How do you calculate #arcsin 2#?

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Answer 1 · 2015-11-11T19:22:30

As a Real valued function #arcsin 2# is undefined, since #sin(x) in [-1, 1]# for all #x in RR#.

Defining #sin(z)# for #z in CC# we find:

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

#sin(x) in [-1, 1]# for all #x in RR#. So there is no Real value of #x# such that #sin(x) = 2#.

However, it is possible to define #sin(z)# for #z in CC# and hence find a workable definition and value for #arcsin(2)#.

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 #sin(z) = 2#, that is:

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

Let #t = i e^(iz)#

Then this equation becomes:

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

Multiply both ends by #-2t# to get:

#t^2+1 = -4t#

Add #4t# to both sides to get:

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