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### Euler's Formula

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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.

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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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Associated Topics:
High School Imaginary/Complex Numbers

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