` be the smallest angle, i.e. the one between the hypotenuse and the longer leg.

Then `cot(theta) = "adjacent"/"opposite" = 4/1 = 4`

and `csc(theta) = "hypotenuse"/"opposite" = sqrt(17)/1 = sqrt(17)`

So:

`csc(arc cot(4)) = color(blue)(sqrt(17))`