2007-11-29 13:51:57 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Just use the expansion for tan(a+b)
tan(π + θ) = [tanπ + tanθ] / (1 - tanπ tanθ)
= tanθ
2007-11-29 13:58:07 · answer #1 · answered by Dr D 7 · 1⤊ 0⤋
Tan Pi-theta
2016-10-13 09:33:53 · answer #2 · answered by Anonymous · 0⤊ 0⤋
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Method 1 (cosx+isinx)^9 = cos9x+isin9x ....(1) ; from de Moivre's theorem (cosx+isinx)^9 = cos^9x +9icos^8xsinx -36cos^7xsin^2x - 84icos^6xsin^3x+126cos^5xsin^4x + 126icos^4xsin^5x - 84cos^3xsin^6x-36icos^2xsin^7x+9cosxsin^... +isin^9x ...(2) form binomial theorem Equating imaginary parts from (1) & (2), sin9x = 9sinxcos^8x - 84sin^3xcos^6x +126sin^5xcos^4x - 36sin^7xcos^2x+sin^9x sin9x = sin^9x - 36sin^7xcos^2x + 126sin^5xcos^4x - 84sin^3xcos^6x+9sinxcos^8x
2016-04-02 09:25:12 · answer #3 · answered by Anonymous · 0⤊ 0⤋
i agree with what the second person said about using the sum identity for tan to prove this equation. The first person doesn't make much sense.
2007-11-29 14:02:36 · answer #4 · answered by Purple 2 · 0⤊ 0⤋
the period of tan is pi, and i think that's the best proof you can get
its similar to prove sin(theta) = sin(2*pi + theta)
2007-11-29 13:55:48 · answer #5 · answered by dohboy000 4 · 1⤊ 0⤋
First, tan(π) = 0
and
tan(π + θ) = (tan π + tan θ)/ (1 - tan π tan θ).
Since tan π = 0, this is just tan θ/1 = tan θ.
2007-11-30 03:58:52 · answer #6 · answered by steiner1745 7 · 0⤊ 0⤋