fx=cos2x-根号三sinxcosx
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![fx=cos2x-根号三sinxcosx](/uploads/image/f/587112-24-2.jpg?t=fx%3Dcos2x-%E6%A0%B9%E5%8F%B7%E4%B8%89sinxcosx)
cos2x/sin{x-π/4}=(cos²x-sin²x)/[根号2/2(sinx-cosx)]=-根号2(cosx+sinx)由已知得,-根号2(cosx+sinx)=-根号2
答案是选A的请看大图
函数f(x)=2sin^2(x+π/4)-√3cos2x-1=-cos(2x+π/2)-√3cos2x=sin2x-√3cos2x=2sin(2x-π/3)1.当x属于R时,函数f(x)的最小正周期T
f(x)=sin(2x+π3)+sin(2x-π/3)f(x)=sin(2x)cos(π/3)+co(2x)sin(π/3)+sin(2x)cos(π/3)-cos(2x)sin(π/3)f(x)=2
f(x)=(1+cos2x+sin2x)/sin(x+π/2)=(1+cos2x+sin2x)/cosx(1)定义域,cosx≠0定义域x≠kπ+π/2,k∈Z(2)f(x)=(1+cos2x+sin
函数f(x)=[2sin^2(x+π/4)-1]-√3cos2x+1=[-cos(2x+π/2)]-√3cos2x+1=sin2x-√3cos2x+1=2(1/2*sin2x-√3/2*cos2x)+
fx=1/2sin2x-√3/2cos2x=sin2xcosπ/3-cos2xsinπ/3=sin(2x-π/3)f(x)最小正周期T=2π/2=π当2x-π/3=2kπ-π/2,即x=kπ-π/12
答案是1/2cos2X可以分解成(cosX+sinX)(cosX-sinX),分母可以化成根2/2乘以(sinX-cosX)追问:分母可以化成根2/2乘以(sinX-cosX)?是如何化的?回答:si
发现你对三角函数公式之间的转化用的不是很熟啊,要努力!不过题目输入的不错,能不能告诉我是在哪里面输入的?我看你的办公软件用的挺好,将2sin^2(π/4+x)化简为1+sin2x,再与后面一项合并化简
(cos2x)/[sin(x-π/4)]=[(cosx)^2-(sinx)^2]/[((根号2)/2)sinx-((根号2)/2)cosx]=(cosx-sinx)(cosx+sinx)/[-((根号
fx=4cos²x-2+1-cos²x-4cosx=3cos²x-4cosx-1令t=cosx则-1≤t≤1即求[3t²-4t-1]的最值
fx=2cosxsin(x+π/3)-√3sin^2x+sinxcosx+1=2cosx(√3/2cosx+1/2sinx)-√3sin^2x+sinxcosx+1=√3cos^2x-√3sin^2x
F(X)=(1+cos2x)sin²x问F(X)是最小正周期为多少的什么函数F(x)=(1+cos2x)sin²x=(1+cos2x)(1-cos2x)/2=(1-cos²
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
f(x)=2sin^2((π/4)-x)-(√3)(cos2x)=1-cos(π/2-2x)-(√3)cos2x=1-sin2x-(√3)cos2x=1-2sin(2x+π/3),(1)f(x)的最小
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
f(x)=2sinx/2cosx/2√3cosx=sin(x/2x/2)√3cosx=sinx√3cosx=√(1^2√3^2)sin(xπ/3)=2sin(xπ/3)函数f(x)的最小正周期T=2π
解:原式=√3sin2x+cos2x+1=2(√3/2sin2x+1/2cos2x+1=2cos(2x-pai/3)+1.