Re: trig question

In article <1141796214.009337.35450@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
jennifer <scrilla_12_1999@xxxxxxxxx> wrote:

I'm trying to prove:

1-cos(2x) + cos(4x)- cos(6x) = 4sin(x)cos(2x)sin(3x)

Which side should i attempt and what trig formulas would you recommend
please.

Ye Gods, where do you get these problems?

Mathematica verifies the identity. I applied TrigExpand to both
expressions (this reduces forms like Sin[n x] and Cos[n x] to sines and
cosines of x). You could do that yourself, by hand, I suppose. Use,
e.g.

sin(3x) = sin(2x) cos(x) + cos(2x) sin(x)

= 2 sin(x) cos^2(x) + (cos^2 x - sin^2 x) sin(x).

But if you're going to do this, at the end I'd be sure to replace all
expressions cos^(2n) x by (1-sin^2 x)^n, and cos^(2n+1) x by (1-sin^2
x)^n cos(x). Otherwise it may get messy (!)

A more clever way might be to use the formula

1 - z^8
1 - z^2 + z^4 - z^6 = -------
1 + z^2

where z = e^(ix), then take the real parts. Frankly, I think this
would be more trouble than it's worth.

Expand both sides and be done with it.

--Ron Bruck
.