设函数f(x)=2cosxsin(x+π/6)+2sinxcos(x+π/6)⑴当x属于0到2分之π的闭区间求f(x)的值域⑵设三角形ABC的三个内角ABC所对的三边依次为abc已知f(A)=1,a=根号7,三角形ABC面积为2分之3倍根号3求b+c.
人气:278 ℃ 时间:2020-10-01 22:15:04
解答
f(x)=2cosxsin(x+π/6)+2sinxcos(x+π/6)=2sin(2x+π/6)(1)∵x∈[0,π/2] ∴2x∈[0,π] ∴2x+π/6∈[π/6,7π/6] ∴sin(2x+π/6)∈[﹣1/2,1]∴f(x)的值域为:f(x)∈[﹣1,2](2)∵f(A)=2sin(2A+π/6)=1 ∴sin...
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