已知向量a=(sinx,3 2),b=(cosx,-1)
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a.b=(sinx-cosx)(sinx+cosx)+2cosxsinx=sin2x-cos2x=3/5=>(sin2x-cos2x)^2=9/251-2sin2xcos2x=9/25sin4x=16
f(x)=(2sinx)×(√3cosx)+(cosx+sinx)×(sinx-cosx)f(x)=2√3sinxcosx-(cos²x-sin²x)f(x)=√3sin2x-co
第一题:(1):f(x)=2倍sinx的平方+2倍根号3cosxsinx-1化简为:f(x)=-2cos(2x+π/3)显然f(x)在x=0处去最小为-1;在x=π/3处取最大为2(2):f(x)=-
解:f(x)=向量a*向量b=1+sin(2x)+(sinx-cosx)(sinx+cosx)=1+sin(2x)-((cosx)^2-(sinx)^2)=1+sin(2x)-cos(2x)-->f(
这个.我还以为什么压轴难题呢.完全口算就可以了嘛(玩笑..)应该是f(x)=sin2x+cos2x然后f(x)=√2sin(2x+π/4)(如果我没记错的话)当2x+π/4=π/2时,f(x)取到最大
1、先把第一题答案给你.a·b=(sinx)^2+sinxcosx=(1-cox2x)/2+(sin2x)/2=(sin2x-cos2x)/2+1/2=(√2/2)[(√2/2)sin2x-(√2/2
1、f(x)=2sin²x+2√3sinxcosx=1-cos2x+√3sin2x=2sin(2x-π/6)+1.当x∈[0,π/2]时,f(x)∈[2,3];若f(x)关于直线x=a对称,
f(x)=a*b=2sinxcosx+2√3(sinx)^2=sin(2x)+√3[1-cos(2x)]=2sin(2x-π/3)+√3,因为y=f(x+φ)=2sin(2x+2φ-π/3)+√3为偶
f(x)=向量a×向量b=(sinx,√3cosx)*(cosx,cosx)=sinxcosx+√3cosxcosx=1/2(2sinxcosx+2√3cosxcosx)=1/2(sin2x+√3co
f(x)=1+sin2x+(sinx)^2-(cosx)^2=1+sin2x-cos2x=1+√2sin(2x-π/4),它的最大值=1+√2,这时,2x-π/4=(2k+1/2)π,k∈Z,∴x=(
1.f(x)=2(√3sinxcosx+(cosx)^2)+2m-1=√3sin2x+cos2x+2m=2sin(2x+pi/6)+2m最小正周期=pi2.x属于[0,pi/2]f(x)最小值=2si
(1)a*b=0sin2x-cos2x=0sqr(2)sin(2x-π/4)=0x=π/8+kπ/2,k∈Z(2)f(x)=sqr(2)sin(2x-π/4)x∈(3π/8+kπ,7π/8+kπ),k
f(x)=cos^2x+sinxcosx=1/2(sin2x+cos2x)+1/2=√2/2sin(2x+π/4)+1/21.增[kπ-π/8,kπ+3π/8]2.f(B)=√2/2sin(2B+π/
f(x)=a·b=sin²x-√3sinxcosx²=1/2-(cos2x+√3sin2x)/2=1/2-sin(2x+π/6)单调递增区间2x+π/6∈[(2n+1/2)π,(2
=(8sinx,2sinx+1)?1f(x)=a·b=(cosx,4sinx-2)·(8sinx,2sinx+1)=8sinxcosx+8sinx^2-2+4sinx-4sinx=4sin(2x)+4
(1)a⊥b则:f(x)=sinxcosx+√3cosxcosx=sin2x/2+√3(1+cos2x)/2=sin(2x+π/3)+√3/2=0∴2x+π/3=2kπ+3π/2±π/6∴x=kπ+7
a*b=sin²x+sinxcosx=sinx(sinx+cosx)=(1/2)(sinx+cosx),所以,sinx+cosx=0或sinx=1/2.1、若sinx+cosx=0,则tan
1)因为X=π/6,所以向量a=(根号3/2,1/2),根据公式a•c=|a|*|c|*cos<a,c>所以向量a与向量c的乘积为cosπ/6*(-1)+sinx*0=负根号3/2,向量a
(1)f(x)=2sin²x+2sinxcosx=1-cos2x+sin2x=1+√2sin(2x-π/4)递减区间为:π/2+2kπ≤2x-π/4≤3π/2+2kπ化简得到:3π/8+kπ