Product Sum and Difference Formulae

Date: 11/20/97 at 02:25:58
From: Saurabh Agarwal
Subject: Trignometry (product sum and difference formulae)

cos6theta+6cos4theta+15cos2theta+10
------------------------------------ = 2 cos theta
cos5theta+5cos3theta+10cos theta

Date: 12/01/97 at 11:47:30
From: Doctor Bruce
Subject: Re: Trignometry (product sum and difference formulae)

Hello Suarabh,

First I want to be sure I understand what you are asking. 
When you write "cos6theta", I assume you mean "cos(6 theta)" and not 
cos^6(theta). If you don't mind, I'll write `t' instead of `theta' for 
easier readability. Then, you want to know how to demonstrate the 
identity

[cos(6t)+6cos(4t)+15cos(2t)+10]/[cos(5t)+5cos(3t)+10cos(t)] = 2cos(t)

Such an identity can always be proved by using formulas like

     cos(2t) = cos^2(t) - sin^2(t),
     cos(3t) = cos(2t)cos(t) - sin(2t)sin(t)

and so on to convert all the trigonometric expressions to sines and 
cosines of the angle t only. The final form of the numerator and 
denominator of your identity turns out to be very simple, but the 
intermediate work is very tedious. I would certainly not want to do 
this for a problem as complicated as yours! 

On the other hand, you may have learned some general recursive 
formulas for the functions cos(nt) in terms of the powers cos^n(t), 
cos^(n-1)(t), and so on, where n is an arbitrary integer. As a 
consequence of such formulas, you can show that

     cos(6t)+6cos(4t)+15cos(2t)+10 = 32cos^6(t)     and
     cos(5t)+5cos(3t)+10cos(t)     = 16cos^5(t)

which proves your identity. But this approach is also not very 
satisfying, since it says, "if you know some special formulas, then 
your problem is easy."

The best way to deal with a complicated expression like yours is to 
use the DeMoivre formula to change the cosines to exponential 
functions. That is, we know

     e^(ix)  =  cos(x) + i sin(x)
     e^(-ix) =  cos(x) - i sin(x)

so we also have

     cos(x)  =  [e^(ix) + e^(-ix)]/2.

If we make substitutions

     cos(6t)  =  [e^(6it) + e^(-6it)]/2
     cos(5t)  =  [e^(5it) + e^(-5it)]/2
     cos(4t)  =  [e^(4it) + e^(-4it)]/2

and so forth, then your identity reduces to an equation looking like

     [e^(6it) + ... ]/[e^(5it) + ... ] = e^(it) + e^(-it)

which you can verify by multiplying it out.

Good luck!

-Doctor Bruce,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/