请教两道数学题

2025-12-13 21:21:11
推荐回答(1个)
回答1:

1.y=-sinxcosx-[(cosx)^2-(sinx)^2]
=-(1/2)sin2x-cos2x
=-((√5)/2)sin(2x+θ)
最小值为-((√5)/2)
2.sinx+(sinx)^2=1=(sinx)^2+(cosx)^2
(cosx)^2=sinx
(cosx)^2+(cosx)^4=sinx+(sinx)^2=1