已知函数f(x)=(3/5)sinx+sinβcosx+1(β为常数)且f(0)=9/5
问题描述:
已知函数f(x)=(3/5)sinx+sinβcosx+1(β为常数)且f(0)=9/5
(1)求sinβ与cos2β的值
(2)求函数f(x)的最大值与最小值
答
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...