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公式multiply from k=1 to n tg kπ/(2n+1)=(2n+1)~(1/2)已有过证明,本文给出一种简单的证法如下: 由棣美弗公式和二项式定理可得 sin(2n+1)θ=C_(2n+1)~1sinθcos~(2n)θ-C_(2n+1)~3sin~3θcos~(2n-2)θ+C_(2n+1)~5sin~5θ·cos~(2n-4)θ-…+(-1)~nC_(2n+1)~(2n+1)sin~(2n+1)θ, cos(2n+1)θ=C_(2n+1)~0cos~(2n+1)θ-C_(2n+1)~2sin~2θcos~(2n-1)θ+C_(2n+1)~4sin~4θcos~(2n-3)θ-…+(-1)~nC_(2n+1)~(2n)sin~(2n)θcosθ。从而有 tg(2n+1)θ=[(2n+1)tgθ-C_(2n+1)~3tg~3θ+C_(2n+1)~5tg~5θ-…+(-1)~nC_(2n+1)~(2n+1)tg~(2n+1)θ]/[1-C_(2n+1)~2tg~2θ+C_(2n+1)~4tg~4θ-C_(2n+1)~6tg~6θ+
The formula multiply from k=1 to n tg kπ/(2n+1)=(2n+1)~(1/2) has been proved. This paper gives a simple proof as follows: The binomial theorem yields sin(2n+1)θ=C_(2n+1)~1sinθcos~(2n)θ-C_(2n+1)~3sin~3θcos~(2n-2)θ+C_(2n+ 1) ~5sin~5θ·cos~(2n-4)θ-...+(-1)~nC_(2n+1)~(2n+1)sin~(2n+1)θ, cos(2n+1) θ=C_(2n+1)~0cos~(2n+1)θ-C_(2n+1)~2sin~2θcos~(2n-1)θ+C_(2n+1)~4sin~4θcos~(2n- 3) θ-...+(-1)~nC_(2n+1)~(2n)sin~(2n)θcosθ. Thus there are tg(2n+1)θ=[(2n+1)tgθ-C_(2n+1)~3tg~3θ+C_(2n+1)~5tg~5θ-...+(-1)~nC_(2n +1)~(2n+1)tg~(2n+1)θ]/[1-C_(2n+1)~2tg~2θ+C_(2n+1)~4tg~4θ-C_(2n+1) ~6tg~6θ+