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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
人气:215 ℃ 时间:2019-08-19 16:47:57
解答
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...
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