角成等差数列的正(余)弦三角函数的求和
2002
对于求:sinα+sin(a+d)+sin(α+2d)+…+sin[α+(n-1)d],如果我们给每一项都乘以sind/2,然后每项积化和差,即有sinαsind/2=1/2[cos(α-d/2)-cos(α+d/2)].sin(α+d)sind/2=1/2[cos(α+d/2)-cos(α+(3d)/2)]
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