Question

How do you evaluate `sin(sin^-1(10/3))`?

Answer 1

For Real valued `sin` this is undefined, but with the Complex definition of `sin z` we find:

`sin(sin^(-1)(10/3)) = 10/3`

as we would hope.

Explanation:

If `sin x` is treated as a Real function of Real values, then its range is `[-1, 1]`. So `sin^(-1)(10/3)` is undefined and the expression `sin(sin^(-1)(10/3))` is consequently undefined.

But...

Note that:

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

Hence we find:

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

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

We can use these to extend the definition of `cos` and `sin` to Complex valued functions of Complex numbers, by defining:

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

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

So what values of `z in CC` satisfy `sin z = 10/3` and which should we consider the principal value?

Suppose:

`10/3 = sin z = (e^(iz) - e^(-iz))/(2i)`

Multiplying both ends by `2i` and transposing, we get:

`e^(iz) - e^(-iz) = 20/3i`

Multiplying through by `e^(iz)` and rearranging slightly, we get:

`(e^(iz))^2-20/3i (e^(iz)) - 1 = 0`

Using the quadratic formula, we find:

`e^(iz) = (20/3i+-sqrt(-400/9+4))/2`

`color(white)(e^(iz)) = 10/3i+-sqrt(91)/3i`

`color(white)(e^(iz)) = 1/3(10+-sqrt(91))i`

Hence:

`z = 1/i(ln(1/3(10+-sqrt(91))i))+2kpi`

`color(white)(z) = 2kpi-i (ln(1/3(10+-sqrt(91))i))`

Rather than spend time figuring out which of these might be the best candidate for the principal value of `arcsin(10/3)` let's see what happens if we substitute any of them:

Then:

`sin z = sin (2kpi-i (ln(1/3(10+-sqrt(91))i)))`

`color(white)(sin z) = sin (-i (ln(1/3(10+-sqrt(91))i)))`

`color(white)(sin z) = (e^(ln(1/3(10+-sqrt(91))i))-e^(-ln(1/3(10+-sqrt(91))i)))/(2i)`

`color(white)(sin z) = (1/3(10+-sqrt(91))i-1/(1/3(10+-sqrt(91))i))/(2i)`

`color(white)(sin z) = (1/3(10+-sqrt(91))i+1/3(10 color(white)(.)bar("+")color(white)(.) sqrt(91))i)/(2i)`

`color(white)(sin z) = 10/3`