已知向量m=(cos2分之x)
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(1)由题意得sinA-2cosA=0两边同除以cosA得tanA-2=0所以tanA=0(2)f(x)=cos2x+tanAsinx(2)f(x)=cos2x+tanAsinx=(1-2sin^2x
(1)f(x)=a•b+1=23sinxcosx−2cos2x+1=3sin2x−cos2x=2sin(2x−π6),∴函数f(x)的最小正周期T=2π2=π,由−π2+2kπ≤2x−π6≤2kπ+π
a=(√3sinx/2,cosx/2),b=(cosx/2,-cosx/2)f(x)=a·b=√3sin(x/2)cos(x/2)-(cosx/2)^2=(√3/2)sinx-(1/2)cosx-1/
y=sin(x/2)+√3cos(x/2)=2sin(x/2+π/3)x=4kπ+π/3(k为整数)时取最大值2
1.cos((A+C)/2)=根号下(1-cos^2(B/2))=三分之根号三,解得cos(B/2)=三分之根号六,cosB=cos^2(B/2)-sin^2(B/2)=三分之根号三由a^2+c^2=
连续使用2sinxcosx=sin2x这个公式再答:再答:采纳一下,好吗?谢谢了
首先,求俩括号里的两个向量的点击a·b=2+cos2θ;c·d=2-cos2θ;再者,看求什么,f(x).最终得根号1+cos2θ<根号1-cos2θ;得cos2θ<0;θ取值范围:π/4+kπ<θ<
f(sin(x/2))=cosx+1=1-2(sin(x/2))^2+1=2-2(sin(x/2))^2令y=sin(x/2)则f(y)=2-2y^2令y=cos(x/2)f(cos(x/2))=2-
1)m*n=0则sinA+2cosA=0tanA=-22)f(x)=cos2x-2sinx=1-2(sinx)^2-2sinx=1-2(sinx+1/2)^2+1/2=3/2-2(sinx+1/2)^
∵向量m=(cosx,2sinx),向量n=(2cosx,-sinx),∴f(x)=向量m*向量n=2cos^2x-2sin^2x=2cos2x(1)f(-3009/3π)=2cos(-2006π)=
(1)a·b=(cos2/3x,sin2/3x)*(cos2/x,-sin2/x)=cos2/3x*cos2/x-sin2/3x*sin2/x=cos(2/3x+2/x)=cos8/3x|a+b|=√
a^2=1,b^2=1,a*b=cos(α/2)cos(β/2)+sin(α/2)sin(β/2)=cos[(α-β)/2],由|a-b|=2√5/5两边平方得a^2-2a*b+b^2=4/5,所以c
向量a乘以向量b=cos(3x/2)乘以cos(x/2)-sin(3x/2)乘以sin(x/2)=cos(3x/2+x/2)(余弦函数两角和公式)=cos2x因为x属于(0,π/2),则2x属于(0,
1f(x)=a·b+2λ|a+b|a·b=(cos(3x/2),sin(3x/2))·(cos(x/2),-sin(x/2))=cos(2x)|a+b|^2=|a|^2+|b|^2+2a·b=2+2c
1)aXb=cos(2/3x)cos(2/x)-sin(2/3x)(sin(2/x)=cos(2/3x+2/x)=cos(8/3x)=0所以8/3x∈π/2+kπ,k∈Z即x∈3π/16+3kπ/8,
a=(cos3x/2,sin3x/2),b=(cosx/2,-sinx/2),(1)a*b=(cos3x/2,sin3x/2)*(cosx/2,-sinx/2)=cos(3x/2)*cos(x/2)-
(1)∵m=(2cos2(x−π6),sinx),n=(1,2sinx),f(x)=m•n=cos(2x-π3)+1+(1-cos2x)=sin(2x-π6)+2,∴T=π;(2)∵0≤x≤5π12,
f(x)=sin(x/2)cos(x/2)+√3*sin²(x/2)+√3/2=1/2*sinx+√3/2*(1-cosx)+√3/2=1/2*sinx-√3/2*cosx+√3=sin(x
1.已知向量a=(sin3分之x,cos3分之x),b=(cos3分之x,根号3cos3分之x),函数f(x)=向量a·向量b则有:f(x)=sin(3分之x)cos(3分之x)+cos(3分之x)*