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