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函数F(x)=sin^2x+根号3sinxCosx在区间[兀/4,兀/2]...

max=3/2,min=1 解: f(x) =sin²x+√3sinxcosx =(1-cos2x)/2+(√3/2)sin2x =sin(2x-π/6)+1/2 ∵x∈[π/4,π/2] ∴2x-π/6∈[π/3,5π/6] ∴ f(x)_max=1+1/2=3/2 f(x)_min=1/2+1/2=1

第一个是sin2x还是sin²x?

f(x)=(1-cos2x)/2+√3/2*sin2x =√3/2sin2x-1/2*cos2x+1/2 =sin2xcosπ/6-cos2xsinπ/6+1/2 =sin(2x-π/6)+1/2 π/4

解: (1) f(x)=cos²x-√3sinxcosx+½ =½[1+cos(2x)]-(√3/2)sin(2x)+½ =½cos(2x)-(√3/2)sin(2x)+1 =cos(2x+π/3)+1 最小正周期T=2π/2=π cos(2x+π/3)=1时,f(x)取得最大值f(x)max=1+1=2 cos(2x+π/3)=-1时,f(x)取得最小值f(...

f(x)= - √3sin^2(x)+sinxcosx = -√3[(1-cos2x)/2]+(1/2)sinxcosx =(1/2)sin2x+(√3/2)cos2x-(√3/2) =sin2xcos60º+cos2xsin60º - (√3/2) f(x)=sin(2x+60º)-(√3/2) f(x/2)=sin[2(x/2)+60º] - (√3/2) =sin(x+60º) - (√3/2)...

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。

(1+根号3/2)sin2x,T=兀

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

f( x)= sin2x +2根号3cosxsinx+sin(x-4分之派)sin(x+4分之派) f( x)= sin2x +根号3sin2x+sin(x-4分之派)cos(x-4分之派) f( x)= sin2x +根号3sin2x+1/2 sin(2x-2分之派) f( x)= (根号3+1)sin2x-1/2 cos2x

解: (1) f(x)=2√3sinxcosx+2cos²x-1 =√3sin(2x)+cos(2x) =2[(√3/2)sin(2x)+(1/2)cos(2x)] =2sin(2x+π/6) 最小正周期T=2π/2=π (2) f(x0)=6/5 2sin(2x0+π/6)=6/5 sin(2x0+π/6)=3/5 x0∈[π/4,π/2] 2π/3≤2x0+π/6≤7π/6 cos(2x0+π/6)

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