cos(a π 6)=根号3 3 ,则sin(2a-π 6)

cos(a+π/6)-sina=cosa*cos(π/6)-sina*sin(π/6)-sina=√3/2cosa-3/2sina=3/5*√3cos(a+π/3)=cosa*cos(π/3)-sin

/>cos(5π/6+a)=cos[π-(π/6-a)]=-cos(π/6-a)=-√3/3cos²(π/3+a)=cos²[(π/2)-(π/6-a)]=sin²(π/

sin(a+π/3）+sina=1/2sina+√3/2cosa+sina=3/2sina+√3/2cosa=√3（√3/2sina+1/2cosa)=√3cos(a-π/3）=-√3cos(a+2π

sin(a-π/4)]/cos(π+2a)=√2,√2/2(sina-cosa)=-√2cos2a,sina-cosa=-2(cos^2a-sin^2a),sina-cosa=2(sina+cosa)

sin(3π-a)=√2cos(3π/2+β)sina=√2cos(-π/2+β)=√2cos(π/2-β)=√2sinβ√3cos(-a)=-√2cos(π+β)√3cosa=-√2cos(π+β)

解cos(4π/3+a)=cos(π+π/3+a)=-cos(π/3+a)=-cos[π/2-(π/6-a)]=-sin(π/6-a)=-√3/3

你可以看做COS（x）=3分之根号3,求COS（π-x）=好像是-3分之根号3

cos(a-π/6)+sina=4√3/5展开cosacosπ/6+sinasinπ/6+sina=4√3/5√3/2*cosa+3/2*sina=4√3/51/2*cosa+√3/2*sina=4/

右边2sin(π/6+A)=2sin(π/6)cosA+2sinAcos(π/6)=cosA+根号3sin(A)=左式.得证#

∵cos（x-π/6）=-√3/3∴cosx+cos（x-π/3）=cosx+cosxcosπ/3+sinxsinπ/3=cosx+（1/2）cosx+（√3/2）sinx=（3/2）cosx+（√3

根据:cos（a-π/6）+sina=（4*根号3）/5有:cosacosπ/6+sinasinπ/6+sina=（4*根号3）/5(根号3)/2cosa+3/2sina=（4*根号3）/5得到:co

变换已知条件,有sinα=[2^(1/2)]*sinβ(1式）,[3^(1/2)]*cosα=[2^(1/2)]*cosβ(2式）,整理（2式）,并且把（1）与整理后的（2）两边平方,相加,化简,整理

cosacosπ/6+sinasinπ/6+sina=4/5√3√3/2cosa+1/2sina+sina=4/5√3√3/2cosa+3/2sina=4/5√31/2cosa+√3/2sina=4/

由和差化积公式：cosa+cosb=2cos{(a+b)／2}cos{(a-b)／2}得：cos2a+cos2b=2cos(a+b)cos(a-b)又由已知条件4sinasinb=根号2,4cosac

/>(sina-sinb)²=2/3sin²a+sin²b-2sinasinb=2/3(1)(cosa+cosb)²=1/2cos²a+cos

cos(π/6+a)=√3/3cos(π-5π/6+a)=√3/3cos[π-(5π/6-a)]=√3/3-cos(5π/6-a)=√3/3cos(5π/6-a)=-√3/3

[sin(-a)+sin(-a-90°)]/[cos(540°-a)-cos(-a-270°)]=(-sina-cosa)/[cos(180°-a)-cos(270°+a)]=-(sina+cosa)

由公式(sina)^2+(cosa)^2=1sin^2(π/6-a)=1-cos^2(π/6-a)=1-(根号3/3)^2=1-1/3=2/3

a是第二象限角,即a属于（π/2+2kπ,π+2kπ）所以a/2属于（π/4+kπ,π/2+kπ）,即a/2是第一象限或第三象限角.又知cos(a/2)值为负,所以a/2只能是第三象限角.填空题只需分

cos(7π/6+a)=cos[π+（π/6+a)]=-cos(π/6+a)=-根号3/3