已知函数f(x)=2sin(1/3x-5π/3),x∈r. 1.求f(0)的值 2.设a.b∈[0,π/2],f(3a+

1,f（0）=2sin(1/3*0-5π/3)=2sin(0-5π/3)=2sin(π/3)=根号32,f(3a+5π)=2sin【1/3（3a+5π）-5π/3】=2sin（a）=10/13,即sin（a）=5/13f(3b+2π）=2sin【1/3(3b+2π）-5π/3】=2sin（b-π）=-2sinb=-6/5即sinb=3/5由sina²+cosa²=1,a。b∈[0,π/2]得 cosa=12/13,cosb=4/5运用两角和公式 sin（a+b）=sinacosb+sinbcosa=5/13*4/5+3/5*12/13=56/65