2019版数学人教A版必修4训练：1-3-1　诱导公式二、三、四 WORD版含解析.docx

1、1.3三角函数的诱导公式第1课时诱导公式二、三、四课时过关能力提升基础巩固1.cos-203等于()A.12B.32C.-12D.-32解析:cos-203=cos203=cos6+23=cos23=-12.答案:C2.sin 600+tan 240的值是()A.-32B.32C.-12+3D.12+3解析:sin 600+tan 240=sin(360+240)+tan(180+60)=sin 240+tan 60=sin(180+60)+tan 60=-sin 60+tan 60=-32+3=32.答案:B3.已知tan 5=t,则tan(-365)=()A.tB.360+tC.-tD.与

2、t无关解析:tan(-365)=-tan 365=-tan(360+5)=-tan 5=-t.答案:C4.已知是三角形的一个内角,cos(+)=23,则tan(-)=()A.-52B.255C.52D.-255解析:cos(+)=23,cos =-23.又是三角形的一个内角,sin =53,tan(-)=-tan =-sincos=52,故选C.答案:C5.已知sin(45+)=513,则sin(135-)=.解析:sin(135-)=sin180-(45+)=sin(45+)=513.答案:5136.已知cos(-)=12,且为第二象限角,则sin =,tan =.解析:cos(-)=-co

3、s =12,cos =-12.又为第二象限角,sin 0,sin =1-cos2=1-122=32,tan =sincos=32-12=-3.答案:32-37.已知tan7+=5,则tan67-=.解析:tan67-=tan-7+=-tan7+=-5.答案:-58.若P(-4,3)是角终边上的一点,则cos(-3)tan(-2)sin2(-)的值为.答案:-539.已知cos =14,求sin(2+)cos(-+)cos(-)tan的值.解:sin(2+)cos(-+)cos(-)tan=sin(-cos)costan=-cos =-14.10.求证:tan(2-)sin(-2-)cos(6-

4、)sin(3+)cos(-3)=tan .证明左边=tan(-)sin(-)cos(-)sin2+(+)cos-2+(-+)=(-tan)(-sin)cossin(+)cos(-+)=sin2-sincos-(-)=sin2-sincos(-)=sin2-sin(-cos)=tan =右边.所以原等式成立.能力提升1.若cos(+)=-12,322,则sin(2-)= ()A.12B.32C.32D.-32解析:cos(+)=-12,cos =12.又322,sin =-1-cos2=-1-122=-32,sin(2-)=-sin =32.答案:C2.cos(2+)tan(+)sin(-)co

5、s2-cos(-)=()A.1B.-1C.tan D.-tan 解析:cos(2+)tan(+)sin(-)cos2-cos(-)=costansinsincos=tan ,故选C.答案:C3.已知02,sin =45,则sin(+)+cos(-)sin(-)+cos(2-)的值为()A.4B.7C.8D.9解析:因为02,sin =45,所以cos =35,tan =43.所以原式=-sin-cos-sin+cos=sin+cossin-cos=tan+1tan-1=7.答案:B4.sin-296+cos125tan 2 018-cos-223+sin152=.解析:原式=-sin 56+c

6、os 250-cos 23-sin 2=-sin 6+cos 3-1=-12+12-1=-1.答案:-15.若点A(x,y)是-300角终边上异于原点的一点,则yx的值为.解析:yx=tan(-300)=tan 60=3.答案:36.cos 1+cos 2+cos 3+cos 180=.解析:cos 1+cos 179=cos 1+(-cos 1)=0,cos 2+cos 178=cos 2+(-cos 2)=0,原式=(cos 1+cos 179)+(cos 2+cos 178)+(cos 89+cos 91)+cos 90+cos 180=-1.答案:-17.已知f()=sin(-)cos

7、(2-)tan(-+2)tan(-+)sin(3-).(1)化简f();(2)若是第三象限角,sin =-35,求f();(3)若=-313,求f().解:(1)f()=-sincostan-tansin=cos .(2)sin =-35,且是第三象限角,f()=cos =-1-sin2=-1-925=-45.(3)f-313=cos-313=cos-3=cos3=12.8.设f(x)=asin(x+)+bcos(x+)+7,均为实数,若f(2 017)=6,求f(2 018)的值.解:f(2 017)=asin(2 017+)+bcos(2 017+)+7=-asin -bcos +7,-asin -bcos +7=6,asin +bcos =1.又f(2 018)=asin(2 018+)+bcos(2 018+)+7=asin +bcos +7,f(2 018)=1+7=8.