高考数学专项练习讲义-- 简单的三角恒等变换

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1C.ᖾcos2D./cos2zᫀD|᪆gN1+cos4=42COS22,cos2<0,ᡠo⌱⚗DT.43.4aR▲Ḅ+tanᐔ+2a=—7ᑣtana!y
)A.B.2C.-zD.-2zᫀD4|᪆ᵫtan7i+2a=,,-4tan2a=—y2tana4tan2a=1-tan%|otana=-1ᡈtana=2,aR▲+ᡠtana=■᪶⚪⚪-ᦪḄᓄ1
1
2021•ᐰᵬᔁa£0,9+tan2a=3Itana!yzᫀAsin2a2sinacosa|᪆gtan2a=cos2a~1—2sin2aa.c____COS(Zgrpi.2sinacosaitctan2a+ᡠ|osina=w.g2—sina1-2sin2a2—sinasinaV15ᡠcostana—cosa~152sina2tanacosa2sinacosa2sinQCOSQcosagtan2a—lsin%cos2a—sin2a1—2sin2actan2a=1-tan2a2—sinacos2a2sinacosacosaᡠ+1—2sin2a2—sina

2|osinot=.gQ£(0,¡,¢2sinaᡠcosa=4,tana==ic-"TLAr£U2COS4X_2cos2ஹ+-ফᓄ:"'7'2tanl7-xl-sinIJ%+4,zᫀ2COS2xcos2x—sin2x=2COS2rூᦟ¦ᜓ⌱௃1.(2020•ᐰI)4]£(0,ᐔ)+c3cos2஺-8cosa=5,ijsina!y()A*B.|C.|D.¨jjjyzᫀA|᪆ᵫ3cos2a—8cosa=5,o3(2cos2a—1)—8cosa=5,ᓽ3cos2a—4cosa—4=0,2ஹ|ocosa=-gᡈcosa=2(ª«).ga£(0,7i),ᡠsina>0,ᡠsina=y]1—cos1a=y\J1—1)2~3,(1+sinJ+cose)(sin?-cos+)2.40v*ᐔ+ij-------------/τஹ--------=________.^2+2cos0zᫀcose

3原式e-2s夕•c:o2一OScOSg0<ᐶᐔ+QTT0ᡠ0<᱇<2+ᡠcos/>0,ᡠ³=-cos9.I´ᓣµ(1)-ᦪḄᓄ⌕⍳·“-¸”³ᑣ:¸+¸¹+-¸ºP᪀ᱯ½.(2)-ᦪḄᓄ⌕¾¿ÀÁᩩÃXÄÅḄὶÇ(ஹஹஹÈஹÈ⊡!)+Ê˺-ᦪÄÅḄὶÇÌ.ÍÎÏÐ1(1Ñ1+sin4+t2+2cos4!y()A.2cos2B.2sin2C.4sin2+2cos2D.2sin2+4cos2zᫀB|᪆2ஹ1+siஹ4+12+2cos4=2^/sin22+2sin2cos2+cos22+^2+2(2cos22-1)=2^(sin2+cos2)2+^4cos22=2|sin2+cos2|+2|cos2|.V1<20,+³=2(sin2+cos2)—2cos2=2sin2.

4C.D.2ᫀBtan27.5°+l᪆=tan27.5°—8sin27.5°+l__________$-27.5஺+8527.5஺________sin27.5°—8sin27.5°cos27.5°+cos27.5°_1_1_2S-l-2sin215o-cos300-3-⚪#$ᦪḄ&'(⚪)1*&'+2(l)sin4(T(tan10,-)/()A.2B.-2C.1D.-1ᫀD᪆sin40°-(tan10°-^3)=sin40஺(123-)sin100—-cos10°=sin40°-^lO32&in10஺3ᙶcos100=sin40°-------------------------------'2(cos60°sin10°-sin60°cos10°)=sin400--cos10°2sin(10°-60°)=sin40°-cos100-2sin50°=sin400-cos10஺-2sin40%cos40°cos10°3sin80஺cos10°(2)cos200-cos40°-cos100஺=.79—-8:8᪆cos20°-cos40°-cos100°=-cos20°-cos40°cos80°sin20°-cos20°-cos40°-cos80°sin20°

5pin40°-cos40°-cos80°sin200|sin80°-cos80°~sin20°|sin160°=-sin20°|sin20°]sin20°Ó⚪Ì2Ôff3পcosea)=g,ᑣcosÕ+2,!y()77c--9D.-9等案c解析=sin(W+a)=7.3'’■2ᐔy+2aj=1-2sin2+acos3_27=-9-9,(2)(2022Ù᧡jÛ)4sin^a—!OSᑣ)A.|B.,C._7gD.-9zᫀDsin(V|᪆:osᐔ.ᐔ_LÜ1•'•sinacosg-cosasm1ᓝÞ3cos4sina--cosa+y[3cosa=^,/.|sin13'

6_7=ßÓ⚪Ì3Ôf2s3s44ã+äᙳg┦+cosa=-'Y~,sinQ=^Cᑣcos2a=,2a—fi=ᫀ11|᪆gcosa=ì+ᡠcos2a=2cos2aí1=z.ga,4ᙳg┦+sin/?=ñ+ᡠSini=^^,COS4=||,4;õsin2a=2sinacosa=~y~,ᡠsin(2a—/?)=sin2acosfi-cos2«sinä=^^\÷ᙶ.gag┦+ᡠ0<2"ᐔcos2a>0,ᡠ0<2a<^,it7T£g┦+ᡠ]<2a£<2+Asin(2a-£)=ý+ᡠ2஺4=?ூᦟ¦ᜓ⌱௃1.--°஺40஺Ḅfgcos25°^/1-Sin40°A.1B.V§C.y[2D.2ᫀC

7cos220O-sin220°᪆=cos25°
cos20°-sin20°cos200+sin20°cos25°V2cos25°r-=cos25°=722.A,3ᙳsin4+cosA+="~*^,sinB=¥j&ᑣA+B"#/\11U3ᐔc5ᐔA.TB%ᫀC᪆&sin4+cosA+g="■(3,ᡠ*ᓃ,+.s4—0inA=12,ᓽ5ᙶsinA=±¥,78sinA=9&A,ᡠ*cos2^55.ᵫsin3=;<B,cosB=-^^1—sin2B=—3V1510-ᡠ*cos
A+8=cosAcosB-sinAsin5=-x-.=A,3>ᓽA,8£?ᐔ@ᡠ*A+B£7i,2TI,ᡠ*A+3=A.

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978sin[=ªH.2@2021•ᐰ¥¬ᔁ@cos°Hcos/"#@A.|BᙶC.ᙶDᙶᫀD7᪆&8$´=$µ¶H·@=sin^,ᓃr[ஹ/7ᐔ஺5ᐔ@ᐔ.஺ᐔ»*cos|2^—cos*^2—cos^-sin^2=COS2X¿@=cosÀ=ᙶবsin(x+¶)=g,ᑣsin2x=,ᫀTzஹ1—cos(2x+5j7᪆Vsin2(x+1)=------——-l+sin2x12=ysin2x=%y⚪Ç\\ឤ"tᣚḄÊᔠÌᵨÎ52022•ÏᓭÑ᪥ὶὃ@]ᦪÔx@=4cosxcosx+"HÕ.পo./U@ḄᓫÙ⌴ÛÜÝফß0,,7஺@=,,ocos2a7(1)«x)=4cosxcos^v+^jHÕ=4cosA^^COSx-ÝsinJHÕ=2*\/3COS2X_2sinxcosx-y[3=^3(1+cos2x)-sin2x~~=V§cos2x-sin2x=2cos(2x+§,jrç2E—7tW2x+%W2E(%eZ),

1077r7E78E—y^WxWEHï(᝕£Z),E—ò(ZWZ).一兀O一2一6兀6而—\-一6--/(«/2COSI2a+-5所以兀Q+一6因为兀Oa4-2兀兀一一776T66则兀兀兀-W一W一62a+-62兀一6一/-兀兀.ICOSI2a一6一Jk+■6一一兀兀K兀+一---66663正4-X5223Õ+4=10-ூᦟ÷ᜓ⌱௃û]ᦪüý,sin£—x@+þcos£—X@.পo]ᦪÿᙠᵨḄᨬফcos6=,,஺2ᐔ/20+1Ḅ.1ᵫ⚪#$

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sin20—cosIffEFᓣH1I?JKឤ*Mᣚ⌕ᢕQRMKஹM7ᦪTUஹMV᪀XᐸZK[Ḅᐵ]^#_`Ḅ⌮ᵨbMcdᵨ.ফc᝞y=asinx+Z?cosxᓄ&y=y
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2022♦¾¿ÀÁ~6»e0,T,tan6»=ᖾᑣcos20*¸A.-ᙶBᙶ13»ᫀCcos%—sin2஺1—tar^e᪆26>=cos26>-sin26>='cos2^+sin2<91+tan2஺3,3.
2022•ÄÅÀÁtan67.5°—ÆᯅÈḄ&A.1BV2C.2D.4»ᫀCÉᱏA7co____1_____sin67.50_1_____sin67.5°cos67.5°᪆tan67.5Zan67.5°=cos67.5°-¬67.5°=cos67.5°-sin67.5°cos67.5°sin267.5°—COS267.5°-cos135°=sin67.5°cos67.5°^7=24.
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30஺-a—sinᑣsin
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30°—a—sinᐜ11.1Ï2cosa-]sin

14ᓽcos(30°+ct)=^,ᡠ(sin(30°-2a)=cos(60°+2a)=2COS2(30°+a)-l=2x1-l7=-5.(Ð⌱)~ᓻ)=/l+cos2r)sin2x(xeR),ᑣÓᑡVÕÖ×ḄZ()A.<)Ḅᨬ<ÖᕜlT=jB.y(x)ZᏔ7ᦪc.y(x)Ḅᨬᜧ&1D.y(x)Ḅᨬ<ÖᕜlT=n»ᫀABC᪆*'/(x)=4(l+cos2x)(1—cos2x)=^(1—COS22X)=^sin22x=^(1-cos4x),2.f(-x)=^[\—cos4(—x)]t=g(l—cos4x)=y(x),7=Ù=14T#x)Ḅᨬᜧ&X2=,ᦑA,B,CÖ×D┯Ü.6.(Ð⌱)Óᑡᔜ`{&gḄZ()A2ᐔ•2ᐔA.cosp-sintan22.5°Bl-tan222.5°C.2sin195°cos195°

15ᦑA┯Ütan22.5°12tan22.5°1—tan222.5°21—tan222.5=^tan45஺=3ᦑBÖ×2sin195°cos195°=2sin(180°+15°)cos(180°+15°)=2sin15°cos15°=sin30஺=1,ᦑCÖ×7fᢣ<-tan12஺_ឤR(2cos212°-l)sin12°~»ᫀ8᪆ß`=cos24%à12஺

16=2cos2^-197,25125,áâãVcos^-=^
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