Question

已知函数f(x)=sin(x-π/2)sin(x-π)

求函数的最值和最小正周期 计算f（π/24）+f(5π/24)

Answer 1

f(x)=sin(x-π/2)sin(x-π)
=-cosx*(-sinx)
=sinxcosx
=1/2*sin2x

所以最大值=1/2，最小值=-1/2
T=2π/2=π

f（π/24）+f(5π/24)
=-1/2sinπ/12-1/2*sin5π/12
=-1/2[sin(π/3-π/4)+sin(π/6+π/4)]
=-1/2[(√6-√2)/4+(√6+√2)/4]
=-√6/4