数理答疑团为您解答,希望对你有所帮助.
函数f(x)=sin(wx+π/6)(w＞0)的最小正周期为π,则w=2
f(a/2+π/6)=3/5,即sin(a+π/3+π/6)=3/5,sin(a+π/2)=3/5,a∈(0,π/2),
则cosa=3/5,sina=4/5,
f(β/2+5π/12)= -12/13,即sin(β+5π/6+π/6)= -12/13,sin(β+π)= -12/13,β∈(π/2,π),
则sinβ= 12/13,cosβ= 5/13,
所以：sin(a+β)=sinacosβ+cosasinβ=56/65谢谢哈。祝越来越帅不好意思，cosβ= -5/13，所以：sin(a+β)=sinacosβ+cosasinβ=16/65