高考數學（理）二輪專題突破演練（浙江專版）第3部分 專題1 第3講 “4道”保分題專練卷4 Word版含解析

1、“4道”保分題專練卷(四)1在abc中，角a、b、c的對邊分別為a、b、c，滿足ac2b，且cos(bc).(1)求cos c的值；(2)若a5，求abc的面積解：(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數列an的前n項和為sn，a11，an12sn1(nn*)，等差數列bn滿足b33，b59.(1)分別求數列an，bn的通項公式；(2)設cn(nn*)，求證cn1cn.解：

2、(1)由an12sn1，得an2sn11，得an1an2(snsn1)，an13an，an3n1.b5b32d6，d3，bn3n6.(2)證明：an23n1，bn23n，cn，cn1cn0，cn1cnc1，cn1cn.3某社區(qū)為豐富居民的業(yè)余文化生活，準備舉行一次趣味運動會在“射擊氣球”這項比賽中，制定的比賽規(guī)則如下：每人只能參加一場比賽，每場比賽中選手依次射擊編號為的5個氣球；在這5次射擊中，若號氣球都被擊中，且號氣球至少有1個被擊中，則此人獲獎；否則不獲獎已知甲每次射擊擊中氣球的概率都為，且各次射擊結果互不影響(1)求甲在比賽中獲獎的概率；(2)設甲在5次射擊中擊中氣球的個數為隨機變量，求

3、的分布列及數學期望解：(1)記甲在5次射擊中，擊中k次獲獎的事件為ak，k3,4,5.a3，a4，a5互斥，甲獲獎的概率pp(a3)p(a4)p(a5)p(a3)c2，p(a4)c2，p(a5)c3，甲在比賽中獲獎的概率p.(2)隨機變量的取值可以為0,1,2,3,4,5.p(0)5，p(1)c4，p(2)c23，p(3)c32，p(4)c4，p(5)5.的分布列為012345pe()012345.4.如圖，在四棱錐pabcd中，pd平面abcd，四邊形abcd是菱形，ac2，bd2，e是pb上任意一點(1)求證：acde；(2)已知二面角apbd的余弦值為，若e為pb的中點，求ec與平面pa

4、b所成角的正弦值解：(1)證明：pd平面abcd，ac平面abcd，pdac.四邊形abcd是菱形，bdac.又bdpdd，ac平面pbd.de平面pbd，acde.(2)連接eo，在pdb中，eopd，eo平面abcd.分別以oa，ob，oe所在直線為x，y，z軸建立空間直角坐標系，設pdt(t0)，則a(1,0,0)，b(0，0)，c(1,0,0)，e，p(0，t)由(1)知，平面pbd的一個法向量為n1(1,0,0)，設平面pab的一個法向量為n2(x，y，z)，且(1, ，0)，(1，t)，則根據得令y1，得平面pab的一個法向量為n2.二面角apbd的余弦值為，|cosn1，n2|，

5、即，解得t2或t2(舍去)，p(0，2)設ec與平面pab所成的角為，(1,0，)，n2(，1,1)，sin |cos，n2|，ec與平面pab所成角的正弦值為.高考資源網版權所有！投稿可聯系qq：1084591801c29c91afe4ced1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91afedd1b6c87d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91af95ad9c91afe4ce

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