已知tanα=2,sin눀α-3sinαcosα+1=

2024-12-21 20:21:11
推荐回答(1个)
回答1:

(sinα)^2-3sinαcosα+1
=[(sinα)^2-3sinαcosα+(sinα)^2+(cosα)^2]/[(sinα)^2+(cosα)^2]
=[2(sinα)^2-3sinαcosα+(cosα)^2]/[(sinα)^2+(cosα)^2] 分子分母同除(cosα)^2
=[2(tanα)^2-3tanα+1]/[(tanα)^2+1]
=(8-6+1)/(4+1)
=3/5