Question

比较大小1: cos((-47/10)π)与cos((-44/9)π) 2: tan((3/2)π-1)与tan((3/2)π+1)

Answer 1

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在区间【0，π】上是减函数
所以 cos8π/9<cos7π/10
所以 cos((-47/10)π)>cos((-44/9)π)
tan((3/2)π-1)=cot1>0
tan((3/2)π+1)=-cot1<0

tan((3/2)π-1)>tan((3/2)π+1)

追问

在问下哈，2问中老师不让用cot怎么解啊

追答

tan((3/2)π-1)=sin((3/2)π-1)/cos((3/2)π-1)=(-cos1)/(-sin1)=sin1/cos1>0
tan((3/2)π+1)=sin((3/2)π+1)/cos((3/2)π+1)=(-cos1)/sin1>0