Proof: (sin2kx)/(2sinx) + cos(2k + 1)x = [sin2(k+1)x]/(2sinx)?

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Proof: (sin2kx)/(2sinx) + cos(2k + 1)x = [sin2(k+1)x]/(2sinx)?

2007-03-15 00:50:26 · 4 answers · asked by Stephen M 1 in Science & Mathematics ➔ Mathematics

4 answers

sin(2(k+1)x)/(2sinx) =
sin(2kx + 2x)/(2sinx) =
(sin(2kx)cos(2x) + cos(2kx)sin(2x))/(2sinx) =
(sin(2kx)(1-2sin²x) + cos(2kx)2sin(x)cos(x)) / (2sinx) =
sin(2kx)/(2sin(x)) - sin(2kx)sin(x) + cos(2kx)cos(x) =
sin(2kx)/(2sin(x)) + cos(2kx+x) =
sin(2kx) / (2sin(x)) + cos((2k+1)x)

2007-03-15 01:21:50 · answer #1 · answered by Deriver 3 · 0⤊ 0⤋

what on earth is that? that looks like pi repeated.....if anybody answers you with a CORRECT answer, that will prove that they dont have a life.

2007-03-15 07:52:59 · answer #2 · answered by jenrulz13 4 · 0⤊ 0⤋

So, what?

2007-03-15 07:52:59 · answer #3 · answered by mrquestion 6 · 0⤊ 0⤋

I agree, dammit!!!

2007-03-15 07:53:20 · answer #4 · answered by Anonymous · 0⤊ 0⤋