已知a属于(π 2,π),且sina 2 cosa 2=根号6 2

f(x)=[sin(x+a/2)+根号3cos(x+a/2)]*cos(x+a/2)=sin(x+a/2)cos(x+a/2)+根号3cos(x+a/2)cos(x+a/2)=1/2sin(2x+a)

2(sina)^2-sinacosa-3(cosa)^2=0,(2sinα-3)(sinα+1)=0,sinα=-1(sinα=3/2舍去)sin(a+π/4)/(sin2a+cos2a+1)=sin

a∈(0,π/2)时,tana-cota=sina/cosa-cosa/sina=(sina+cosa)(sina-cosa)/(sinacosa)因为sina>0,cosa>0,若使cota0,这与

1>cosa>sinb=cos(π/2-b)>0;a

1）向量a+b=(cosx-siny,sinx+cosy)=(cosx-sin3x,sinx+cos3x)|a+b|²=(cosx-sin3x)²+(sinx+cos3x)

∵cosa=-1/3,且a属于（π/2,π）∴sina>0,sina=√(1-cos²a)=2√2/3∴tana=sina/cosa=2√2sin2a=2sinacosa=2*2√2/3*(

/>因为sina=3/5,a属于(π/2,π)所以cosa=-4/5(1)cos(a-π/4)=cosa*cosπ/4+sina*sinπ/4=-根号2/10；(2)sin(a/2)的平方+tan(a

解.2sin²a-sinacosa-3cos²a=(2sina-3cosa)(sina+cosa)=0∵a∈(0,π/2）∴2sina=3cosa即sina=3/√13,cosa=

(1)|a+b|^2=|a|^2+2a•b+|b|^2,将|a|=13,|b|=19代入得2a•b=46又因为|a-b|^2=|a|^2-2a•b+|b|^2=|a

tana=-2所以sina=2/根号5cosa=-1/根号5所以,原式=根号3/2*sina+1/2*cosa=（2根号3-1）/（2根号5）=（2根号15-根号5）/10稍稍说一下：先不管a的象限,

提醒你修改,2sin²α-sinαcosα-3cos²α.这少个等号在哪里?还有,这是我们那一届做过的某地的高考题,你搜一下哟.

a向量*b向量=cos2xcosx+sin2xsinx=cos(2x-x)=cosx|a向量+b向量|=根号下(cos2x+cosx)^2+(sin2x+sinx)^2=根号下2+2cosx=根号下2

sina+cosa=tana-1/tanasina+cosa=sina/cosa-cosa/sinasina+cosa=(sin²a-cos²a)/sinacosa1=(sina-

tana=tan(a-b+b)=[tan(a-b)+tanb]/[1-tan(a-b)*tanb]=(1/2-1/7)/(1+1/2*1/7)=1/3所以tan(2a-b)=tan(a+a-b)=[t

a属于（0,π/4）,sina-cosa

f(x)=cosx-(-sinx)=sinx+cosx=√2(√2/2*sinx+√2/2cosx)=√2(sinxcosπ/4+cosxsinπ/4)=√2sin(x+π/4)所以最大值=√2f(a

cosa=1/3+sina平方cos²a=1-sin²a=1/9+2/3*sina+sin²asin²a+1/3*sina-4/9=0由a范围则sina

sina>0,且cosa

(1)sin(2A+π/6)+sin(2A-π/6)+2cos^2A>=2(sin2Acosπ/6+cos2Asinπ/6)+(sin2Acosπ/6-cos2Asinπ/6)+cos2A+1≥2(√

解由sina=4/5,a属于(0,2分之π)则cosa=√1-sin^2a=3/5故sin2a=2sinacosa=2×4/5×3/5=24/25cos2a=2cos^a-1=2（3/5)^2-1=-