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+5π)=10/13,f(3b+2π)=-6/5,求sin(a+b)的值

问题描述:

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+5π)=10/13,f(3b+2π)=-6/5
,求sin(a+b)的值

1,f(0)=2sin(1/3*0-5π/3)=2sin(0-5π/3)=2sin(π/3)=根号3
2,f(3a+5π)=2sin【1/3(3a+5π)-5π/3】=2sin(a)=10/13,
即sin(a)=5/13
f(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