已知tanx=2,求2sin2x-sinxcosx cos2x

數(shù)學(xué)人氣：525 ℃時(shí)間：2020-04-08 13:35:18

優(yōu)質(zhì)解答

分兩部分求
2sin2x=4sinxcosx 注：sin2x=2sinxcosx
=4sinxcosx / {(cosx)^2+(sinx)^2} 注：{(cosx)^2+(sinx)^2=1
=4tanx / {1+(tanx)^2}注：分子分母同時(shí)除以(cosx)^2

sinxcosx cos2x=1/2sin2xcos2x 注：sinxcosx=1/2sin2x
=1/2sin2xcos2x / {(cos2x)^2+(sin2x)^2} 注：{(cos2x)^2+(sin2x)^2=1
=1/2tan2x / {1+(tanx)^2}注：分子分母同時(shí)除以(cos2x)^2

tan=2 tan2x=（2tanx）/{1-(tanx)^2}=-(4/3)數(shù)字代入計(jì)算一下就可以了
2sin2x-sinxcosx cos2x=26/15