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《【金版新学案】高考数学总复习课时作业19简单的三角恒等变换试题文新人教A版.docx》由会员上传分享,免费在线阅读,更多相关内容在教育资源-天天文库。
1、课时作业(十九)简单的三角恒等变换A级771.如果aCi~
2、~,兀i!,且sin4a=5,那么sincos1a+—
3、=()A.452B.4;25D.2.(2012•山东卷)若。C片,7ty,,sin22」37nt—则sin9=(83A.5C.D.33.已知tan1a+-42sin<0,则——2.a+sin25住-7-寺i兀'A.C.3,:1010cosaD*4.(2013•中山模拟)已知角A为△ABC勺内角,且sinA.I7B・532A=—4,则sinA—cosA=(71D.26.化简㈤n际—1—tan215
4、'sin2cosacosa2a一sina7.已知a、3均为锐角,且cos(a+3)=sin(a—3),则tan8.若锐角a、3满足(1+^3tana)(1+^3tan3)=4,则a+1c.-2abcd=ad—bc,若cos-1,sinasin37
5、cosacos35.定义运算33^47,0<36、均不为0,asina+bcosar-t=tanB,JeLB—aacosa—bsina”,”A.3B.l3C.—32.计算:cos10°+^/3sin10小一cos80fC2f(x)=cos^2x+—i+sinx.⑴求函数f(x)的最大值;7(2)设A,B,C为△ABC勺三个内角,若cosB=;,fC;=一;,且C为锐角,求sinA答案:课时作业(十九)1.4兀5,y7、a+cosacos7t4=2啦sina2.55.4.AA为^ABC勺内角且sin2A=2sin3AcosA=--<0,4sinA>0,cosA<0,•1-sinA—cosA>0.又(sinA—cosA2=1—2sin7AcosA=1.477sin5.D依题意有sinacos3—cosasin=sin(a—3)=第3,7又0<3故cos(13有而cos7,,sin笨,是sin3=sin[a—(a—3)]=sinacos(a—3)—cosasin(x£」x双也14714选D.6.解析:…、1。原式=2tan(90—8、2a),1-sin2a2cos2a1SI-1—2a一2(3-2a12sin2acos2a1cos2a2sin21sin2a2cos2a答案:7.解析:根据已知条件:cosacos3—sinasin3=sinacos3—cosssin3,cos3(cosa—sina)+sin3(cosa—sina)=0,即(cos3+sin3)(cosa—sina)=0.3为锐角,则sin3+cos3>0,一sina=0,•1-tana=1.答案:8.解析:由(1+#tana)(1+馅tan3)=4,可得tana+tan31—t9、anatan3=*^3,即tan(a+3)=1/3.一一一,一、一兀又a+3C(0,兀),,a+3=w3答案:9.解析:原式=1+2sin0-cos0—1—2sin201+2sin0-cos0+2COS20—12sin20-cos0+2sin02sin01COS0+sin0~-2-=二;一■二7二~;=tan0.2sin0-cos0+2cos02cos01S..10+cos07答案:tan010.解析:.tana12=2,,tana2tan-21+sin2a-1+sin2a+cos2a1-tan2^2sin210、a+cos2a+2sinasin民+cos5tan民2cosa11.证明:sina左边二cosFsin民sin7tcoscos兀ai十5一cosasinsin11-4cosa+17一=6cos彳十万+sin卜万a।2—11、+cosacos.一a।4+万7ti兀a।sin彳十万cos兀a।—+-[cosasin7t—443.兀a।了十万一=右边a・•・原式得证.1-3tana+23-3tana+J33sin5+也853—,3tana3cosa—3sin巴与已知比较可设a=3t,b=J3t,tw0故2=3?选a3B.12、2.解析:cos10°+Msin10sin2sin40小一cos8030°cusnr3U°sin1(T.2sin240°2sin40=2.答案:,27…L兀Tt1-cos2x3.解析:(1)f(x)=cos2xcos--sin2xsin—H2131113=2cos2x--2-sin2x+2—2cos2x=2一上sin2x.一一兀所以,当2x=-y+2kTt,kez,即x=—1+kTt(
6、均不为0,asina+bcosar-t=tanB,JeLB—aacosa—bsina”,”A.3B.l3C.—32.计算:cos10°+^/3sin10小一cos80fC2f(x)=cos^2x+—i+sinx.⑴求函数f(x)的最大值;7(2)设A,B,C为△ABC勺三个内角,若cosB=;,fC;=一;,且C为锐角,求sinA答案:课时作业(十九)1.4兀5,y7、a+cosacos7t4=2啦sina2.55.4.AA为^ABC勺内角且sin2A=2sin3AcosA=--<0,4sinA>0,cosA<0,•1-sinA—cosA>0.又(sinA—cosA2=1—2sin7AcosA=1.477sin5.D依题意有sinacos3—cosasin=sin(a—3)=第3,7又0<3故cos(13有而cos7,,sin笨,是sin3=sin[a—(a—3)]=sinacos(a—3)—cosasin(x£」x双也14714选D.6.解析:…、1。原式=2tan(90—8、2a),1-sin2a2cos2a1SI-1—2a一2(3-2a12sin2acos2a1cos2a2sin21sin2a2cos2a答案:7.解析:根据已知条件:cosacos3—sinasin3=sinacos3—cosssin3,cos3(cosa—sina)+sin3(cosa—sina)=0,即(cos3+sin3)(cosa—sina)=0.3为锐角,则sin3+cos3>0,一sina=0,•1-tana=1.答案:8.解析:由(1+#tana)(1+馅tan3)=4,可得tana+tan31—t9、anatan3=*^3,即tan(a+3)=1/3.一一一,一、一兀又a+3C(0,兀),,a+3=w3答案:9.解析:原式=1+2sin0-cos0—1—2sin201+2sin0-cos0+2COS20—12sin20-cos0+2sin02sin01COS0+sin0~-2-=二;一■二7二~;=tan0.2sin0-cos0+2cos02cos01S..10+cos07答案:tan010.解析:.tana12=2,,tana2tan-21+sin2a-1+sin2a+cos2a1-tan2^2sin210、a+cos2a+2sinasin民+cos5tan民2cosa11.证明:sina左边二cosFsin民sin7tcoscos兀ai十5一cosasinsin11-4cosa+17一=6cos彳十万+sin卜万a।2—11、+cosacos.一a।4+万7ti兀a।sin彳十万cos兀a।—+-[cosasin7t—443.兀a।了十万一=右边a・•・原式得证.1-3tana+23-3tana+J33sin5+也853—,3tana3cosa—3sin巴与已知比较可设a=3t,b=J3t,tw0故2=3?选a3B.12、2.解析:cos10°+Msin10sin2sin40小一cos8030°cusnr3U°sin1(T.2sin240°2sin40=2.答案:,27…L兀Tt1-cos2x3.解析:(1)f(x)=cos2xcos--sin2xsin—H2131113=2cos2x--2-sin2x+2—2cos2x=2一上sin2x.一一兀所以,当2x=-y+2kTt,kez,即x=—1+kTt(
7、a+cosacos7t4=2啦sina2.55.4.AA为^ABC勺内角且sin2A=2sin3AcosA=--<0,4sinA>0,cosA<0,•1-sinA—cosA>0.又(sinA—cosA2=1—2sin7AcosA=1.477sin5.D依题意有sinacos3—cosasin=sin(a—3)=第3,7又0<3故cos(13有而cos7,,sin笨,是sin3=sin[a—(a—3)]=sinacos(a—3)—cosasin(x£」x双也14714选D.6.解析:…、1。原式=2tan(90—
8、2a),1-sin2a2cos2a1SI-1—2a一2(3-2a12sin2acos2a1cos2a2sin21sin2a2cos2a答案:7.解析:根据已知条件:cosacos3—sinasin3=sinacos3—cosssin3,cos3(cosa—sina)+sin3(cosa—sina)=0,即(cos3+sin3)(cosa—sina)=0.3为锐角,则sin3+cos3>0,一sina=0,•1-tana=1.答案:8.解析:由(1+#tana)(1+馅tan3)=4,可得tana+tan31—t
9、anatan3=*^3,即tan(a+3)=1/3.一一一,一、一兀又a+3C(0,兀),,a+3=w3答案:9.解析:原式=1+2sin0-cos0—1—2sin201+2sin0-cos0+2COS20—12sin20-cos0+2sin02sin01COS0+sin0~-2-=二;一■二7二~;=tan0.2sin0-cos0+2cos02cos01S..10+cos07答案:tan010.解析:.tana12=2,,tana2tan-21+sin2a-1+sin2a+cos2a1-tan2^2sin2
10、a+cos2a+2sinasin民+cos5tan民2cosa11.证明:sina左边二cosFsin民sin7tcoscos兀ai十5一cosasinsin11-4cosa+17一=6cos彳十万+sin卜万a।2—
11、+cosacos.一a।4+万7ti兀a।sin彳十万cos兀a।—+-[cosasin7t—443.兀a।了十万一=右边a・•・原式得证.1-3tana+23-3tana+J33sin5+也853—,3tana3cosa—3sin巴与已知比较可设a=3t,b=J3t,tw0故2=3?选a3B.
12、2.解析:cos10°+Msin10sin2sin40小一cos8030°cusnr3U°sin1(T.2sin240°2sin40=2.答案:,27…L兀Tt1-cos2x3.解析:(1)f(x)=cos2xcos--sin2xsin—H2131113=2cos2x--2-sin2x+2—2cos2x=2一上sin2x.一一兀所以,当2x=-y+2kTt,kez,即x=—1+kTt(
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