【名师一号】2020-2021学年新课标A版高中数学必修4-第一章三角函数双基限时练6

双基限时练(六)1．cos300°＝()A．－eq\f(\r(3),2) B．－eq\f(1,2)C.eq\f(1,2) D.eq\f(\r(3),2)答案C2．若sin(3π＋α)＝－eq\f(1,2)，则coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(7π,2)－α))等于()A．－eq\f(1,2) B.eq\f(1,2)C.eq\f(\r(3),2) D．－eq\f(\r(3),2)解析∵sin(3π＋α)＝sin(π＋α)＝－sinα＝－eq\f(1,2)，∴sinα＝eq\f(1,2).∴coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(7π,2)－α))＝coseq\b\lc\[\rc\](\a\vs4\al\co1(4π－\f(π,2)＋α))＝coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))＝－sinα＝－eq\f(1,2).答案A3．sin(π－2)－coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－2))化简的结果是()A．0 B．－1C．2sin2 D．－2sin2解析sin(π－2)－coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－2))＝sin2－sin2＝0.答案A4．若tan(7π＋α)＝a，则eq\f(sinα－3π＋cosπ－α,sin－α－cosπ＋α)的值为()A.eq\f(a－1,a＋1) B.eq\f(a＋1,a－1)C．－1 D．1解析由tan(7π＋α)＝a，得tanα＝a，∴eq\f(sinα－3π＋cosπ－α,sin－α－cosπ＋α)＝eq\f(－sin3π－α－cosα,－sinα＋cosα)＝eq\f(sinα＋cosα,sinα－cosα)＝eq\f(tanα＋1,tanα－1)＝eq\f(a＋1,a－1).答案B5．已知sineq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,4)))＝eq\f(1,3)，则coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,4)＋α))的值等于()A.eq\f(2\r(2),3) B．－eq\f(2\r(2),3)C.eq\f(1,3) D．－eq\f(1,3)解析∵eq\f(π,4)＋α－eq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,4)))＝eq\f(π,2)，∴coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,4)＋α))＝coseq\b\lc\[\rc\](\a\vs4\al\co1(\f(π,2)＋\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,4)))))＝－sineq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,4)))＝－eq\f(1,3).故选D.答案D6．A，B，C为△ABC的三个内角，下列关系式中不成立的是()①cos(A＋B)＝cosC②coseq\f(B＋C,2)＝sineq\f(A,2)③tan(A＋B)＝－tanC④sin(2A＋B＋C)＝sinA．①② B．③④C．①④ D．②③解析由于cos(A＋B)＝－cosC，所以①错；coseq\f(B＋C,2)＝coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－\f(A,2)))＝sineq\f(A,2)，所以②正确；tan(A＋B)＝tan(π－C)＝－tanC，故③正确；sin(2A＋B＋C)＝sin(π＋A)＝－sinA，故④错．所以选C.答案C7．若θ∈(0，π)，cos(π＋θ)＝eq\f(3,5)，则sinθ＝__________.解析∵cos(π＋θ)＝eq\f(3,5)，∴cosθ＝－eq\f(3,5)，故θ∈eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)，π))，∴sinθ＝eq\f(4,5).答案eq\f(4,5)8．化简：sin(450°－α)－sin(180°－α)＋cos(450°－α)＋cos(180°－α)＝________.解析原式＝sin(90°－α)－sinα＋cos(90°－α)－cosα＝cosα－sinα＋sinα－cosα＝0.答案09．化简：sin(－eq\f(23,6)π)＋coseq\f(13π,7)·tan4π－coseq\f(13,3)π＝________.解析原式＝－sineq\b\lc\(\rc\)(\a\vs4\al\co1(4π－\f(π,6)))＋coseq\b\lc\(\rc\)(\a\vs4\al\co1(2π－\f(π,7)))·0－coseq\b\lc\(\rc\)(\a\vs4\al\co1(4π＋\f(π,3)))＝sineq\f(π,6)＋0－coseq\f(π,3)＝eq\f(1,2)－eq\f(1,2)＝0.答案010．已知coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))＝2sineq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,2)))，则eq\f(sinπ－α＋cosπ＋α,5cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(5π,2)－α))＋3sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(7π,2)－α)))＝________.解析∵coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))＝2sineq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,2)))，∴sinα＝2cosα.原式＝eq\f(sinα－cosα,5sinα－3cosα)＝eq\f(2cosα－cosα,10cosα－3cosα)＝eq\f(1,7).答案eq\f(1,7)11．已知sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,3)－α))＝eq\f(1,2)，求coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,6)＋α))·sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(2π,3)＋α))的值．解coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,6)＋α))·sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(2π,3)＋α))＝coseq\b\lc\[\rc\](\a\vs4\al\co1(\f(π,2)－\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,3)－α))))·sineq\b\lc\[\rc\](\a\vs4\al\co1(π－\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,3)－α))))＝sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,3)－α))·sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,3)－α))＝eq\f(1,2)×eq\f(1,2)＝eq\f(1,4).12．在△ABC中，sineq\f(A＋B－C,2)＝sineq\f(A－B＋C,2)，试推断△ABC的外形．解∵A＋B＋C＝π，∴A＋B－C＝π－2C，A－B＋C＝π－2B∵sineq\f(A＋B－C,2)＝sineq\f(A－B＋C,2)，∴sineq\f(π－2B,2)＝sineq\f(π－2C,2).∴sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－B))＝sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－C)).∴cosB＝cosC.∴B＝C.∴△ABC为等腰三角形．13．已知α是第三象限的角，f(α)＝eq\f(sin\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,2)))cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋α))tanπ－α,tan－α－πsin－α－π)(1)化简f(α)；(2)若coseq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(3π,2)))＝eq\f(1,5)，求f(α)的值．解(1)f(α)＝eq\f(－cosα·sinα·－tanα,－tanα·sinα)＝－cosα.(2)∵cose