【名師一號】2020-2021學年新課標B版數(shù)學必修4-雙基限時練8

雙基限時練(八)基礎強化1．已知coseq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,2)))＝eq\f(\r(6),3)，則sinα的值為()A.eq\f(\r(3),3) B．－eq\f(\r(3),3)C.eq\f(\r(6),3) D．－eq\f(\r(6),3)解析coseq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,2)))＝－sinα，∴sinα＝－eq\f(\r(6),3).答案D2．已知sineq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,3)))＝eq\f(1,3)，則coseq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,6)))的值為()A．－eq\f(1,3) B.eq\f(1,3)C.eq\f(2\r(3),3) D．－eq\f(2\r(3),3)解析∵eq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,6)))－eq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,3)))＝eq\f(π,2).∴coseq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,6)))＝coseq\b\lc\[\rc\](\a\vs4\al\co1(\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,3)))＋\f(π,2)))＝－sineq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,3)))＝－eq\f(1,3).答案A3．已知cosα＝eq\f(2,3)，α是第四象限角，則coseq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,2)))的值為()A.eq\f(2,3) B.eq\f(\r(5),3)C．－eq\f(2,3) D．－eq\f(\r(5),3)解析∵α是第四象限角，∴sinα＝－eq\f(\r(5),3).coseq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,2)))＝coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α))＝sinα＝－eq\f(\r(5),3).答案D4．若f(cosx)＝cos2x，則f(sin150°)的值為()A.eq\f(1,2) B．－eq\f(1,2)C.eq\f(\r(3),2) D．－eq\f(\r(3),2)解析f(sin150°)＝f(sin30°)＝f(cos60°)＝cos120°＝－cos60°＝－eq\f(1,2).答案B5．已知tanθ＝2，則eq\f(sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋θ))－cosπ－θ,sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－θ))－sinπ－θ)＝()A．2B．－2C．0D.eq\f(2,3)解析原式＝eq\f(cosθ＋cosθ,cosθ－sinθ)＝eq\f(2,1－tanθ)＝eq\f(2,1－2)＝－2.答案B6．已知sineq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,12)))＝eq\f(1,3)，則coseq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(7π,12)))的值為()A.eq\f(1,3) B．－eq\f(1,3)C．－eq\f(2\r(2),3) D.eq\f(2\r(2),3)解析∵eq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(7π,12)))－eq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,12)))＝eq\f(π,2).∴coseq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(7π,12)))＝coseq\b\lc\[\rc\](\a\vs4\al\co1(\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,12)))＋\f(π,2)))＝－sineq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,12)))＝－eq\f(1,3).答案B7．若cos(π＋α)＝－eq\f(1,3)，則sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)－α))＝________.解析cos(π＋α)＝－cosα，∴cosα＝eq\f(1,3).sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)－α))＝－cosα，∴sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)－α))＝－eq\f(1,3).答案－eq\f(1,3)8．sin21°＋sin22°＋sin23°＋…＋sin288°＋sin289°＋sin290°＝________.解析設A＝sin21°＋sin22°＋…sin289°＋sin290°，則A＝cos289°＋cos288°＋…＋cos21°＋sin290°＝cos21°＋cos22°＋…cos289°＋sin290°.∴2A＝(sin21°＋cos21°)＋(sin22°＋cos22°)＋…＋(sin289°＋cos2∴2A∴A＝45.5，∴sin21°＋sin22°＋…＋sin289°＋sin290°＝45.5.答案45.5能力提升9．coseq\f(α,2)＝－eq\r(1－cos2\b\lc\(\rc\)(\a\vs4\al\co1(\f(π－α,2))))，則eq\f(α,2)是第________象限角(設α是其次象限角)．解析由coseq\f(α,2)＝－eq\r(1－cos2\b\lc\(\rc\)(\a\vs4\al\co1(\f(π－α,2))))，得coseq\f(α,2)＝－eq\r(sin2\b\lc\(\rc\)(\a\vs4\al\co1(\f(π－α,2))))，即coseq\f(α,2)＝－eq\b\lc\|\rc\|(\a\vs4\al\co1(sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－\f(α,2)))))＝－eq\b\lc\|\rc\|(\a\vs4\al\co1(cos\f(α,2)))，∴coseq\f(α,2)<0，即eq\f(α,2)為其次、三象限角．∵α為其次象限角，∴eq\f(α,2)為第一、三象限角．∴eq\f(α,2)為第三象限角．答案三10．已知sin(5π－θ)＋sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(5π,2)－θ))＝eq\f(\r(7),2)，求sin3eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋θ))－cos3eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)－θ))的值．解析∵sin(5π－θ)＋sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(5π,2)－θ))＝eq\f(\r(7),2)，∴sinθ＋cosθ＝eq\f(\r(7),2).∴sin3eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋θ))－cos3eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)－θ))＝cos3θ＋sin3θ＝(sinθ＋cosθ)(sin2θ－sinθcosθ＋cos2θ)＝(sinθ＋cosθ)eq\b\lc\[\rc\](\a\vs4\al\co1(1＋\f(1－sinθ＋cosθ2,2)))＝eq\f(5\r(7),16).11．若sinθ＝eq\f(\r(3),3)，求eq\f(cosπ－θ,cosθ\b\lc\[\rc\](\a\vs4\al\co1(sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(3,2)π－θ))－1)))＋eq\f(cos2π－θ,cosπ＋θsin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋θ))－sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋θ)))的值．解析cos(π－θ)＝－cosθ，sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3,2)π－θ))＝sineq\b\lc\[\rc\](\a\vs4\al\co1(π＋\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－θ))))＝－sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－θ))＝－cosθ，cos(2π－θ)＝cosθ，cos(π＋θ)＝－cosθ，sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋θ))＝cosθ，sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋θ))＝sineq\b\lc\[\rc\](\a\vs4\al\co1(π＋\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋θ))))＝－sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋θ))＝－cosθ.∴原式＝eq\f(－cosθ,cosθ－cosθ－1)＋eq\f(cosθ,－cosθcosθ＋cosθ)＝eq\f(1,1＋cosθ)＋eq\f(1,1－cosθ)＝eq\f(2,sin2θ)＝eq\f(2,\b\lc\(\rc\)(\a\vs4\al\co1(\f(\r(3),3)))2)＝6.12．化簡：(1)eq\f(sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(3,2)π＋α))cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α)),cos10π＋α)＋eq\f(sin11π－αcos\b\lc\(\rc\)(\a\vs4\al\co1(\f(5,2)π＋α)),sinπ＋α)；(2)coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3k＋1,3)·π＋α))＋coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3k－1,3)·π－α))(k∈Z)．解析(1)原式＝eq\f(－cosαsinα,cosα)＋eq\f(sinαcos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α)),－sinα)＝－sinα＋sinα＝0.(2)當k＝2n，n∈Z時，原式＝coseq\b\lc\(\rc\)(\a\vs4\al\co1(kπ＋\f(π,3)＋α))＋coseq\b\lc\(\rc\)(\a\vs4\al\co1(kπ－\f(π,3)－α))＝coseq\b\lc\(\rc\)(\a\vs4\al\co1(2nπ＋\f(π,3)＋α))＋coseq\b\lc\(\rc\)(\a\vs4\al\co1(2nπ－\f(π,3)－α))＝coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,3)＋α))＋coseq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(π,3)－α))＝coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,3)＋α))＋coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,3)＋α))＝2coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,3)＋α))；當k＝2n＋1，n∈Z時，原式＝coscoseq\b\lc\[\rc\](\a\vs4\al\co1(2n＋1π－\f(π,3)－α))＝coseq\b\lc\(\rc\)(\a\vs4\al\co1(π＋\f(π,3)＋α))＋coseq\b\lc\(\rc\)(\a\vs4\al\co1(π－\f(π,3)－α))＝－coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,3)＋α))－coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,3)＋α))＝－