高三数学答案第1页共10页2020届高三第一学期期末考试数学试题参考答案及评分标准2020．1一、单项选择题：本大题共8小题，每小题5分，共40分．14−：BCDA58−：CDBC二、多项选择题：本大题共4小题，每小题5分，共20分．9．AB10．ABD11．BCD12．AC三、填空题：本题共4小题，每小题5分，共20分．13．12，13，1614．π615．216．45°，6π四、解答题：本题共6小题，共70分．17．在横线上填写“3(cos)sinbCacB−=”．解：由正弦定理，得3(sincossin)sinsinBCACB−=．·········································2分由sinsin()sincoscossinABCBCBC=+=+，得3cossinsinsinBCCB−=．由0πC<<，得sin0C≠．所以3cossinBB−=．··········································4分又cos0B≠（若cos0B=，则sin0B=，22sincos0BB+=．这与22sincos1BB+=矛盾），所以tan3B=−．又0πB<<，得2π3B=．·····················································································6分由余弦定理及23b=，得2222π(23)2cos3acac=+−，································8分即212()acac=+−．将4ac+=代入，解得4ac=．········································9分所以113sin43222ABCSacB==××=△．························································10分在横线上填写“22cosacbC+=”．解：由22cosacbC+=及正弦定理，得2sinsin2sincosACBC+=．···············································································2分又sinsin()sincoscossinABCBCBC=+=+，所以有2cossinsin0BCC+=．············································································4分因为(0,π)C∈，所以sin0C≠．从而有1cos2B=−．又(0,π)B∈，所以2π3B=．···············································6分由余弦定理及23b=，得2222π(23)2cos3acac=+−，································8分即212()acac=+−．将4ac+=代入，解得4ac=．········································9分高三数学答案第2页共10页所以113sin43222ABCSacB==××=△．························································10分在横线上填写“sin3sin2ACbAa+=”．解：由正弦定理，得πsinsin3sinsin2BBAA−=．·················································2分由0πA<<，得sin0A≠，所以sin3cos2BB=．············································4分由二倍角公式，得2sincos3cos222BBB=．由π022B<<，得cos02B≠，所以3sin22B=．所以π23B=，即2π3B=．·······6分由余弦定理及23b=，得2222π(23)2cos3acac=+−，································8分即212()acac=+−．将4ac+=代入，解得4ac=．········································9分所以113sin43222ABCSacB==××=△．························································10分18．解：（1）设{}na的公比为q．因为1a，2a，31aa−成等差数列，所以21312()aaaa=+−，即232aa=．因为20a≠，所以322aqa==．··········...
