已知函数y=根号2cos
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y=1/2cos²x+(根号3)/2sinxcosx+1=(1/2)*(cos2x+1)/2+(根号3)/4*sin2x+1=(1/2)*[cos2x*(1/2)+sin2x*根号3/2]+
y=根号3cos²x+sinxcosx-根号3/2=√3(1/2cos2x+1/2)+1/2sin2x-√3/2=√3/2cos2x+√3/2+1/2sin2x-√3/2=√3/2cos2x
1/2=sin(π/6),√3/2=cos(π/6),因此可对表达式化简:y=(1/2)(cosx)^2+(√3/2)sinxcosx+1=cosx[sin(π/6)cosx+cos(π/6)sinx
y=2√3sinxcosx+2cos²x=√3sin2x+cos2x+1=2(√3/2sin2x+1/2*cos2x)+1=2(sin2xcosπ/6+cos2xsinπ/6)+1=2sin
y=根号2cos(2x+π/4)由2x+π/4=kπ+π/2,k∈Z得x=kπ/2+π/8,k∈Z∴函数的对称轴中心坐标为(kπ/2+π/8,0)k∈Z由2x+π/4=kπ,k∈Z得x=kπ/2-π/
y=(sinx+cosx)^2+2cos^x=2+sin2x+cos2x=2+√2sin(2x+π/4)ymax=2+√2,ymin=2-√2.2kπ+π/2≤2x+π/4≤2kπ+3π/22kπ+π
y=1/2cos²x+√3/2sinxcosx+1=1/2(1/2+1/2cos2x+√3/2sin2x)+1=1/2(sin2xcosπ/6+cos2xsinπ/6)+1+1/4=1/2s
1.y=1/4(1+cos2x)+√3/4sin2x+1=1/2sin(2x+π/6)+5/4当2x+π/6=2kπ+π/2,即x=kπ+π/6时,ymax=7/4.自变量x的集合{x│x=kπ+π/
(1)原式=2[(1/2)sin(x/2)+(根号3/2)cos(x/2)]=2sin[(x/2)+pi/3]所以当[(x/2)+pi/3]=2kpi+pi/2时,y最大值为2解得x=4kpi+pi/
y=2根号3sinxcosx-2cos^2x+1=根号3sin2x-cos2x=2sin(2x-π/6)T=2π/2=π2x-π/6=2kπ+π/2当x=kπ+π/3时,函数的最大值=22x-π/6=
y=sinxcosx+√3cos²x-√3/2=1/2*(2sinxcosx)+√3/2(2cos²x-1)=1/2*sin2x+√3/2cos2x.正弦余弦的二倍角公式=sin(
y=2(2cos²x-1)+2倍根号三sin2xy=2cos2x+2倍根号三sin2xy=4(1/2倍cos2x+根号三/2倍sin2x)y=4sin(π/6+2x)三角函数解析式有了想要什
Y=4(cosx)^2+4√3sinxcosx-2=2cos2x+4√3sinxcosx=4sin〖(
已知函数y=sin^2x+2根号3sinxcosx-cos^x=根号3sin2x-cos2x=2sin(2x-π/6)(1)求函数y的最大值,并求y取最大值时,自变量x的集合ymax=22x-π/6=
解y=2cos^2x+2根号3sinxcosx-1=(1+cos2x)+√3sin2x-1=cos2x+√3sin2x=2(√3/2sin2x+1/2cos2x)=2sin(2x+π/6)即T=2π/
是求两个函数(1)y=√(sinx)(2)y=√(cosx)的定义域吧还是求(3)y=√sin(cosx)定义域(1)要使y=√(sinx)有意义,须令sinx≥0所以2kπ≤x≤π+2kπ,k∈z即
y=12cos2x+32sinxcosx+1=14cos2x+34sin2x+54=12sin(2x+π6)+54,y取最大值,只需2x+π6=π2+2kπ(k∈Z),即x=kππ6(k∈Z),∴当函
y=2√3*sinxcosx+2cos^2x=√3sin2x+cos2x+1=sin(2x+π/6)+1∴最小正周期:t=2π/2=π