Question

已知tanα,tanβ是方程x^2-4x-2=0的两个根,求sinαcosβ+cosαsinβ+2sinαsinβ的值

尽快！！！

Answer 1

sinαcosβ+cosαsinβ+2sinαsinβ
tanatanb=-2
tana+tanb=-(-4)/1=4

sinasinb/cosacosb=-2
sinasinb=-2cosacosb..........................1

sina/cosa+sinb/cosb=(sinacosb+sinbcosa)/(cosacosb)=4
sinacosb+sinbcosa=4cosacosb.............2

1，2代入：sinαcosβ+cosαsinβ+2sinαsinβ得：
4cosacosb+2*(-2cosacosb)=0