求证tanα+2tan2α+4tan4α+……+2^ntan2^nα=cotα-2^(n+1)cot2^(n+1)α请尽量具体,3Q!
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![求证tanα+2tan2α+4tan4α+……+2^ntan2^nα=cotα-2^(n+1)cot2^(n+1)α请尽量具体,3Q!](/uploads/image/z/8571820-4-0.jpg?t=%E6%B1%82%E8%AF%81tan%CE%B1%2B2tan2%CE%B1%2B4tan4%CE%B1%2B%E2%80%A6%E2%80%A6%2B2%5Entan2%5En%CE%B1%3Dcot%CE%B1-2%5E%28n%2B1%29cot2%5E%EF%BC%88n%2B1%EF%BC%89%CE%B1%E8%AF%B7%E5%B0%BD%E9%87%8F%E5%85%B7%E4%BD%93%2C3Q%21)
求证tanα+2tan2α+4tan4α+……+2^ntan2^nα=cotα-2^(n+1)cot2^(n+1)α请尽量具体,3Q!
求证tanα+2tan2α+4tan4α+……+2^ntan2^nα=cotα-2^(n+1)cot2^(n+1)α
请尽量具体,3Q!
求证tanα+2tan2α+4tan4α+……+2^ntan2^nα=cotα-2^(n+1)cot2^(n+1)α请尽量具体,3Q!
cot2^nα-tan2^nα=cos2^nα/sin2^nα-sin2^nα/cos2^nα
= [(cos2^nα)²-( sin2^nα) ²]/[sin2^nαcos2^nα]
= cos2^(n+1)α/[1/2 sin2^(n+1)α]
=2 cos2^(n+1)α/sin2^(n+1)α
=2 cot2^(n+1)α.
即cot2^nα-tan2^nα=2 cot2^(n+1)α.
tan2^nα= cot2^nα-2 cot2^(n+1)α.
2^ntan(2^nα)= = 2^n cot2^nα- 2^(n+1)cot2^(n+1)α.
令n=0,1,2,3……得:
tanα= cotα-2 cot2α,
2tan2α=2 cot2α-2^2 cot2^2α,
2^2tan(2^2α)= 2^2 cot2^2α-2^3 cot2^3α,
……………………
2^ntan(2^nα)= = 2^n cot2^nα- 2^(n+1)cot2^(n+1)α,
以上各式相加得:
tanα+2tan2α+2^2tan(2^2α)+……+2^ntan(2^nα)=cotα-2^(n+1)cot2^(n+1)α.
用公式tana=cota-2cot2a,裂项求和