2019_2020学年高中数学第五章诱导公式（第2课时）诱导公式五、六课后篇巩固提升（含解析）新人教A版.docx

第2课时诱导公式五、六课后篇巩固提升基础巩固1.若,32,则1-sin232-=()A.sin B.-sin C.cos D.-cos 解析,32,sin0.1-sin232-=1-cos2=sin2=-sin.答案B2.已知P(sin 40,-cos 140)为锐角终边上的点,则=()A.40B.50C.70D.80解析P(sin40,-cos140)为角终边上的点,因而tan=-cos140sin40=-cos(90+50)sin(90-50)=sin50cos50=tan50,又为锐角,则=50,故选B.答案B3.已知sin(-)=-2sin2+,则sin cos =()A.25B.-25C.25或-25D.-15解析sin(-)=-2sin2+,sin=-2cos.再由sin2+cos2=1可得sin=255,cos=-55,或sin=-255,cos=55,sincos=-25.故选B.答案B4.在ABC中,若sinA+B2=45,则cosC2=()A.-35B.-45C.35D.45解析A+B+C=,A+B2=2-C2.sinA+B2=sin2-C2=cosC2=45.答案D5.已知cos(60+)=13,且-180-90,则cos(30-)的值为()A.-223B.223C.-23D.23解析由-180-90,得-12060+0,所以-9060+-30,即-150-90,所以12030-180,cos(30-)0,所以cos(30-)=sin(60+)=-1-cos2(60+)=-1-(13)2=-223.答案A6.(一题多空题)若cos =13,且是第四象限的角,则sin =,cos+32=.解析因为是第四象限的角,所以sin=-1-cos2=-223.于是cos+32=-cos+2=sin=-223.答案-223-2237.若sin2+=37,则cos22-=.解析sin2+=cos=37,则cos22-=sin2=1-cos2=1-949=4049.答案40498.求值:sin24-+sin24+=.解析4-+4+=2,sin24+=sin22-4-=cos24-.sin24-+sin24+=sin24-+cos24-=1.答案19.化简:sin-32sin32-tan2(2-)cos2-cos2+cos2(-).解原式=sin-+2-sin2-tan2(2-)cos2-cos2+cos2(-)=cos(-cos)tan2sin(-sin)cos2=tan2sin2=1cos2.10.已知角的终边经过点P45,-35.(1)求sin 的值;(2)求sin2-tan(-)sin(+)cos(3-)的值.解(1)P45,-35,|OP|=1,sin=-35.(2)sin2-tan(-)sin(+)cos(3-)=costan-sin(-cos)=1cos,由三角函数定义知cos=45,故所求式子的值为54.能力提升1.已知0,cos0,所以cos1|cos|+sin1|sin|=-1+1=0,即原式等于0.答案04.sin21+sin22+sin23+sin289=.解析sin21+sin22+sin23+sin289=sin21+sin22+sin23+sin245+cos244+cos21=(sin21+cos21)+(sin22+cos22)+(sin244+cos244)+sin245=44+12=892.答案8925.已知函数f(x)=2cosx-12,xR.若cos =35,32,2,则f-512=.解析f-512=2cos-512-12=2cos-2=2cos2-=2sin,由已知可得为第四象限角,所以sin0,故sin=-1-cos2=-45,f-512=2sin=2-45=-425.答案-4256.是否存在角,-2,2,(0,),使等式sin(3-)=2cos2-,3cos(-)=-2cos(+)同时成立?若存在,求出,的值;若不存在,请说明理由.解由条件,得sin=2sin,3cos=2cos,2+2得sin2+3cos2=2,sin2=12.又