角a:角B=4:5

sin（a-b)cosa-cos(a-b)sina=3/5sin(a-b-a)=3/5sin(-b)=3/5sinb=-3/5cosb=-4/5sin(b+5/4π）=sinbcos1/4π+cosb

a*a+b*b+a-2b+5/4=0a*a+a+1/4+b*b+-2b+1=0(a+1/2)²+(b-1)²=0a=-1/2,b=1【a-b】/【a+b】=(-3/2)/(1/2)

sin(a-b)cosa-cos(b-a)sina=sin(a-b-a)=sin(-b)=-sin(b)=3/5;sin(b)=-3/5;b是第三象限角,cos(b)=-4/5;sin(b+5pi/4

(b+c)比(c+a)比(a+b)=4:5:6b+c=8则c+a=10a+b=12三式相加得2a+2b+2c=8+10+12=30a+b+c=15分别减上面三式得a=7b=5c=3cosA=(b^2+

5(a+b)²-(a+b)+2(a+b)²-(a+b)³+4(a+b)=-(a+b)³+7(a+b)²+3(a+b)a+b=-0.1原式=-0.001

设向量a=（3,5,-4）,b=（2,1,8）,计算2a+3b,3a-2b,a*b以及a与b所成角的余弦值,并确定γ,μ应满足的条件,使γa+μb与z轴垂直解析：∵向量a=(3,5,-4）,b=(2,

a+b分之a-b=4∴(a+b)/(a-b)=1/4a+b分之5（a-b）-3（a-b）分之a+b=5×[(a-b)/(a+b)]-3×[(a+b)/(a-b)]=5×4-3×1/4=20-0.75=

∵a/sinA=b/sinB∴b*b=4a*a*sinB*sinB化为b^2/(sinB^2)=4a^2a^2/(sinA^2)=4a^2sinA^2=1/4sinA=1/2或sinA=-1/2(舍)

a为第二象限角,cosa=-3/5b为第四象限角,sinb=-4/5cos(a+b)=cosacosb-sinasinb=(-3/5)(3/5)-(4/5)(-4/5)=7/25sin(a-b)=si

（1）因为在△中,所以sinA=√（1-16/25）=3/5,因为∠B=60度,所以sinB=√3/2,cosB=1/2,所以sin（A+B）=sinC=（4√3+3）/10（2）根据正弦定理得：a=

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cos(a-b)cosa+sin(a-b)sina=cos[(a-b)-a]=cos(-b)=cosb=-4/5b是第三象限角2kπ+π

tan2A=tan[(A+B)+(A-B)]=[tan(A+B)+tan(A-B)]/[1-tan(A+B)tan(A-B)]=(3+5)/(1-3×5）=-4/7tan2B=tan[(A+B)-(A

[(4a+3b)(4a-3b)-(3a+3b)(2a-3b)]/5a=[16a^2-9b^2-6a^2+9ab-6ab+9b^2]/5a=[10a^2+3ab]/5a=2a+3b/5

sin（c）＝sin（π-π/3-A）＝sin（2/3π-A）=sin2/3πcosA＋cos2/3πsinA=√3/2*4/5+1/2*sinAsinA可由cosA=4/5求得3/5,上述等式得=2

(1):由题意得：因为cosA=4/5又因为A、B、C是三角形ABC的内角.所以sinA=[根号下(5^2-4^2)]/5=3/5又因为角B=60度所以sinB=(根号3)/2,B=1/2所以可得si

cos（a+B）×cosa+sin（a+B）×sina=-4/5,cos（a+B-a)=-4/5,cosB=-4/5,sinB=-3/5,cos（90°+B）=-sinB=-(-3/5)=3/5第二题

1sinB=√3/2,cosB=1/2,cosA=4/5,sinA=3/5sinC=sin(A+B)=sinAcosB+cosAsinB=(3+4√3)/102S=b*sinC/sinB*sinA*b

a*a+b*b-2a+4b+5=0.即（a-1)^2+(b+2)^2=0.∴a=1,b=-2.代入原式=-2.

sin(a-b)cosa-cos(b-a)sina=-[sin(b-a)cosa+cos(b-a)sina]=-sin(b-a+a)=-sinb=>sinb=-3/5=>cosb=-4/5sin(b+