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已知函数Fx=2Cosxsin(x加三分之派)%根号三sin平...

fx=2cosxsin(x+π/3)-√3sin^2x+sinxcosx+1 =2cosx(√3/2cosx+1/2sinx)-√3sin^2x+sinxcosx+1 =√3cos^2x-√3sin^2x+2sinxcosx+1 =√3cos2x+sin2x+1 =2(√3/2cos2x+1/2sin2x)+1 =2sin(2x+π/3)+1 T=2π/2=π

f(x) = sin(π/2-x)sinx - √3cos²x = cosxsinx - √3cos²x = 1/2sin2x - √3/2cos2x - √3/2 = sin2xcosπ/3-cos2xsinπ/3 - √3/2 = sin(2x-π/3) - √3/2 最小正周期:2π/2 = π 最大值:1 - √3/2 = (2-√3)/2

f(x)=4cosxsinx-4sin(x +派/3)cosx +根号3 =2sin2x-4(1/2*sinx+根号3/2*cosx)cosx+根号3 =2sin2x-(2sin2x+2根号3cos平方x)+根号3 =根号3-2根号3cos平方x

因为:函数f(x)=2cosx·sin(x+π/3)-√3(sinx)^2+sinx·cosx =2cosx·sin(x+∏/3)-√3(sinx)^2+sinx·cosx =2cosx·sinx·cos∏/3+2cosx·cosx·sin∏/3-√3(sinx)^2+sinx·cosx =cosx·sinx+√3(cosx)^2-√3(sinx)^2+sinx·cosx =2cosx·sinx+√3cos2x =sin2x+√3cos2...

f(x)=√3/2sin2x-(cosx)²-½ =√3/2sin2x-1/2cos2x-1 =sin(2x-π/6)-1 T=2π/2=π 2x-π/6在[2kπ+π/2,2kπ+3π/2]是调递减 x在[kπ+π/3,kπ+5π/6]是调递减

求采纳,

解: 已知△ABC中,AB=AC,BD⊥AC,且BD=1/2AB 求∠BAC的度数 解:作BD⊥AC,交直线AC于点D (1)当点D在AC上时 ∵BD=1/2AB ∴∠BAC=30° (2)当点D在CA的延长线上时, ∵BD=1/2AB ∴∠BAD=30° ∴∠BAC=150° f'(x)=2x-m/x, h'(x)=2x-1, 取f'(x)=0,得m=2x^2;x=√m/2,...

已知函数f(x)=sin(x-3派/2)cos(派/2-x)+cosxcos(派-x). (1)求函数f(x)的最小正周期 f(x)=sin(x-3派/2)cos(派/2-x)+cosxcos(派-x). =cosxsinx-cosx平方 =1/2*(sin2x-2cosx平方+1-1) =1/2*(sin2x-cos2x)-1/2 =根号2/2*(根号2/2sin2x-根号2/...

fx=4√3sinxcosx-4sin2x+1 =2√3sin2x-4sin2x+1 =(2√3-4)sin2x+1 f'x=(4√3-8)cos2x 单调增区间:f'(x)>0 ∴cos2x

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