求函数y=cos^2x 4sinx的最值及到最大值和最小值的x的集合
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(1)∵-π/6<x<π/6∴-π/3<2x<π/3∴0<2x+π/3<2π/3当t∈(0,2π/3)时,y=sint的取值范围是(0,1]∴y=2sint的取值范围是(0,2]即y=2sin(2x+
y=2cos(x+π4)cos(x−π4)+3sin2x=2(12cos2x−12sin2x)+3sin2x=cos2x+3sin2x=2sin(2x+π6)∴函数y=2cos(x+π4)cos(x−
y=cos²x+asinx-a²+2a+5=1-sin²x+asinx-a²+2a+5=-sin²x+asinx-a²+2a+6=-(sin
y=sinxcosx-cos^2x=1/2sin2x-1/2(1+cos2x)=1/2(sin2x-cos2x-1)=1/2[√2*sin(2x-派/4)-1]=√2/2*sin(2x-派/4)-1/
∵y=cos(x2-π3)的单调递减区间即为y=-cos(x2-π3)的单调递增区间,由2kπ≤x2-π3≤2kπ+π(k∈Z)得:2π3+4kπ≤x≤8π3+4kπ(k∈Z),∴函数y=-cos(x
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y=(cosx+2)/(sinx-1)ysinx-y=cosx+2ysinx-cosx=y+2√(y²+1)sin(x-t)=y+2,t=arctan(1/y)sin(x-t)=(y+2)/
就是简单的复合函数求导问题嘛.1.y'=[1/cos(10+2x)]*[-sin(10+2x)]*2=[-2sin(10+2x)]/cos(10+2x)2.y'=[1/cos(3+x²)]*
y=cos^2x+sinx=1-2(sinx)^2+sinx=-2(sinx-1/4)^2+9/8因为|x|
u=2sinx,v=2cosx+3,y=u/vu^2+(v-3)^2=4(vy)^2+(v-3)^2=4(y^2+1)v^2-6v+5=0Δ=36-4*5*(y^2+1)>=0y^2+1
y=2cos^2x+sin2x=cos2x+sin2x+1=√2sin(2x+pi/4)+11.最小正周期T=2pi/2=pi2.最大值就是当sin(2x+pi/4)=1时,得到y=√2+1
你的X呢?如果是COSX,那么答案是R
y=cos^2x-sinx=1-sin²x-sinx=-(sinx+1/2)²+5/4所以当sinx=-1/2时,有最大值=5/4当sinx=1时,有最小值=1-1-1=-1值域为
y=[cosx-1-1]/(cosx-1)=1-1/(cosx-1)=1-1/(1-2sin^2(x/2)-1)=1+1/(2*sin^2(x/2))故其周期是T=2π
y=cos(x²-2x+2)+(-x)²求y的导函数y=cos(x²-2x+2)+x²dy/dx=-[sin(x²-2x+2)](2x-2)+2x再问
y=3cosX-cos(2X)=3cosx-(2*(cosx^)2-1)=-2(cosx)^2+3cosx+1=-2(cosx-3/4)^2+17/8当cos=3/4时,y有最大值,为17/8当cos
函数y=cos2(x+π4)−sin2(x+π4)=cos2(x+π4)=-sin2x,∴T=2π2=π.故答案为π.
y=cos(2x-派/2)=sin2xA最小正周期为派的奇函数
y=(sinx+cos)^2+2cos^2x=1+2sinxcosx+cos2x-1=sin2x+cos2x=√2sin(2x+π/4)
y=cos²x-sin²x+2sinxcosx=cos2x+sin2x=√2sin(2x+π/4)所以值域为【-√2,√2】