已知函数f(x)=(3/5)sinx+sinβcosx+1(β为常数)且f(0)=9/5

1f(x)=(3/5)sinx+sinβcosx+1f(0)=9/5∴sinβcos0+1=9/5∴sinβ=4/5cos2β=1-2sin²β=1-2(4/5)²=-7/252f(x)=(3/5)sinx+(4/5)cosx+1设sinφ=4/5,cosφ=3/5 f(x) =sinxcosφ+cosxsinφ+1=sin(x+φ)+1∵sin(x...