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

##### 1 Answer

Dec 20, 2016

#### 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#