Sin阿法=3/5

显示全部楼层 · 2012-4-27 15:49:02

Sin阿法=3/5阿法属于(派/2，派) 所以cos阿法=-4/5sin2阿法=2sin阿法cos阿法=2*(3/5)*(-4/5)=-24/25cos2阿法=cos^2阿法-sin^2阿法=(-4/5)^2-(3/5)^2=7/25tan2阿法=sin2阿法/cos2阿法=(-24/25)/(7/25)=-24/7...

千问 · 2012-4-27 15:49:02

按条件知 2a属于(pai,1.5pai)sina=3/5cosa=-4/5cos(2a)=sin(2a)=2*sina*cosa=-24/25sin(2a)=-7/25tan(2a)=7/24...

千问 · 2012-4-27 15:49:02

解：∵α∈（π/2,π）∴sinα>0,cosα<0cosα=-4/5sin2α=2sinacosα=-24/25cos2α=2cos^2a-1=7/25tan2a=sin2a/cos2a=-24/7...

千问 · 2012-4-27 15:49:02

sinα=3/5α∈(π/2,π)cosα=-根号(1-sin2α)=-4/5sin2α=2sinαcosα=2×3/5×(-4/5)=-24/25cos2α=1-2sin2α=1-18/25=7/25tan2α=sin2α/cos2α=-24/7...

千问 · 2012-4-27 15:49:02

Sin阿法=3/5阿法属于(派/2，派) cosa=-4/5sin2a=2sinacosa=-24/25cos2a=cos^2a-sin^2a=7/25tan2a=sin2a/cos2a=-24/7...