Date: 01/27/98 at 21:33:23 From: Anonymous Subject: Trig id's; Euler's equation I have been using trig for some time now, but the identities are still one of my weaknesses. I read the answer to a related question you posted. A few years ago I saw a electrical engineer use a technique that involved a combination of Euler's equation and the unit circle. He could quickly derive whatever identity he needed from this simple set-up. I took notes at that time but only was shown two examples and never understood exactly how to do it. Do you know what I am referring to? If so, please explain it to me.
Date: 01/31/98 at 14:24:10 From: Doctor Luis Subject: Re: Trig id's; Euler's equation If by the trigonometric identities you mean the addition formulas and the like, I can give you a few hints.. Of course, you know that Euler's formula is exp(i*t) = cos(t) + i*sin(t) where exp(w) = e^w Now, interesting things happen when you start doing regular operations with the exponential function and then applying the identity.. exp(i*a)*exp(i*b) = exp(i*(a+b)) exponent law Now, substitute the identities, exp(i*(a+b)) = cos(a+b)+i*sin(a+b) and exp(i*a)*exp(i*b) = (cos(a)+i*sin(a))*(cos(b)+i*sin(b)) multiply these two complex no. = (cos(a)*cos(b)-sin(a)*sin(b)) + i*(cos(a)*sin(b)+sin(a)*cos(b)) Therefore, cos(a+b)+i*sin(a+b) = (cos(a)*cos(b)-sin(a)*sin(b)) + i*(cos(a)*sin(b)+sin(a)*cos(b)) Since two complex numbers are equal only when their real parts are equal to each other, and their imaginary parts as well, that means (real part) cos(a+b) = cos(a)*cos(b)-sin(a)*sin(b) and also, (imaginary parts) sin(a+b) = cos(a)*sin(b)+sin(a)*cos(b) Notice, that all I needed to remember was Euler's identity, and how to multiply complex numbers.. Need an expression for cos(3x) ? No problem: exp(i*3t) = (exp(i*t))^3 cos(3t)+i*sin(3t) = (cos(t)+i*sin(t))^3 with a little bit of algebra, you can see that (x+y)^3 = x^3 + 3(x^2)y + 3x(y^2) + y^3 And so, cos(3t)+i*sin(3t) = (cos(t)+i*sin(t))^3 = (cos(t))^3 + 3((cos(t))^2)*(i*sin(t)) + 3(cos(t))*(i*sin(t))^2 + (i*sin(t))^3 = cos^3(t) + (3cos^2(t)sin(t))*i -3cos(t)sin^2(t) - i*sin^3(t) Taking the real part of both sides, we find, cos(3t) = cos^3(t) - 3cos(t)sin^2(t) We obtain a similar identity for sin(3t) by taking the imaginary part, sin(3t) = 3cos^2(t)sin(t) - sin^3(t) If you know the binomial theorem you can quickly come up with an identity for cos(n*x). -Doctor Luis, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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