> 数学 >
已知cosα-sinα=3√ 2/5,且π<α<3π/2,求sin2α+2sin^2α/1tanα
人气:110 ℃ 时间:2020-03-25 01:57:08
解答
cosα-sinα=3√ 2/5(cosα-sinα)^2 = 18/251 -2sinαcosα = 18/252sinαcosα = 1 - 8/25 = 7/25sin2α+2sin^2α/tanα= 2sinαcosα+2sin^2α/(sinα/cosα)= 2sinαcosα+2sinαcosα= 2*sinαcosα= 2*7/25= 1...sin2α+2sin^2α/1-tanα怎么化简....题目不是【sin2α+2sin^2α/1tanα】吗?打错了 不好意思 是sin2α+2sin^2α/1-tanα分子分母个包括哪些内容,加上括号(或说明一下)分子是sin2α+2sin^2α 分母是1-tanαcosα-sinα=3√2/5cosα=sinα+3√2/5cos^2α=sin^2α+6√2/5sinα+18/251-sin^2α=sin^2α+6√2/5sinα+18/252sin^2α+6√2/5sinα-7/25 = 0(sinα+7√2/10)(2sinaα-√2/5)=0∵π<α<3π/2∴sinα<0,cosα<0∴2sinaα-√2/5<0∴sinα+7√2/10=0∴sinα= - 7√2/10∴cosα = -√(1-sin^2α) = -√(1-98/100) = -√2/10sin2α=2sinαcosα=2*(- 7√2/10)*(-√2/10)=7/25sin^2α=(-7√2/10)^2=49/50tanα=sinα/cosα=(- 7√2/10)/(-√2/10)=7(sin2α+2sin^2α)/(1-tanα) = (7/25+2*49/50)/(1-7) = (56/25)/(-6) = -28/75
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版