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Deriving the Arcsin Formula


Date: 01/24/99 at 22:15:37
From: Michael Suzio
Subject: Inverse sin of numbers greater than 1

Dear Dr. Math,

In physics recently we were studying waves, and in a form of Snell's 
law, we used sin(theta) = n1/n2. If n1 is greater than n2, then light 
cannot be refracted because sin(theta) cannot ever be greater than 1. 
But on my calculator if I ask for inverse sin(4/3), it gives me 
(1.57079632679, -.795365461224). This is equivalent in A + BI form to
1.57-.795i. I was wondering why this was true and why the real part 
of the imaginary number is always 1.57...

Thank you.


Date: 01/25/99 at 12:21:17
From: Doctor Peterson
Subject: Re: Inverse sin of numbers greater than 1

Hi, Michael. I had to work out the formula for the inverse sine in this 
case, because I didn't recall ever seeing it, but it's not too hard, 
given that you are familiar with trig and exponential functions of 
complex numbers. Start with the fact that

   exp(x + iy) = exp(x) (cos(y) + i sin(y))

and derive from this (if you don't already know it) that

   sin(y) = (exp(iy) - exp(-iy)) / (2i)

Now we're looking for a complex number whose sine is, in your example, 
4/3. To get the general formula for the inverse sine of a real A > 1, 
let's say

   sin(x + iy) = A

Put x + iy in for y in my sine formula:

   sin(x + iy) = (exp(i(x + iy)) - exp(-i(x + iy))) / (2i)
               = (exp(-y + ix) - exp(y - ix)) / (2i)
               = (exp(-y) * (cos(x) + i sin(x)) -
                  exp(y) * (cos(x) - i sin(x)) / (2i)
               = cos(x) * (exp(-y) - exp(y)) / (2i) +
                 i sin(x) * (exp(-y) + exp(y)) / (2i)
               = i cos(x) * (exp(y) - exp(-y))/2 +
                   sin(x) * (exp(y) + exp(-y))/2

Setting this equal to our real number A, 

   sin(x) * (exp(y) + exp(-y))/2 + i cos(x) * (exp(y) - exp(-y))/2 = A

Since A is real, the imaginary part of this is zero, which means either

   cos(x) = 0                so  x = pi/2   or
   (exp(y) - exp(-y))/2 = 0  so  y = 0

The second case just gives A = sin(x), and only works for x <= 1. So 
the first case must be true, and we have the real part x = pi/2 of the 
answer. For the imaginary part y, we set the real parts of the two 
sides above equal, remembering that sin(x) = sin(pi/2) = 1:

   (exp(y) + exp(-y))/2 = A

Solve this for A (using a quadratic equation in exp(y)), and you get

   y = ln(A +- sqrt(A^2 - 1))

With a little thought you'll see that the +- can be moved outside, 
because

   ln(A - sqrt(A^2 - 1)) = -ln(A + sqrt(A^2 - 1))

(If you are familiar with hyperbolic functions, this is just

   cosh(y) = A
   y = arccosh(A)

so you could save some work if you knew these formulas.)

So our solution is

   arcsin(A) = pi/2 + i ln(A + sqrt(A^2 - 1))

For your A = 4/3, this gives

   arcsin(4/3) = pi/2 + i ln(4/3 + sqrt(7)/3)
               = 1.570796326795 + 0.7953654612239i

By the way, notice that if A = 1, both formulas apply:

  arcsin(1) = pi/2
  pi/2 + i ln(1 + sqrt(1^2 - 1)) = pi/2

As A increases, its arcsin moves along the real axis to pi/2, then 
turns 90 degrees and moves away parallel to the imaginary axis.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Physics/Chemistry
High School Trigonometry

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