【文档说明】2021学年数学高中必修4人教A版:1.3 第2课时 诱导公式(二).docx,共(5)页,98.771 KB,由小赞的店铺上传
转载请保留链接:https://www.doc5u.com/view-525d6ef738c1bff81f0afa7c46eafe4d.html
以下为本文档部分文字说明:
(本栏目内容,在学生用书中以独立形式分册装订!)一、选择题(每小题5分,共20分)1.已知sinα+π2=13,α∈-π2,0,则sinα等于()A.-22B.22C.-223D.223解析:sinα+π2=cosα=13,又α∈
-π2,0,所以sinα=-1-cos2α=-223.答案:C2.如果cos(π+A)=-12,那么sinπ2+A等于()A.-12B.12C.-32D.32解析:cos(π+A)=-cosA=-12,∴cosA=12,∴sinπ2+A=cosA=12.答案:B
3.已知cosα+π4=23,则sinπ4-α的值等于()A.23B.-23C.53D.±53解析:因为α+π4+π4-α=π2.所以sinπ4-α=sinπ2-α+π4=cosα+π4=23.答
案:A4.若sin(π+α)+cosπ2+α=-m,则cos3π2-α+2sin(6π-α)的值为()A.-23mB.-32mC.23mD.32m解析:∵sin(π+α)+cosπ2+α=-m,即-sinα-sinα=-2sinα=-m,从而sinα=m
2,∴cos3π2-α+2sin(6π-α)=-sinα-2sinα=-3sinα=-32m.答案:B二、填空题(每小题5分,共15分)5.若sinπ2+θ=35,则cos2θ-sin2θ=________.解析:sinπ2+θ=cosθ=35
,从而sin2θ=1-cos2θ=1625,所以cos2θ-sin2θ=-725.答案:-7256.已知sinφ=611,则cos11π2+φ+sin(3π-φ)的值为________.解析:∵sinφ=6
11,∴cos11π2+φ=cos6π-π2+φ=cos-π2+φ=cosπ2-φ=sinφ=611,sin(3π-φ)=sin(2π+π-φ)=sin(π-φ)=sinφ=611,∴cos11π2+φ+sin(3π-φ)=611+611=121
1.答案:12117.已知cos(75°+α)=13,且-180°<α<-90°,则cos(15°-α)=________.解析:∵-180°<α<-90°,∴-105°<75°+α<-15°,∴sin(75°+α)=-1-cos2(75°+α)=-223,cos(15°-α)=cos[90°-
(75°+α)]=sin(75°+α)=-223.答案:-223三、解答题(每小题10分,共20分)8.求证:cos(π-θ)cosθsin3π2-θ-1+cos(2π-θ)cos(π+θ)sinπ2+θ-sin
3π2+θ=2sin2θ.证明:因为左边=-cosθcosθ(-cosθ-1)+cosθ-cosθcosθ+cosθ=11+cosθ+11-cosθ=1-cosθ+1+cosθ(1+cosθ)(1-cosθ)=21-cos2θ=2sin2θ=右边.所以等式成立.
9.化简:(1)cos(α-π)sin(π-α)·sinα-π2cosπ2+α;(2)sin(-α-5π)cosα-π2-sin3π2+αcos(α-2π).解析:(1)原式=cos[-(π-α)]sinα·sin
-π2-α(-sinα)=cos(π-α)sinα·-sinπ2-α(-sinα)=-cosαsinα·(-cosα)(-sinα)=-cos2α.(2)原式=sin(-α-π)cos-π2-α+
cosαcos[]-(2π-α)=sin[-(α+π)]cosπ2-α+cosαcos(2π-α)=-sin(α+π)sinα+cosαcosα=sin2α+cos2α=1.尖子生题库☆☆☆10.已知f(α)=sin(α-3π)co
s(2π-α)sin-α+3π2cos(-π-α)sin(-π-α).(1)化简f(α);(2)若α是第三象限角,且cosα-3π2=15,求f(α)的值.解析:(1)f(α)=sin(α-3π)cos(2π-α)sin-α+3
π2cos(-π-α)sin(-π-α)=(-sinα)·cosα·(-cosα)(-cosα)·sinα=-cosα.(2)因为cosα-3π2=-sinα,所以sinα=-15,又α是第三象限角,所以
cosα=-1--152=-256.所以f(α)=256.获得更多资源请扫码加入享学资源网微信公众号www.xiangxue100.com