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已知F 2Cosxsin

(1) f(x)=2cosxsin(x-π/3)+√3 sin^2x+sinxcosx =2cosx[1/2sinx-√3/2cosx)+√3/2[1-cos2x]+1/2sin2x =1/2sin2x-√3cos^2x-√3/2cos2x+1/2sin2x+√3/2 =sin2x-√3/2cos2x-√3/2cos2x+√3/2-√3/2 =sin2x-√3cos2x =2sin(2x-π/3) 函数y=f(x)图像的对称中心...

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: 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,...

解:由已知得 ,(1)∴f(x)的最小正周期T= =π;(2)∵ ∴ ∴ ∴f(x)的值域为[-1,2]。

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)=2cosx(sinx-cosx)+1 =2sinxcosx-2(cosx)^2+1 =sin2x-cos2x =根号2sin(2x-Pai/4) 故最小正周期T=2Pai/2=Pai 最大值=根号2,最小值=-根号2

f(x)=sinx/2cosx/2-sin^2x/2 =1/2*sinx-1/2(1-cosx) =√2/2(√2/2sinx+√2/2cosx)-1/2 =√2/2sin(x+π/4)-1/2 T=2π 因x在[-π,0] , 所以有:x+π/4在[-3π/4,π/4] sin(x+π/4)在[-3π/4,π/4]上最小值=-1 所以,f(x)在[-π,0]上最小值=-(1+√2)/2

(1)∵函数f(x)=2cosxsin(x+π3)?3sin2x+sinxcosx=sinxcosx+3cos2x-3sin2x+sinxcosx=sin2x+3cos2x=2sin(2x+π3).∵x∈[?π12,π6],∴π6≤2x+π3≤2π3,∴12≤sin(2x+π3)≤1,故函数f(x)的最大值为2,最小值为1.(2)锐角△ABC中,由f(A)=0 可得 s...

(1) f(x)=2cosxsin(x+ π 3 )- 3 si n 2 x+sinxcosx+2 =2sin(2x+ π 3 )+2∴最小正周期T= 2π 2 =π,当2kπ- π 2 ≤2x+ π 3 ≤2kπ+ π 2 时,即kπ- 5π 12 ≤x≤kπ+ π 12 ,函数单调增∴函数的单调增区间为:[kπ- 5π 12 ,kπ+ π 12 ](k∈Z)(2)由函数...

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