Deriving Lagrange's Trig Identity

Date: 01/09/2004 at 20:07:45
From: Mike
Subject: complex analysis--Lagrange's trig identity

Using the identity, 1 + z + z^2 +...+ z^n = (1 - z^(n+1))/(1 - z), z
not = 1, derive Lagrange's trig identity:

1 + cosx + cos(2x) +...+ cos(nx) = 1/2 + (sin[(2n+1)x/2])/(2sin(x/2))
where 0 < x < 2*pi.  

I think I start by plugging z = e^(ix) in the first identity but I'm 
not sure what to do from there.

Date: 01/11/2004 at 23:12:27
From: Doctor Fenton
Subject: Re: complex analysis--Lagrange's trig identity

Hi Mike,

Thanks for writing to Dr. Math.  Each of the terms on the left side,
cos(kx), is the real part of z^k, where

    z = e^(ix), since

    e^(ikx) = cos(kx) + i*sin(kx).

The left side 

  1 + cos(x) + cos(2x) + ... + cos(nx) 

is the real part of

  1 + z + z^2 + ... + z^n ,

for z = e^(ix) or z = cos(x) + i*sin(x).

From the geometric series summation formula, the left side must be the
real part of

   1 - z^(n + 1)   1 - (cos((n + 1)x) + i*sin((n + 1)x)
   ----------- = --------------------------------------
      1 - z             1 - (cos(x) + i*sin(x))  

If you "rationalize" the expression on the right side by multiplying
by the complex conjugate of the denominator,

   (1 - cos(x)) + i*sin(x) ,

the denominator becomes

   (1 - cos(x))^2 + sin^2(x) = 2(1 - cos(x))
                             = 4 sin^2(x/2) .

The real part of the product in the numerator (you don't need to worry
about the imaginary part) is

  (1 - cos((n + 1)x))(1 - cos(x)) + sin((n + 1)x)*sin(x)

or

  (1 - cos((n + 1)x))(2sin^2(x/2)) + sin((n + 1)x)(2sin(x/2)cos(x/2))

which equals

2sin^2(x/2) - 2sin(x/2)[cos((n + 1)x)sin(x/2) - sin((n + 1)x)cos(x/2)] 

Putting this expression over 4sin^2(x/2) and simplifying with an
identity (the difference formula) will give you the expression you are
looking for.

If you have any questions or need more help, please write back and
show me what you have been able to do, and I will try to offer further
suggestions.

- Doctor Fenton, The Math Forum
  http://mathforum.org/dr.math/