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已知函数F(x)=2根号3sinxCosx+2Cos²x.(1...

f(x)=2√3sinxcosx+2cos²x =√3sin2x+cos2x+1 =2sin(2x+π/6)+1 f(4π/3)=2sin(8π/3+π/6)+1=2sin5π/6+1=2 x∈[0,π/2] 2x+π/6∈[π/6,2π/3] f(x)的值域:[2,3]

(1) f(x)=2√3sinxcosx+2cos²x-1=√3sin2x+cos2x=2(√3/2sin2x+1/2cos2x)=2sin(2x+π/6) 函数的最小正周期T=π (2) f(x)=2sin(2x+π/6)=6/5 sin(2x+π/6)=3/5 x∈[π/4,π/2], cos(2x+π/6)=-4/5 cos2x=cos(2x+π/6-π/6)=cos(2x+π/6)cosπ/6...

f(x)=2√3sinxcosx+1-2(sinx)^2 =√3sin2x+cos2x =2sin(2x+π/6) 所以最小正周期=2π/2=π -π/2+2kπ

f(x)=根号3sin2x+cos2x=2sin(2x+π/6) 1) T=2π/2=π x∈[0,π/2] 2x+π/6[π/6,7π/6] f(x)小=2sin(π/6)=1 f(x)大=2sin(π/2)=2 2) sin(2x0+π/6)=3/5 co2x0=cos(2x0+π/6-π/6)=(3-4根号3)/10 字限制

解: f(x)=2√3sinxcosx+1 -2sin²x =√3sin(2x)+cos(2x) =2[(√3/2)sin(2x)+(1/2)cos(2x)] =2sin(2x+π/6) -1≤sin(2x+π/6)≤1 -2≤2sin(2x+π/6)≤2 -2≤f(x)≤2 函数的值域为[-2,2]

你题目的第一项与第三项, 可以改化为: cos2x, 第二项可以改写为 (√3)×sin2x, 于是, f(x)=2×sin{(2x)+(π/6)}, 然后自己可以完成任务啦! ——题目的a是啥意思?

f(x)=√3sinxcosx-cos²x-1/2 =(√3/2)sin2x-(1+cos2x)/2-1/2 =(√3/2)sin2x-(1/2)cos2x-1 =sin(2x-π/6)-1. 早小正周期T=2π/2=π. sin(2x-π/6)=-1, 即x=kπ-π/6时, 最小值f(x)|min=-2。

答: f(x)=2√3sinxcosx+2cos²x+1 f(x)=√3sin2x+cos2x+1+1 f(x)=2×[(√3/2)sin2x+(1/2)cos2x] +2 f(x)=2sin(2x+π/6)+2 最小正周期T=2π/2=π

f(x)=2cos²x+2根号3sinxcosx = (cos2x+1) + √3sin2x = √3sin2x + cos2x + 1 = 2(sin2xcosπ/6+cos2xsinπ/6) +1 = 2sin(2x+π/6) + 1 最小正周期 = 2π/2 = π 2x+π/6∈(2kπ+π/2,2kπ+3π/2),其中k∈Z时单调减 所以单调减区间:x∈(kπ+π/6,kπ+2π/...

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