数学，三角函数

∵1+(tanα)^2=(secα)^2=1/[(cosα)^2]
∴[1-(tanα)^2]/(2tanα)]
={2-1/[(cosα)^2]}/(2tanα)
=[2-2tanα/(sin2α)]/(2tanα)
=[1-tanα/(sin2α)]/(tanα)
∴1/(tan2α)=1/(tanα)-1/(sin2α)
∴1/(sin2α)=1/(tanα)-1/(tan2α)
同理有1/(sin4α)=1/(tan2α)-1/(tan4α)
1/(sin8α)=1/(tan4α)-1/(tan8α)
…
1/{sin[(2^n)α]}=1/{tan[2^(n-1)]α}-1/{tan[(2^n)α]}
各式相加可以得到
(1/sin2α)+(1/sin4α)+（1/sin8α)+......+1/{sin[(2^n)α]}
=1/(tanα)-1/{tan[(2^n)α]}

等与一些数字和字母组成的式子`！