2019-2020學(xué)年新教材高中數(shù)學(xué)課時(shí)素養(yǎng)評(píng)價(jià)四十五誘導(dǎo)公式一新人教A版必修第一冊(cè)

PAGEPAGE1課時(shí)素養(yǎng)評(píng)價(jià)四十五誘導(dǎo)公式(一)(20分鐘·40分)一、選擇題(每小題4分,共16分,多項(xiàng)選擇題全選對(duì)的得4分,選對(duì)但不全的得2分,有選錯(cuò)的得0分)1.cos-16π3A.-1+32C.3-1【解析】選C.原式=cos16π3-sin16π3=cos4π3-sin4π【加練·固】tan(5π+α)=m,則sin(A.m+1mC.-1 D.1【解析】選A.因?yàn)閠an(5π+α)=tanα=m,所以原式=sinα+cosαsinα-cosα2.(多選題)若cos(π+α)=-12,則sin(α-2πA.12 B.-C.32 D.-【解析】選C、D.由cos(π+α)=-12得cosα=12故sin(α-2π)=sinα=±1=±1-1223.在△ABC中,cos(A+B)的值等于 ()A.cosC B.-cosCC.sinC D.-sinC【解析】選B.由于A+B+C=π,所以A+B=π-C.所以cos(A+B)=cos(π-C)=-cosC.4.已知cos(508°-α)=1213,則cos(212°+αA.-1213 B.1213 C.-5【解析】選B.方法一:由于cos(508°-α)=cos(360°+148°-α)=cos(148°-α)=1213所以cos(212°+α)=cos(360°+α-148°)=cos(α-148°)=cos(148°-α)=1213方法二:cos(212°+α)=cos720=cos(508°-α)=1213二、填空題(每小題4分,共8分)5.cos(-585°)sin【解析】原式=cos=cos=-=-222答案:2-26.已知f(x)=sinπx(x<0),f(【解析】因?yàn)閒-116=sin-=sinπ6=12;f116=f5=sin-π6-2=-12所以f-116+f答案:-2三、解答題7.(16分)化簡(jiǎn)下列各式.(1)sin-193πcos(2)sin(-960°)cos1470°-cos(-240°)sin(-210°).【解析】(1)sin-193π=-sin6π+π3cosπ+π6=sin(2)sin(-960°)cos1470°-cos240°sin(-210°)=-sin(180°+60°+2×360°)cos(30°+4×360°)+cos(180°+60°)sin(180°+30°)=sin60°cos30°+cos60°sin30°=1.【加練·固】求證:1cos(-α)【證明】左邊=1=1=1-cos2αcosα1所以原等式成立.(15分鐘·30分)1.(4分)已知tanπ3-α=1A.13 B.-C.233【解析】選A.因?yàn)閠an4=tanπ+π3-α2.(4分)已知n為整數(shù),化簡(jiǎn)sin(A.tannα B.-tannαC.tanα D.-tanα【解析】選C.當(dāng)n=2k,k∈Z時(shí),sinnπ+αcosnπ+α當(dāng)n=2k+1,k∈Z時(shí),sinnπ+αcos=-sinα-cosα【加練·固】計(jì)算sin21°+sin22°+sin23°+…+sin289°= ()A.89B.90C.89【解析】選C.原式=sin21°+sin22°+sin23°+…+sin244°+sin245°+sin2(90°-44°)+…+sin2(90°-3°)+sin2(90°-2°)+sin2(90°-1°)=sin21°+sin22°+sin23°+…+sin244°+sin245°+cos244°+…+cos23°+cos22°+cos21°=(sin21°+cos21°)+(sin22°+cos22°)+(sin23°+cos23°)+…+(sin244°+cos244°)+sin245°=44+12=893.(4分)已知cos(α-55°)=-13,且α為第四象限角,則sin(α+125°)=________,cos(235°-α)=________. 【解析】因?yàn)閏os(α-55°)=-13<0且α是第四象限角所以α-55°是第三象限角.所以sin(α-55°)=-1=-22因?yàn)棣?125°=180°+(α-55°),所以sin(α+125°)=sin[180°+(α-55°)]=-sin(α-55°)=22又因?yàn)?35°-α=180°-(α-55°),所以cos(235°-α)=cos[180°-(α-55°)]=-cos(α-55°)=13答案:2234.(4分)已知sin(π-α)=log814,且α∈-π2,0,則tan(2π-α【解析】因?yàn)閟in(π-α)=sinα=log814=log232-2=-23,即sin又因?yàn)棣痢?π2,0,所以cosα=1-sin2α所以tan(2π-α)=tan(-α)=-tanα=25答案:2【加練·固】若sin(-α)=log279,且α∈-π2,0,則cos(π+α)A.53 B.-C.±53 D.【解析】選B.因?yàn)閟in(-α)=-sinα=log3332=所以sinα=-23,所以cos(π+α)=-cosα=-1-sin25.(14分)在△ABC中,若sin(2π-A)=-2sin(π-B),3cosA=-2cos(π-B),求△ABC的三個(gè)內(nèi)角. 【解析】由條件得sinA=2sinB,3cosA=2cosB,平方相加得2co