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/
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