fx=sin2x-根号3cos2x 1
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![fx=sin2x-根号3cos2x 1](/uploads/image/f/587193-33-3.jpg?t=fx%3Dsin2x-%E6%A0%B9%E5%8F%B73cos2x+1)
f(x)=2(cosx)^2+√3*sin2x[利用cos2x=2(cosx)^2-1化简]=1+cos2x+√3*sin2x=1+2[(1/2)*cos2x+(√3/2)*sin2x]=1+2[si
f(x)=1+cos2x+根号3sin2x+a=2sin(2x+π/6)+a+11、若f(x)max=2,则sin(2x+π/6)=1,即2+a+1=2,得a=-12、正弦的单调减区间在第二和第三象限
tanx=根号3,那么x=60度sin2x=sin120度=sin60度=根号3/2cos^2*x=(-1/2)^2=1/4那么sin2x/1+cos^2*x=2*根号3/5
令t=sin2x/1+cos²x=2sinxcosx/sin^x+2cos^x,则有1/t=sin^x+2cos^x/2sinxcosx=(sinx/2cosx)+cosx/sinx=tan
f(x)=(√3/2)sin2x-cos²x-1/2=(√3/2)sin2x-(1+cos2x)/2-1/2=(√3/2)sin2x-(1/2)cos2x-1=sin(2x-π/6)-1从而
化简得:y=sin(2x+π/3)+根号3/2周期是π,增区间是(kπ-5π/12,kπ+π/12)(k属于z)
/>y=sin2x+2√2cos(π/4+x)+3=cos(2x-π/2)+2√2cos(π/4+x)+3=1-2sin²(x-π/4)-2√2sin(x-π/4)+3=4-2[sin
fx=sin2x-根号3*(1+cos2x)+a+根号3=2sin(2x-60°)+aT=pi,增区间[k*pi-pi/6,k*pi+5pi/12],k属于Z 2.由题意得-5pi/6<
F(X)=cos(√3x+t)F'(X)=-√3sin(√3x+t)F(X)+F'(X)=cos(√3x+t)-√3sin(√3x+t)是奇函数所以F(0)+F'(0)=0即cost-√3sint=0
y=√3cos²x+1/2sin2x=√3/2(cos2x+1)+1/2sin2x=√3/2cos2x+1/2sin2x+√3/2=sin(π/3)cos2x+cos(π/3)sin2x+√
f(x)=sin2x-2√3(cosx)^2+√3=sin2x-√3(1+cos2x)+√3=sin2x-√3cos2x=2sin(2x-π/3)π/4=再问:π/6=
再问:得数再答:最后的不是得数?你这是有多差呀再问:?。。。再问:给我吧再问:采纳了,再问:我懂了谢谢,采纳了
一样是先化简的问题f(x)=2cos^2x+√3*sin2x-1=2(cosx)^2-1+√3*sin2x=cos2x+√3*sin2x=2(1/2*cos2x+√3/2*sin2x)=2(sinπ/
(1)f(x)=sinx(cosx-√3sinx)=sinxcosx-√3sin²x=1/2sin2x-√3/2(1-cos2x)=1/2sin2x+√3/2cos2x-√3/2=sin(2
f(x)=[1-cos(2x)]/2+sin(2x)+3[1+cos(2x)]/2=sin(2x)+cos(2x)+2=√2sin(2x+π/4)+2.周期T=kπ,k∈Z且k≠0.最小正周期为π.
cos^2x=cos2x+1所以原式等于二分之根号三倍的(cos2x+1)+1/2sin2x化简得sin(2x+60)+二分之根号三最大值就是1+二分之根号三
fx=1/2sin2x-根号3/2cos2x+1=sin2xcosπ/3-cos2xsinπ/3+1=sin(2x-π/3)+1最小正周期=2π÷2=π增区间:2kπ-π/2≤2x-π/3≤2kπ+π
1,f(x)=√3sin2x+cos2x+1-m=2sin(2x+π/6)+1-m.若0再问:2A+π/6=π/2,为什么不行再答:sin(2A+π/6)=1/2,怎么可能是2A+π/6=π/2呢?再
tanx=-4/3tanx=sinx/cosx3sinx=-4cosxsin²x+cos²x=1∴sin2x=2sinxcosx=-24/25cos2x=-7/25cosx=-3/