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已知F 2Cosxsin

f(x)=2cosxsin(x+π/3)-√3sin²x+sinxcosx =2cosx(1/2*sinx+√3/2*cosx) -√3sin²x+sinxcosx = sinxcosx+√3cos²x-√3sin²x+sinxcosx =2 sinxcosx+√3(cos²x-sin²x) =sin2x+√3cos2x =2sin(2x+π/3) (1) 最小正周期=2π/2=π...

因为已知f′(2+cosx)=sin2x+tan2x,所以f′(2+cosx)=1?cos2x+1?cos2xcos2x.设u=2+cosx,则f′(u)=1?(u?2)2+1(u?2)2?1.故 f′(x)=-(x-2)2+1(x?2)2.f(x)=∫ [?(x?2)2+1(x?2)2]dx=?(x?2)23?1x?2+C.故答案为:?(x?2)23?1x?2+C.

解: 分式有意义,cosx≠0 f(x)=[sin(2x)+cos(2x)+1]/(2cosx) =(2sinxcosx+cos²x-sin²x+cos²x+sin²x)/(2cosx) =(2sinxcosx+2cos²x)/(2cosx) =2cosx(sinx+cosx)/(2cosx) =sinx+cosx =√2sin(x+π/4) x∈[0,π/3],则x+π/4∈[...

f(x)=sinxcosx+sqrt(3)(cosx)^2-sqrt(3)(sinx)^2+sinxcosx+2=2sinxcosx+sqrt(3)[(cosx)^2-(sinx)^2]=sin2x+sqrt(3)cos2x=2sin(2x+π/3)所以最小正周期是π单调增区间[-5π/12+kπ,π/12+kπ],k是整数。 ______________ 施主,我看你骨骼清奇,器宇轩...

公式:2sinAcosB = sin(A+B) +sin(A-B) f(x)=2cosxsin(x+π/3)-√3/2 = sin(x+π/3+x) + sin(x+π/3-x) - √3/2 = sin(2x+π/3) + √3/2 - √3/2 = sin(2x+π/3) 最小正周期:2π/2 = π

f(sinx/2)=cosx+1 =1-2(sin(x/2))^2 +1 =2- 2(sin(x/2))^2 f(x)=2-2x^2 f(cos(x/2)^2=2-2(cos(x/2))^2 =2-(1+cosx) =1-cosx

解答: 换元,令t=cosx ∵ -π/4

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