几道数学题(给出正确答案和过程的立刻采纳)

2024-11-05 17:26:28
推荐回答(3个)
回答1:

sin4α=2sin2acos2a
=2sinacosa(1-2sin^2a)
带入sina,cosa即可(cosa可以通过sina利用辅助三角形法求出)
y=-2sin²x+2sinxcosx+3
=cos2x-1+sin2x+3
=根号2sin(2x+π/4)+2
所以值域是[2-根号2+,根号2]

回答2:

sin4a=2sin2acos2a.sin2a=2sinacosa.cos2a=1-2sin^a.代如即可,望采纳.再回答第二个

回答3:

sin2α=2sinα·cosα=,Cos2α=1-2Sin^2(α),,sin4α=2sin2α·cos2α=4sinα·cosα·(1-2Sin^2(α))