The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Euler's Formula

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)


 = (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))


  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!   
Associated Topics:
High School Imaginary/Complex Numbers

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.