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2Cos^2x

2cos2X=2(cosX)^2-2(sinX)^2 =4*(cosX)^2-2 =2-4*(sinX)^2

1.将分母变为sin2x即原式为∫[(4cos2x/sin^2(2x))]dx 2.进行换元即2x变为t,原式变为∫[(2cos2x/sin^2t)]dt. 3继续换元,可观察到(sin t)'=cost.所以原式等于2∫[(1/sin^2t]d(sint). 4.得出答案为:(-2/sint)+c 5.将t换回为2x有(-2/sin...

cos2x=cos^2x-sin^2x=cos^2x-1+cos^2x=2cos^2x-1 同理。

解y=2cos^2x+5sinx-4 =2(1-sin^2x)+5sinx-4 =-2sin^2x+5sinx-2 令t=sinx 则t属于[-1,1] 则y=-2t^2+5t-2 =-2(t-5/4)^2-21/8 故t=1时,y有最大值y=-2+5-2=1 t=-1时,y有最小值y=-2-5-2=-9 故函数的值域为[-9,1]

cosx=cos2(x/2)=2cos^2(x/2)-1 所以, cos^2(x/2)=[1+cosx]/2

解:cos^2x-sin^2x=cos x.cos x-sin x.sin x =cos(x+x) =cos 2x

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)

dy= d(cos 2x)^2 令cos2x=u d(cos 2x)^2 =du² =2udu 把u=cos2x代入,得 原式=2cos2xdcos2x

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