高考数学（理）二轮专题突破演练（浙江专版）第3部分 专题1 第3讲 “4道”保分题专练卷4 Word版含解析

1、“4道”保分题专练卷(四)1在abc中，角a、b、c的对边分别为a、b、c，满足ac2b，且cos(bc).(1)求cos c的值；(2)若a5，求abc的面积解：(1)ac2b，且abc，b.cos(bc)，sin(bc)，cos ccos(bc)bcos(bc)cos bsin(bc)sin b.(2)由(1)可得sin c，sin asin (bc).在abc中，由正弦定理，得c8.sabcacsin b5810.2数列an的前n项和为sn，a11，an12sn1(nn*)，等差数列bn满足b33，b59.(1)分别求数列an，bn的通项公式；(2)设cn(nn*)，求证cn1cn.解：

2、(1)由an12sn1，得an2sn11，得an1an2(snsn1)，an13an，an3n1.b5b32d6，d3，bn3n6.(2)证明：an23n1，bn23n，cn，cn1cn0，cn1cnc1，cn1cn.3某社区为丰富居民的业余文化生活，准备举行一次趣味运动会在“射击气球”这项比赛中，制定的比赛规则如下：每人只能参加一场比赛，每场比赛中选手依次射击编号为的5个气球；在这5次射击中，若号气球都被击中，且号气球至少有1个被击中，则此人获奖；否则不获奖已知甲每次射击击中气球的概率都为，且各次射击结果互不影响(1)求甲在比赛中获奖的概率；(2)设甲在5次射击中击中气球的个数为随机变量，求

3、的分布列及数学期望解：(1)记甲在5次射击中，击中k次获奖的事件为ak，k3,4,5.a3，a4，a5互斥，甲获奖的概率pp(a3)p(a4)p(a5)p(a3)c2，p(a4)c2，p(a5)c3，甲在比赛中获奖的概率p.(2)随机变量的取值可以为0,1,2,3,4,5.p(0)5，p(1)c4，p(2)c23，p(3)c32，p(4)c4，p(5)5.的分布列为0123待添加的隐藏文字内容345pe()012345.4.如图，在四棱锥pabcd中，pd平面abcd，四边形abcd是菱形，ac2，bd2，e是pb上任意一点(1)求证：acde；(2)已知二面角apbd的余弦值为，若e为pb的

4、中点，求ec与平面pab所成角的正弦值解：(1)证明：pd平面abcd，ac平面abcd，pdac.四边形abcd是菱形，bdac.又bdpdd，ac平面pbd.de平面pbd，acde.(2)连接eo，在pdb中，eopd，eo平面abcd.分别以oa，ob，oe所在直线为x，y，z轴建立空间直角坐标系，设pdt(t0)，则a(1,0,0)，b(0，0)，c(1,0,0)，e，p(0，t)由(1)知，平面pbd的一个法向量为n1(1,0,0)，设平面pab的一个法向量为n2(x，y，z)，且(1, ，0)，(1，t)，则根据得令y1，得平面pab的一个法向量为n2.二面角apbd的余弦值为，

5、|cosn1，n2|，即，解得t2或t2(舍去)，p(0，2)设ec与平面pab所成的角为，(1,0，)，n2(，1,1)，sin |cos，n2|，ec与平面pab所成角的正弦值为.高考资源网版权所有！投稿可联系qq：108459180129c91afe4ced1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91afedd1b6c87d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91af95ad

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