Question

What is `cos(ln(x))`?

Answer 1

`cos(ln(x)) = (x^i+x^(-i))/2`

Explanation:

Given that:

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

`cos (-theta) = cos(theta)`

`sin (-theta) = -sin(theta)`

we can deduce that:

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

So:

`cos(ln(x)) = (e^(ilnx)+e^(-ilnx))/2`

`color(white)(cos(ln(x))) = ((e^(lnx))^i+(e^(lnx))^(-i))/2`

`color(white)(cos(ln(x))) = (x^i+x^(-i))/2`