数学解答题，题目（图片），急急急！！！ 18，请高手速解答高一数学题目，，，答案详细速采纳，急急急！...
\u5404\u4f4d\u5927\u4f6c\u4eec\u6c42\u89e3\u6570\u5b66\u89e3\u7b54\u9898\uff0c\u9898\u76ee\uff08\u56fe\u7247\uff09\u3001\u6025\u6025\u6025\uff01(1)\u516c\u5deed=(a9-a4)/(9-4)=(-6-9)/5=-3\uff0c\u90a3\u4e48a1=a4-3d=9+9=18
\u6240\u4ee5Sn=na1+n(n-1)d/2=18n-3n(n-1)/2=54
\u6574\u7406\u540e\u5f97\uff1an^2-13n+36=0\uff0c(n-4)(n-9)=0\uff0c\u6240\u4ee5n=4\uff0c\u6216n=9
(2)\u7b49\u6bd4\u6570\u5217\u4e2da2*a(n-1)=a1*an=128\uff0c\u800ca1+an=66
\u6240\u4ee5a1=2\uff0can=64\uff1b\u6216a1=64\uff0can=2
Sn=a1(1-q^n)/(1-q)=(a1-anq)/(1-q)
\u5f53a1=2\uff0can=64\u65f6\uff0cSn=(2-64q)/(1-q)=126\uff0c\u89e3\u5f97\uff1aq=2\uff0c\u90a3\u4e48an=a1*2^(n-1)=64\uff0c\u6240\u4ee5n=6\uff1b
\u5f53a1=64\uff0ca1=2\u65f6\uff0can=(64-2q)/(1-q)=126\uff0c\u89e3\u5f97\uff1aq=1/2\uff0c\u90a3\u4e48an=a1*(1/2)^(n-1)=2\uff0c\u6240\u4ee5n=6
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\u628a\u4e0d\u7b49\u5f0f\u62c6\u6210\u4e24\u7ec4 \u4e00\u7ec4f(x)\u5927\u4e8e0 \u4e00\u7ec4f(x)\u5c0f\u4e8e...\u518d\u89e3\uff0c
sin(α+β)=sinαcosβ+sinβcosα=4/5（由已知α+β在第二象限）
cos(α+β)=cosαcosβ-sinαsinβ=-3/5
2√2cosβ/3+sinβ/3=4/5(i)
cosβ/3-2√2sinβ/3=-3/5(ii)
联立(i)(ii)解方程组消去sinβ就可以得到cosβ的值
因为cosa=1/3
所以sina=√（1-1/9）=√8/9=2√2/3
cosb
=cos(a+b-a)
=cos(a+b)cosa+sin(a+b)sina
=-3/5*1/3+4/5*2√2/3
=(-3+8√2)/15
由题知α+β 是钝角，故：
cosβ = cos(α+β - α) = cos(α+β)cosα + sin(α+β)sinα = (-3/5)×(1/3)+(4/5)×(2√2/3) = (8√2 - 3)/15
sina=2√2/3
sin(a+b)=4/5
cosb
=cos(a+b-a)
=cos(a+b)cosa+sin(a+b)sina
=-3/5*1/3+4/5*2√2/3
=(-3+8√2)/15
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