新版高中數(shù)學(xué)人教A版必修4習(xí)題第一章三角函數(shù)1.3.1

1.3三角函數(shù)的誘導(dǎo)公式第1課時誘導(dǎo)公式二、三、四課時過關(guān)·能力提升基礎(chǔ)鞏固1cos-A.解析:cos答案:C2sin600°+tan240°的值是()A.-C.-解析:sin600°+tan240°=sin(360°+240°)+tan(180°+60°)=sin240°+tan60°=sin(180°+60°)+tan60°=sin60°+tan60°=-答案:B3已知tan5°=t,則tan(365°)=()A.t B.360+tC.t D.與t無關(guān)解析:tan(365°)=tan365°=tan(360°+5°)=tan5°=t.答案:C4將cos(π+2)化為某個銳角的三角函數(shù)值為()A.cos2 B.cos2C.cos(π2) D.cos(π2)解析:cos(π+2)=cos2=cos[π(π2)]=cos(π2).又0<π2<π2答案:D5已知sin(45°+α)=5解析:sin(135°α)=sin[180°(45°+α)]=sin(45°+α)=答案:56已知cos(πα)=1解析:cos(πα)=cosα=12,又α為第二象限角,∴sinα>0,∴sinα=∴tanα=答案:37已知tanπ解析:tan答案:58若P(4,3)是角α終邊上一點,則cos答案:-9已知cosα=1解==cosα=-10求證:tan證明左邊==si=右邊.所以原等式成立.能力提升1若cos(π+α)=-1A.解析:∵cos(π+α)=-∴cosα=又∴sinα=-∴sin(2πα)=sinα=答案:C2若sin(180°+α)+cos(180°α)=a,則cos(540°+α)+sin(360°α)的值是()A.a B.a C.解析:由sin(180°+α)+cos(180°α)=a,得sinαcosα=a,cos(540°+α)+sin(360°α)=cosαsinα=a.答案:B3在平面直角坐標系中,若α與β的終邊關(guān)于y軸對稱,則下列等式恒成立的是()A.sin(α+π)=sinβ B.sin(απ)=sinβC.sin(2πα)=sinβ D.sin(α)=sinβ解析:∵α與β的終邊關(guān)于y軸對稱,∴β=πα+2kπ,k∈Z,∴sinβ=sin(πα+2kπ)=sin(πα)=sinα.又sin(α+π)=sinα,sin(απ)=sinα,sin(2πα)=sinα,sin(α)=sinα,∴sin(2πα)=sinβ恒成立.答案:C4若cos(80°)=k,則tan100°=.

解析:∵cos(80°)=k,∴cos80°=k,sin80°=∴cos100°=k,sin100°=故tan100°=-答案:-5若點A(x,y)是300°角終邊上異于原點的一點,則y解析:yx=tan(答案:3★6cos1°+cos2°+cos3°+…+cos180°=.

解析:∵cos1°+cos179°=cos1°+(cos1°)=0,cos2°+cos178°=cos2°+(cos2°)=0,…∴原式=(cos1°+cos179°)+(cos2°+cos178°)+…+(cos89°+cos91°)+cos90°+cos180°=1.答案:17已知f(α)=sin(1)化簡f(α);(2)α是第三象限角,若sinα=-(3)若α=-解(1)f(α)=-sin(2)∵sinα=-35,∴f(α)=cosα=-=-(3)f=cos★8設(shè)f(x)=asin(πx+α)+bcos(πx+β)+7,α,β均為實數(shù),若f(2017)=6,求f(2018)的值.解∵f(2017)=asin(2017π+α)+bcos(2017π+β)+7=asinαbcosβ+7,∴asinαbcosβ+7=6