Question

把下列三角函数值从小到大排列起来：sin4/5π，-cos5/4π，sin32/5π，cos5/12π。

Answer 1

由于[-π/2,π/2]是函数y=sinx的一个单调增区间,[π/2,3π/2]为函数y=sinx的一个单调减区间，可以把上面几个函数化为一个区间内的函数再比较其大小
sin(4π/5)=sin(π-4π/5)=sin(π/5)
-cos(5π/4)=-sin(π/2-5π/4)=-sin(-3π/4)=sin(3π/4)=sin(π-3π/4)=sin(π/4)
sin(32π/5)=sin(6π+2π/5)=sin(2π/5)
cos(5π/12)=sin(π/2-5π/12)=sin(π/12)
因为-π/2<π/12<π/5<π/4<2π/5<π/2 (即—30π/60<5π/60<12π/60<15π/60<24π/60<30π/60)
[-π/2,π/2]是函数y=sinx的一个单调增区间,所以sin(π/12)<sin(π/5)<sin(π/4)<sin(2π/5)
即它们从小到大排列为cos(5π/12)，sin(4π/5),-cos(5π/4),sin(32π/5)

Answer 2

这个，都忘了！！！

追问

求求大家，帮我个忙嘛！谢谢了！

Answer 3

由于[-π/2,π/2]是函数y=sinx的一个单调增区间,[π/2,3π/2]为函数y=sinx的一个单调减区间，可以把上面几个函数化为一个区间内的函数再比较其大小
sin(4π/5)=sin(π-4π/5)=sin(π/5)
-cos(5π/4)=-sin(π/2-5π/4)=-sin(-3π/4)=sin(3π/4)=sin(π-3π/4)=sin(π/4)
sin(32π/5)=sin(6π+2π/5)=sin(2π/5)
cos(5π/12)=sin(π/2-5π/12)=sin(π/12)
因为-π/2<π/12<π/5<π/4<2π/5<π/2
(即—30π/60<5π/60<12π/60<15π/60<24π/60<30π/60)
[-π/2,π/2]是函数y=sinx的一个单调增区间,所以sin(π/12)<sin(π/5)<sin(π/4)<sin(2π/5)
即它们从小到大排列为cos(5π/12)，sin(4π/5),-cos(5π/4),sin(32π/5)