设fx 2根号3sin(π-x)sinx-(sinx-cosx)的平方
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f(x)=根号3cos²ωx+sinωxcosωx=根号3/2(cos²2ωx)+1/2(sin2ωx)=sin(2ωx+π/3);y轴右侧的第一个最高点的横坐标为π/6,π/3ω
17/4根号3/2-321/64a/2带入后,平方,化简
点击[http://pinyin.cn/1fSnrPTivQa]查看这张图片.
因为:f(π/6)=sinπ/2+2sin(-π/6)-根号3cos(π/2)=1-1-0=0所以:f(π/6)f(π/3)=0再问:原式=sin(x+π/3)+√3cos(x+π/3)+2sin(x
2sin^2[(π/4)+x]+根号3(sin^x-cos^x)-1=-(1-2sin^2[(π/4)+x)-√3cos2x=-cos(π/2+2x)-√3cos2x=sin2x-√3cos2x=2[
f(x)=√2/2×cos[2x+(π/4)]+sin²x.=√2/2cos2x*√2/2-√2/2*sin2x*√2/2+sin²x=1/2cos2x-1/2sin2x+sin&
由题意可得:f(x)=sin(2x+π/3)-√3sin^2x+sinxcosx+√3/2=sin(2x+π/3)-√3(1/2-1/2cos2x)+1/2sin2x+√3/2=2sin(2x+π/3
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
∵cosπ/6=√3/2sinπ/6=1/2cosπ/6sinx+sinπ/6cosx=sin(x+π/6)∴√3sinX+cosX=2(√3/2sinX+1/2cosX)=2(cosπ/6sinx+
f(x)=2cosx*sin(x+π/3)-√3sinx^2+sinx*cosx=2cosx*(sinxcosπ/3+cosxsinπ/3))-√3sinx^2+sinx*cosx=sinxcosx+
1)由三角函数和差化积公式:f(x)=2sin(x+x+π/3)/2cos(x-x-π/3)/2=2sin(x+π/6)cos(π/6)=√3sin(x+π/6)f(x)的最小值为-√3.当x+π/6
f(x)=2√3sin²x-sin(2x-π/3)=√3-√3cos2x-1/2sin2x+√3/2cos2x=√3-(1/2sin2x+√3/2cos2x)=√3-sin(2x+π/3)T
函数y=2sin(2x+π3)的图象关于点P(x0,0)成中心对称,所以2x+π3=kπ,k∈Z;所以x=kπ2−π6 k∈Z,因为x0∈[−π2,0],所以x0=−π6;故答案
题目是f(x=sin(x+π/3)+2sin(x+π/3)-√3cos(2π/3-x)!再问:设函数f(x)=sin(x+π/3)+2sin(x+π/3)-根号3cos(2π/3-x)再答:f(x)=
f'(x)=(sinθ)x^2+((根号3)cosθ)x+tanθf'(π/4)=(sinθ)(π^2)/16+[(根号3)*(π/4)*cosθ]+tanθ题目很阴险啊,想让别人把θ和x弄混.
原式=sin(x+π/3)+√3cos(x+π/3)+2sin(x-π/3)=2[1/2sin(x+π/3)+√3/2cos(x+π/3)]+2sin(x-π/3)=2sin(x+π/3+π/6)+2
∵√3/2-√3sin²x+sinxcosx=(√3/2)(1-2sin²x)+(1/2)sin2x=sin(π/3)cos2x+cos(π/3)sin2x=sin(2x+π/3)
根据诱导公式进行化简这个要利用和角和倍角公式逆用
f(x)=2cosxsin(x+π/3)-√3sin²x+sinxcosx=2cosx[sinxcos(π/3)+cosxsin(π/3)]-√3sin²x+sinxcosx=2c
f(x)=-根号3sin^2x+sinxcosx=-√3/2(1-cos2x)+1/2sin2x=√3/2cos2x+1/2sin2x-√3/2=sin(2x+π/3)-√3/2(1)f((23π)/