已知tanα=1/2,求1+2sin(π-α)cos(-2π-α)/sin2(-α)-sin2(5π/2 -α)的值三角函数

tanα=1/2
sina/cosa =1/2 cosa=2sina
sina²+cos²=1 5sina²=1
1+2sin(π-α)cos(-2π-α)/sin2(-α)-sin2(5π/2 -α)
=1+2sinαcosα/(-sin2α)-sin(5π -2α)
=1+2sinαcosα/(-2sinacosa)-sin(5π -2α)
=-sin(5π -2α)
=-sin2α
=-2sinacosa
=-4(sina )²
=-4/5那个 题目错了，已知tanα=1/2，求1+2sin(π-α)cos(-2π-α)/sin*2(-α)-sin2(5π/2 -α)的值1+2sin(π-α)cos(-2π-α)/(sin(-α))²-sin2(5π/2 -α)=1+2sinαcosα/sinα²-sin(5π -2α)=1+2cosα/sinα-sin2a=1+2/tanα-4/5=1+4-4/5=3.81+2cosα/sinα-sin2a这一步怎么到 1+2/tanα-4/5？sin2a上面算出来了 sin2a=4/5cosα/sinα=cota=1/tana1+2cosα/sinα-sin2a=1+2/tanα-4/5