题目内容：

帮忙解决几道三角函数题（要过程）
1.△ABC中,sinBsinC=cos²A/2,判断△ABC的形状.
2.若cos²（a--b）—cos²（a+b）=1/2,（1+cos2α）（1+cos2b）=1/3,求tanαtanb.
3.化简cos²a+cos²（a-π/3）+cos²（a+π/3）.
4.已知,0≤a＜b＜r＜2π,cosa+cosb+cosr=0,sina+sinb+sinr=0,求b-a.
5.已知a,b为锐角,且3sin²a+2sin²b=1,3sin²a=2sin²b,求a+2b.
6.已知,a,b,r为锐角,tana/2=tan三次方r/2,2tana=tanr,求证,a,b,r成等差数列.
7.已知,sina/sinb=cos（a+b）,其中a,b为锐角,求tanb最大值.

优质解答

1.sinBsinC=cos²(A/2)
sinBsinC=(1+cosA)/2
2sinBsinC=1+cosA
2sinBsinC=1-cos(B+C)
2sinBsinC=1-cosBcosC+sinBsinC
1=cosBcosC+sinBsinC=cos(B-C)
∴B-C=0 => B=C,即△ABC为等腰△
2.cos²(a-b)-cos²(a+b)
=[cos(a-b)+cos(a+b)][cos(a-b)-cos(a+b)]
=2cosacosb*2sinasinb
=2sinacosa*2sinbcosb
=sin2asin2b
=2tana/(1+tan²a) * 2tanb/(1+tan²b)=1/2
∴8tanatanb=(1+tan²a)(1+tan²b)
(1+cos2a)(1+cos2b)
=[1+(1-tan²a)/(1+tan²a)][1+(1-tan²b)/(1+tan²b)]
=2/(1+tan²a) * 2/(1+tan²b)
=4/(1+tan²a)(1+tan²b)
=1/(2tanatanb)
=1/3
∴tanatanb=3/2
3.cos²a+cos²(a-π/3)+cos²(a+π/3)
=cos²a+[cos(a-π/3)+cos(a+π/3)]²-2cos(a-π/3)cos(a+π/3)
=cos²a+[2cosacos(π/3)]²-2[cosacos(π/3)+sinasin(π/3)][cosacos(π/3)-sinasin(π/3)]
=cos²a+4cos²acos²(π/3)-2[cos²acos²(π/3)-sin²asin²(π/3)]
=cos²a+2cos²acos²(π/3)+2sin²asin²(π/3)
=3cos²a/2+3sin²a/2
=3(cos²a+sin²a)/2
=3/2
4.sina+sinb+sinr=0
sinr=-(sina+sinb)
cosa+cosb+cosr=0
cosr=-(cosa+cosb)
sin²r+cos²r=(sina+sinb)^2+(cosa+cosb)^2
=2+2(sinasinb+cosacosb)
=2+2cos(b-a)
=1
cos(b-a)=-1/2
∵0≤a＜b＜r＜2π,
∴b-a=2π/3
5.题目打错了,是3sin2a=2sin2b吧
3sin²a+2sin²b=1
3sin²a=1-2sin²b
3sin²a=cos2b
3sin2a=2sin2b
sin2b=3sin2a/2
sin2b=3sinacosa
cos(a+2b)
=cosacos2b-sinasin2b
=cosa*(3sin²a)-sina*3sinacosa
=0
∵0