Product Sum and Difference FormulaeDate: 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/ |
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