2024-2025學(xué)年新教材高中數(shù)學(xué)第一章三角函數(shù)1.4.3誘導(dǎo)公式與對(duì)稱1.4.4誘導(dǎo)公式與旋轉(zhuǎn)課時(shí)作業(yè)含解析北師大版必修第二冊(cè)

PAGE課時(shí)分層作業(yè)(五)誘導(dǎo)公式與對(duì)稱誘導(dǎo)公式與旋轉(zhuǎn)(建議用時(shí)：40分鐘)一、選擇題1．已知sineq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,4)))＝eq\f(1,3)，則coseq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,4)))的值等于()A．－eq\f(1,3) B．eq\f(1,3)C．－eq\f(2\r(2),3) D．eq\f(2\r(2),3)A[coseq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,4)))＝sineq\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(\f(π,4)－α))＝－sineq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,4)))＝－eq\f(1,3).]2．若sin(θ＋π)<0，cos(θ－π)>0，則θ在()A．第一象限 B．其次象限C．第三象限 D．第四象限B[∵sin(θ＋π)＝－sinθ<0，∴sinθ>0.∵cos(θ－π)＝cos(π－θ)＝－cosθ>0，∴cosθ<0，∴θ為其次象限角．]3．已知sineq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,6)))＝eq\f(1,3)，則coseq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,3)))的值為()A．－eq\f(2\r(3),3) B．eq\f(2\r(3),3)C．eq\f(1,3) D．－eq\f(1,3)D[coseq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,3)))＝coseq\b\lc\[\rc\](\a\vs4\al\co1(\f(π,2)＋\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,6)))))＝－sineq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,6)))＝－eq\f(1,3).]4．若sin(π＋α)＋coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))＝－m，則coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)－α))＋2sin(2π－α)的值為()A．－eq\f(2m,3) B．eq\f(2m,3)C．－eq\f(3m,2) D．eq\f(3m,2)C[∵sin(π＋α)＋coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))＝－sinα－sinα＝－m，∴sinα＝eq\f(m,2).故coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)－α))＋2sin(2π－α)＝－sinα－2sinα＝－3sinα＝－eq\f(3,2)m.]5．已知sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(5π,4)＋α))＝eq\f(\r(3),2)，則sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,4)－α))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)D[sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,4)－α))＝sineq\b\lc\[\rc\](\a\vs4\al\co1(π－\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,4)－α))))＝sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,4)＋α))＝－sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(5π,4)＋α))＝－eq\f(\r(3),2).]二、填空題6．cos660°＝________.eq\f(1,2)[cos660°＝cos(360°＋300°)＝cos300°＝cos(180°＋120°)＝－cos120°＝－cos(180°－60°)＝cos60°＝eq\f(1,2).]7．cos1°＋cos2°＋cos3°＋…＋cos179°＋cos180°＝________.－1[cos179°＝cos(180°－1°)＝－cos1°，cos178°＝cos(180°－2°)＝－cos2°，……cos91°＝cos(180°－89°)＝－cos89°，∴原式＝(cos1°＋cos179°)＋(cos2°＋cos178°)＋…＋(cos89°＋cos91°)＋(cos90°＋cos180°)＝cos90°＋cos180°＝0＋(－1)＝－1.]8．已知f(x)＝asin(πx＋α)＋bcos(πx＋β)＋2，其中a、b、α、β為常數(shù)．若f(2)＝1，則f(2024)＝________.1[∵f(2)＝asin(2π＋α)＋bcos(2π＋β)＋2＝asinα＋bcosβ＋2＝1，∴asinα＋bcosβ＝－1.f(2020)＝asin(2020π＋α)＋bcos(2020π＋β)＋2＝asinα＋bcosβ＋2＝－1＋2＝1.]三、解答題9．已知角α終邊經(jīng)過(guò)點(diǎn)P(－4,3)，求eq\f(cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))sin－π－α,cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(11π,2)－α))sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(9π,2)＋α)))的值．[解]∵角α終邊經(jīng)過(guò)點(diǎn)P(－4,3)，∴sinα＝eq\f(3,5)，cosα＝－eq\f(4,5)，∴eq\f(cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))sin－π－α,cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(11π,2)－α))sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(9π,2)＋α)))＝eq\f(－sinαsinα,－sinαcosα)＝－eq\f(3,4).10．求證：eq\f(2sin\b\lc\(\rc\)(\a\vs4\al\co1(θ－\f(3π,2)))cos\b\lc\(\rc\)(\a\vs4\al\co1(θ＋\f(π,2)))－1,1－2cos2\b\lc\(\rc\)(\a\vs4\al\co1(θ＋\f(3,2)π)))＝eq\f(sinθ＋cosθ,sinθ－cosθ).[證明]∵左邊＝eq\f(－2sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(3,2)π－θ))－sinθ－1,1－2sin2θ)＝eq\f(－2sin\b\lc\[\rc\](\a\vs4\al\co1(π＋\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－θ))))－sinθ－1,1－2sin2θ)＝eq\f(2sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－θ))－sinθ－1,1－2sin2θ)＝eq\f(－2sinθcosθ－1,sin2θ＋cos2θ－2sin2θ)＝eq\f(sinθ＋cosθ2,sin2θ－cos2θ)＝eq\f(sinθ＋cosθ,sinθ－cosθ)＝右邊．∴原式成立．11．若cos(π＋α)＝－eq\f(1,2)，eq\f(3,2)π<α<2π，則sin(2π＋α)等于()A．eq\f(1,2) B．±eq\f(\r(3),2)C．eq\f(\r(3),2) D．－eq\f(\r(3),2)D[由cos(π＋α)＝－eq\f(1,2)，得cosα＝eq\f(1,2)，∵eq\f(3,2)π＜α＜2π，∴α＝eq\f(5π,3).故sin(2π＋α)＝sinα＝sineq\f(5π,3)＝－sineq\f(π,3)＝－eq\f(\r(3),2)(α為第四象限角)．]12．(多選)在△ABC中，給出下列四個(gè)式子：①sin(A＋B)＋sinC；②cos(A＋B)＋cosC；③sin(2A＋2B)＋sin2C；④cos(2A＋2B其中為常數(shù)的是()A．① B．②C．③ D．④BC[①sin(A＋B)＋sinC＝2sinC；②cos(A＋B)＋cosC＝－cosC＋cosC＝0；③sin(2A＋2B)＋sin2C＝sin[2(π－C＝－sin2C＋sin2④cos(2A＋2B)＋cos2C＝cos[2(π－C＝cos2C＋cos2C＝2cos213．已知cos(75°＋α)＝eq\f(1,3)，則sin(α－15°)＋cos(105°－α)的值是()A．eq\f(1,3) B．eq\f(2,3)C．－eq\f(1,3) D．－eq\f(2,3)D[sin(α－15°)＋cos(105°－α)＝sin[(75°＋α)－90°]＋cos[180°－(75°＋α)]＝－sin[90°－(75°＋α)]－cos(75°＋α)＝－cos(75°＋α)－cos(75°＋α)＝－2cos(75°＋α)＝－eq\f(2,3).]14．已知f(x)＝eq\b\lc\{\rc\(\a\vs4\al\co1(sinπx，x<0，,fx－1－1，x>0，))則feq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(11,6)))＋feq\b\lc\(\rc\)(\a\vs4\al\co1(\f(11,6)))＝________.－2[feq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(11,6)))＝sineq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(11π,6)))＝sineq\f(π,6)＝eq\f(1,2)，feq\b\lc\(\rc\)(\a\vs4\al\co1(\f(11,6)))＝feq\b\lc\(\rc\)(\a\vs4\al\co1(\f(5,6)))－1＝feq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(1,6)))－2＝sineq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(π,6)))－2＝－eq\f(5,2)，∴feq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(11,6)))＋feq\b\lc\(\rc\)(\a\vs4\al\co1(\f(11,6)))＝eq\f(1,2)－eq\f(5,2)＝－2.]15．化簡(jiǎn)：eq\f(sinkπ－αcos[k－1π－α],sin[k＋1π＋α]coskπ＋α)(k∈Z)．[解]當(dāng)k＝2n(n∈Z)時(shí)，原式＝eq\f(sin2nπ－αcos[2