cos((-47/10)π)=cos(-4π-7π/10)=cos(-7π/10)=cos7π/10
cos((-44/9)π) =cos(-4π-8π/9)=cos(-8π/9)=cos8π/9
y=cosx在區(qū)間【0,π】上是減函數(shù)
所以 cos8π/9cos((-44/9)π)
tan((3/2)π-1)=cot1>0
tan((3/2)π+1)=-cot1tan((3/2)π+1)在問下哈，2問中老師不讓用cot怎么解啊tan((3/2)π-1)=sin((3/2)π-1)/cos((3/2)π-1)=(-cos1)/(-sin1)=sin1/cos1>0tan((3/2)π+1)=sin((3/2)π+1)/cos((3/2)π+1)=(-cos1)/sin1>0