高中数学第三章三角恒等变换3.1.1两角差的余弦公式课后习题新人教A版必修4

1、高中数学第三章三角恒等变换3.1.1两角差的余弦公式课后习题新人教A版必修41.cos(-40°)cos20°-sin(-40°)sin(-20°)等于()101A.-2B.-2C.2解析：cos(-40°)cos20°-sin(-40°)sin(-20°)=cos(-40°)cos20°+sin(-40°)sin20°1=cos(-40-20)=cos(-60)=cos602.答案：C8sM72.计算Bina+cow。的值是()D.-nncos-rcosa+sin-rsin

2、a44解析：原式=sin优+cos优2(cosa + sinK乙：sinu 4- cosa答案：C3.若a=(cos60°,sin60°),a.Eb.2解析：a-b=cos60°cos15°b=(cos15°,sin15C，贝Ua-b=11D.-+sin60°sin15=cos(60-15)=cos45答案：A4.若 sinb叫的值为(3F5"八a二,aC，贝2COS，2A.-57；2C.-=-2X故选A答案：AB.-1"解析：:sin:cos:2cos+2X'=-|S.IT2f1TIT7Tcostrcos

3、+sinasin-445.若sin（兀+8是（）=_5,e是第二象限角,sin=-_5，6是第三象限角，则cos（e-6）的值A.-B.C.ii425解析：由sin（7t3+0)=-5，得sin又由sin2寸55，得3e=,2<5cos6=-5:0是第二象限角，巾是第三象限角，4.cose=-5,sin6=-答案：B6.若cos(A.'解析：,a,§C：-2<a-(3<0,aa,（3均为锐角，且a<§,则+3的值为（5nD.6,/2=-.TTa sin卜1I*sin(a-B)=-JlCOSg-伊=-5.TT.又ae1乙，IL3网：2aC(0,

4、兀),sin2a=J一10:cos(a+B)=cos2a-(a-B)=cos2acos(a-(3)+sin2asin(a-3)Jio43。02J5=l?xT+-xl.-Tl又a+§e(o,兀)，3n-a+3=生.答案：Cla-刍7 .已知cos'=cosa,贝Utan("3解析：:cos'=cosacos+sin答案：128 .若cosacos(3-sinasin§=S,cos(“-§)=S,贝tjtanatan3解析：cos(oc-3)=cosacos(3+sinasin3二可，1又cosacos3-sinasin3=,2+®

5、，得cosacos3=5,1-，得sinocsin§=S,1，得tanatan§=2.1答案：59 .忖学号08720082已知cos(a+30°)=13,30°<a<90°，贝Ucos解析：,.,30°<a<90°,：60°<a+30°<120°又 cos( a+30°sin( a+30°:cos = =cos ( a +30° )-30° =cos( + +30° )cos 30 ° +sin(

6、a +30° )sin 3011的值.10 .已知tana=4型,cos(“+§)=-14,a,§均为锐角，求cos(3铲(°,mj斛：ae,tana=4t£,.sina=4、3cosa,sin2+cos2a=1.4731,由得sina=7,cosa=7.11.a+Be(0,Tt),cos(a+B)=-必，场:sin(a+§)=I".:cos3=cos(a+3)-a=cos(a+3)cosa+sin(a+3)sinOC11.若xCrK”且sin一IXx=5,求2cos%Io的估+2cosx的值.rir解：,xC,sinx=：:cosx=":2cos+2cosx2IT27Tcosxcos+sinxsin+2cosx1祖J-cosx+ysinxl+2cosx4334s,3-3=sinx+cosx=12.导学号08720083已知,sin( a + § ) =-5,sin13r+?l,求 cos的值.3n. 3tt(T