已知f(x)=(2cos²x-1)sin2x+1 2cos4x
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/05 18:43:31
f(x)=cos(-2x+a)(-π
∵f(x)=cos(2x-π/3)+(sinx)^2-(cosx)^2=cos(2x-π/3)-cos2x=2sin(2x-π/6)sin(π/6)=sin(2x-π/6).∴g(x)=[sin(2x
f(x)=[1+cos2x)/2]/[1+(1-cos2x)/2]=(1+cos2x)/(3-cos2x)=-1+4/(3-cos2x)f'(x)=-4/(2-cos2x)^2*(2-cos2x)'=
f(x)=(cosx)^2+sinxcosx=(1/2)sin2x+(1/2)cos2x+1/2=(√2/2)sin(2x+π/4)+1/22kπ-π/2
f(x)=cos^4x-2sinxcosx-sin^4x=(cos^2x+sin^2x)(cos^2x-sin^2x)-sin2x=cos2x-sin2x=根号2*cos(2x+л/4)(1)f(x)
f(x)可化为=1/2+根号3*sin(2x+pai/2)(2x+pai/2)=π/2时有最大值1/2+根号3=1/2+根号3*sin(2x+pai/3)然后(2x+pai/3)=pai/2然后sin
f(x)=sin2x-2cos^2x=sin2x-cos2x-1=√2(sin2x-π/4)-1∴T=2π/2=π函数f(X)最小正周期是π
根据公式:COS^2a=(1+COS2a)/2a=(X+π/12)说句不好听的,你还是基本知识没掌握好,不会活用知识.希望你多背多看,看清题,把公式活用.
f(x)=√2sin(2 x-π/4)+2f(x)max=√2+2;此时X=3π/8令(2 x-π/4)∈【-π/2,π/2】,解得x∈【-π/8,3π/8】,因为x∈(0&nbs
f(x)=sin2x+2cos²x-1+1=sin2x+cos2x+1=√2(√2/2*sin2x+√2/2cos2x)+1=√2(sin2xzosπ/4+cos2xsinπ/4)+1=√2
值域为R,因为这是个连续函数,如果保证cosx=土1了,x可以取得非常大,从而值域可以达到非常大.用matlab画图发现不是单调的.其实可以求导后取几个点的值来验证.比如取x=pi/6pi/3,导数正
两倍角公式2cos²x=cos2x+12sinxcosx=sin2xf(x)=-4cos²x+4√(3)sinxcosx+5=-2(cos2x+1)+2√3sin2x+5=2√3s
y=sinx^2+根3sinxcosx+2cosx^2=-1/2(1-2sinx^2)+1/2根3*2sinxcosx+2cosx^2-1+3/2=-1/2cos2x+二分之根3倍sin2x+cos2
f(x)=cos^2x+(1/2)sin2x.f(x)=(1+cos2x)/2+(1/2)sin2x.=(1/2)(sin2x+cos2x)+1/2.∴f(x)=(1/2)√2sin(2x+π/4)+
f'(x)=-1/2*sin(x/2)*[sin(x/2)-cos(x/2)]+cos(x/2)[1/2cos(x/2)+1/2sin(x/2)]=-1/2*sin²(x/2)+1/2sin
f(x)=cos(2x-π/3)-(cos^2x-sin^2x)=cos(2x-π/3)-cos2x=2sin(2x-π/6)sinπ/6=sin(2x-π/6)因为y=sinx的单减区间为[π/2+
f(x)=cos(2x-π\3)+sin²x-cos²x=1/2cos2x+√3/2sin2x-cos2x=√3/2sin2x-1/2cos2x=-cos(2x+π\3)-1
第一问可以把f(x)看作是一个二次函数-cos^2x+cosx+aa=-1,b=1,c=a只需b^2-4ac≥0即1+4a≥0解得a的取值范围为(-1/4,+∞)第二问我还在想.再问: 谢谢,,,呃
你确定第一个符号是加号不是乘号?
f(x)=2cosx*sinx-2cosx^2+1f(x)=sin2x-cos2xf(x)=根号2*sin(2x-45)周期T=π