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### 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

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/
```
Associated Topics:
High School Trigonometry

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