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已知F 2Cosxsin

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.

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...

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(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

f(x)=√2*cosx*sin(x+π/4)-1/2 =√2*cosx*(√2/2*sinx+√2/2*cosx)-1/2 =sinxcosx+(cosx)^2-1/2 =1/2*sin2x+1/2*(1+cos2x)-1/2 =1/2*sin2x+1/2*cos2x =√2/2*(√2/2*sin2x+√2/2*cos2x) =√2/2*sin(2x+π/4) 令2x+π/4=kπ+π/2,所以x=kπ/2+π/8,所以对称轴...

f(x)=2cosx*(sinx*根号3/2-cosx*1/2)-1/2 =根号3/2sin2x-1/2(1+cos2x)-1/2 =sin2xcosPai/6-sinPai/6cos2x-1 =sin(2x-Pai/6)-1 函数的最小值是:-1-1=-2,最小正周期T=2Pai/2=Pai 2. f(C)=sin(2C-π/6)-1=0 2C-π/6=π/2 C=π/3 2sinA=sinB 正弦定理 ...

(1)f(x)=根号3sinxcosx cos²x-½=根号3/2sin2x ½cos2x=sin(2x π/6) ∵f(x)max=1∴2x π/6=π/2 2kπ ∴{x|x=π/6 kπ} (2)∵cos(α π/6)=4/5且α为锐角∴sin(α π/6)=3/5 f(α/2-π/12)=sin[(α-π/6) π/6]=sin(α-π/6)cosπ/6 cos(α-π/6)sinπ/6=3/5×...

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