Deriving Lagrange's Trig IdentityDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/