a属于(π,3π 2) tana=2 则cosa=

答：π/4.因为3sinb=3sin[(a+b)-a]=3sin(a+b)cosa-3cos(a+b)sinasin(2a+b)=sin[(a+b)+a]=sin(a+b)cosa+cos(a+b)s

sin²2a+sin2acosa-cos2a=1(2sinacosa)^2+2sina(cosa)^2-2(cosa)^2+1=12(cosa)^2[2(sina)^2+sina-1]=02

tan(π/4-a)=1/3-tan(a-π/4)=1/3tan(a-π/4)=-1/3(tana-tanπ/4)/(1+tanπ/4tana)=-1/3(tana-1)/(1+tana)=-1/33

f(x)=√2sin(x-a)+cos(x+b)tana=-1/3cosa=-1/√[1+(-1/3)^2]=-3/√10sina=1/√10cosb=√5/5,sinb=2√5/5f(x)=√2si

tan(π/4+A)=sin(π/4+A)/cos(π/4+A)=(sinπ/4*cosA+cosπ/4*sinA)/(cosπ/4*cosA-sinπ/4*sinA)=(tanπ/4*cosA+si

tana+tanb=-3√3tana*tanb=4tan(a+b)=(tana+tanb)/(1-tana*tanb)=√3接下来判断a,b范围,根据tana*tanb=4>0,那么tana和tanb

∵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*(

将sina=2cosa平方,就是sin²a=4cos²a.再带入第二步的式子,就是4cos²a+cos²A=1.解这个式子得cosa=正负5分之根号5.由于a属

4tana/2=1-tan^2a/22tana/2/(1-tan^2a/2)=1/2=tanasina=(1/2)/√(1+1/4)=√5/5cosa=√(20/25)=2√5/5sin2a=2*5*

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,

由题意可知：左侧=(sinθ)^2+2sinθcosθ+(cosθ)^2=1+sin2θ右侧≥2√[(3^x)*(3^-x)]=2√[3^(x-x)]=2而易知左侧最大值为2,此时sin2θ=1且θ∈

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

tana=3tanb,且a,b属于[0,π/2）,a>b,tanb>0tan(a-b)=(tana-tanb)/(1+tanatanb)=2tanb/(1+3tan²b)1+3tan&sup

cosa=-4/5,a属于(π/2,π)sina=3/5sin2a=2sinacosa=-24/25tana=-4/3=2tan(a/2)/[1-tan^2(a/2)]-2[1-tan^2(a/2)]

解；：(tana/2)/(1-tana/2的平方)=1/4∴2tana/2/(1-tan²a/2)=1/2∴tana=1/2∵a,b∈(0,π/4)∴2sina=cosa∴sina=√5/5

sin2a=2sinacosa=-sina则cosa=-1/2所以a=2π/3所以tana=-√3

a属于(π,3π/2)所以tana>01/sin2a=-13/5(sin²a+cos²a)/2sinacosa=-13/5sin²a/sinacosa+cos²

sin2a=-sina公式得2*sina*cosa=-sina2*cosa=-1cosa=-1/2当a属于(π/2,π)时a=2/3π所以tan2/3π=-根号3

a∈(2kπ,2kπ+3/2π)a终边在第1,2,3象限tana=3说明a在第3象限sina<0cosa<0sina=-3/√10=3√10/10cosa=-1/√10=-√10/10

（cosa+1/tana)（sina+tana）cosa=sina/tanasina=cosa*tana代入,得（sina+1）（cosa+1）-1