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Power Series for Sine and Cosine

Date: 10/12/2000 at 15:51:03
From: Mark
Subject: The expansion of trigonometric functions

Recently it was brought to my attention that 

    e^(pi*i) = -1
I have had numerous teachers try to explain it to me, but all of their 
proofs involve power series. Once you get to Euler's formula, I 
understand how the math works out, and getting Euler's formula from 
the power series, I understand, but I don't understand why the series

    x + x^3/3! + x^5/5! + ... 

or whatever it is, is applicable to the sine or cosine function. How 
do you get that series?

A confused 10th-grade student

Date: 10/12/2000 at 16:20:29
From: Doctor Rob
Subject: Re: The expansion of trigonometric functions

Thanks for writing to Ask Dr. Math, Mark.

The answer to this question involves calculus. The parts you need to 
know are about derivatives of functions. The derivative of f(x) 
evaluated at x = x0 gives the slope of the tangent line to the curve 
y = f(x) at the point (x0,f[x0]). The key facts are these:

   1. The derivative of sin(x) is cos(x).
   2. The derivative of cos(x) is -sin(x).
   3. The derivative of x^n is n*x^(n-1) for any integer n.
   4. The derivative of c*f(x) is c times the derivative of f(x), for
      any constant c.
   5. The derivative of f(x)+g(x) is the sum of the derivatives of 
      f(x) and g(x).

Now assume that sin(x) (for example) has a convergent power series 
expansion, that is,

    sin(x) = a(0) + a(1)*x + a(2)*x^2 + ... + a(n)*x^n + ...

Substitute x = 0 in this equation, and you get a(0) = 0. Now take the 
derivative of both sides:

    cos(x) = a(1) + 2*a(2)*x + 3*a(3)*x^2 + ... + n*a(n)*x^(n-1) + ...

Substitute x = 0 in this new equation, and you get a(1) = 1. Again 
take the derivative of both sides:

    -sin(x) = 2!/0!*a(2) + 3!/1!*a(3)*x + 4!/2!*a(4)*x^2 + ...

Substitute x = 0 in this new equation, and you get a(2) = 0. Again 
take the derivative of both sides:

    -cos(x) = 3!/0!*a(3) + 4!/1!*a(4)*x + 5!/2!*a(5)*x^2 + ...

Substitute x = 0 in this new equation, and you get a(3) = -1/3!. Again 
take the derivative of both sides:

    sin(x) = 4!/0!*a(4) + 5!/1!*a(5)*x + 6!/2!*a(6)*x^2 + ...

Substitute x = 0 in this new equation, and you get a(4) = 0.

If you repeat this, you will see (and can prove by induction) that the 
coefficients are given by

    a(2*k) = 0
    a(2*k+1) = (-1)^k/(2*k+1)!

for every integer k. That means that the series has form

    sin(x) = x - x^3/3! + x^5/5! - x^7/7! + x^9/9! - ...

By taking the derivative of this, you get the series for cosine:

    cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! + x^8/8! - ...

Now I haven't proved that these series converge to the values of sine 
and cosine, just that if there is a convergent series, it must have 
those coefficients. Actually, it isn't too hard to prove that these 
series converge to some limit, no matter the value of x. The method of 
construction is what guarantees that the limit is, in fact, the value 
of sin(x) (or cos(x)), respectively.

The rest of this development you will encounter when you study 
infinite power series in calculus. It is studied under the name 
Maclaurin's Series.

If you need more help, feel free to write back.

- Doctor Rob, The Math Forum   
Associated Topics:
High School Calculus
High School Sequences, Series

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