# How do you calculate #arcsin 2#?

##### 1 Answer

As a Real valued function

Defining

#arcsin(2) = pi/2+ln(2+sqrt(3)) i#

#### Explanation:

However, it is possible to define

Use the following:

#e^(ix) = cos(x) + i sin(x)#

#cos(-x) = cos(x)#

#sin(-x) = -sin(x)#

to find:

#sin(x) = (e^(ix)-e^(-ix))/(2i)# for all#x in RR#

Define:

#sin(z) = (e^(iz)-e^(-iz))/(2i)# for all#z in CC#

We want to solve

#(e^(iz)-e^(-iz))/(2i) = 2#

Let

Then this equation becomes:

#2 = (-it-i/t)/(2i) = -(t+1/t)/2 = -(t^2+1)/(2t)#

Multiply both ends by

#t^2+1 = -4t#

Add

#t^2+4t+1 = 0#

Using the quadratic formula, this has roots:

#t = (-4+-sqrt(4^2-4))/2 = -2+-sqrt(3)#

So:

#i e^(iz) = -2+-sqrt(3)#

#e^(iz) = (-2+-sqrt(3))/i = (2+-sqrt(3))i = (2+-sqrt(3))e^(i pi/2)#

#e^(i(z-pi/2)) = 2+-sqrt(3)#

Taking natural logs:

#i(z-pi/2) = ln(2+-sqrt(3))#

Hence:

#z = pi/2 +-ln(2+sqrt(3)) i#

The principal value in Q1 is