高中數(shù)學(xué)必修4配人教A版-課后習(xí)題Word版第三章　3.1.2　兩角和與差的正弦、余弦、正切公式

1、第三章三角恒等變換3.1兩角和與差的正弦、余弦和正切公式3.1.2兩角和與差的正弦、余弦、正切公式課后篇鞏固探究基礎(chǔ)鞏固1.已知a=(2sin 35,2cos 35),b=(cos 5,-sin 5),則ab=()A.12B.1C.2D.2sin 40解析ab=2sin 35cos 5-2cos 35sin 5=2sin(35-5)=2sin 30=1.答案B2.若sin6-=cos6+,則tan =()A.-1B.0C.12D.1解析由已知得12cos -32sin =32cos -12sin ,因此1-32sin =3-12cos ,于是tan =-1.答案A3.若tan(+)=25,ta

2、n(-)=14,則tan 2=()A.16B.2213C.322D.1318解析tan 2=tan (+)+(-)=tan(+)+tan(-)1-tan(+)tan(-)=25+141-2514=1318.答案D4.sin(+75)+cos(+45)-3cos(+15)的值等于()A.1B.1C.-1D.0解析原式=sin (+45)+30+cos(+45)-3cos (+45)-30=32sin(+45)+12cos(+45)+cos(+45)-332cos(+45)+12sin(+45)=32sin(+45)+32cos(+45)-32cos(+45)-32sin(+45)=0.答案D5.

3、設(shè)0,2,0,2,且tan =1+sincos,則()A.3-=2B.3+=2C.2-=2D.2+=2解析由tan =1+sincos,得sincos=1+sincos,得sin cos -cos sin =cos ,sin(-)=sin2-.又0,2,0,2,故-=2-,即2-=2.答案C6.化簡(jiǎn):sin(-150)+cos(-120)cos=.解析原式=sincos150-cossin150+coscos120+sinsin120cos=-32sin-12cos-12cos+32sincos=-1.答案-17.已知銳角,滿(mǎn)足(tan -1)(tan -1)=2,則+的值為.解析因?yàn)?tan

4、 -1)(tan -1)=2,所以tan +tan =tan tan -1.因此tan(+)=tan+tan1-tantan=-1,因?yàn)?(0,),所以+=34.答案348.已知0,2,tan =2,則cos-4=.解析由tan =2,得sin =2cos .又sin2+cos2=1,0,2,cos =55,sin =255.cos-4=cos cos4+sin sin4=5522+25522=31010.答案310109.tan 23+tan 37+3tan 23tan 37的值是.解析tan 60=3=tan23+tan371-tan23tan37,tan 23+tan 37=33tan

5、23tan 37,tan 23+tan 37+3tan 23tan 37=3.答案310.化簡(jiǎn)求值:(1)sin(+)cos(-)+cos(+)sin(-);(2)cos(70+)sin(170-)-sin(70+)cos(10+);(3)cos 21cos 24+sin 159sin 204.解(1)原式=sin(+-)=sin 2.(2)原式=cos(70+)sin(10+)-sin(70+)cos(10+)=sin (10+)-(70+)=sin(-60)=-32.(3)原式=cos 21cos 24+sin(180-21)sin(180+24)=cos 21cos 24-sin 21s

6、in 24=cos(21+24)=cos 45=22.11.已知cos =-55,tan =13,32,02,求-的值.解(方法一)由cos =-55,32,得sin =-255,tan =2,又tan =13,于是tan(-)=tan-tan1+tantan=2-131+213=1.又由32,02,可得-2-0,2-32,因此-=54.(方法二)由cos =-55,32,得sin =-255.由tan =13,02,得sin =110,cos =310.所以sin(-)=sin cos -cos sin =-255310-55110=-22.又由32,02,可得-2-0,2-32,因此,-=

7、54.能力提升1.已知-2,32,tan-4=-3,則sin =()A.55B.-55C.255D.55解析tan =tan -4+4=tan-4+tan 41-tan-4tan 4=-12,因?yàn)?,32,所以2,故sin =15=55.答案A2.設(shè),都為銳角,且cos =55,sin(+)=35,則sin 等于()A.2525B.11525C.55D.-55或11525解析為銳角,cos =55,sin =255.,都為銳角,0+.sin(+)=35,cos(+)=45.當(dāng)cos(+)=-45時(shí),sin =sin(+)-=sin(+)cos -cos(+)sin =3555+45255=11

8、525;當(dāng)cos(+)=45時(shí),sin =sin(+)-=sin(+)cos -cos(+)sin =355545255=-55,與已知為銳角矛盾.sin =11525.答案B3.若將函數(shù)f(x)=sin 2x+cos 2x的圖象向右平移個(gè)單位長(zhǎng)度,所得圖象關(guān)于y軸對(duì)稱(chēng),則的最小正值是()A.8B.4C.38D.34解析由題意f(x)=sin 2x+cos 2x=2sin2x+4,將其圖象向右平移個(gè)單位長(zhǎng)度,得函數(shù)y=2sin2(x-)+4=2sin2x-2+4的圖象,要使圖象關(guān)于y軸對(duì)稱(chēng),則4-2=2+k,解得=-8k2,kZ.當(dāng)k=-1時(shí),取最小正值38.答案C4.已知cos(+)=45,

9、cos(-)=-45,則cos cos =.解析由已知得cos cos -sin sin =45,cos cos +sin sin =-45,兩式相加得2cos cos =0,故cos cos =0.答案05.已知ABC中,3tan Atan B-tan A-tan B=3,則C的大小為.解析依題意,tanA+tanB1-tanAtanB=-3,即tan(A+B)=-3,又0A+B,所以A+B=23,故C=-A-B=3.答案36.已知,均為銳角,且tan =cos-sincos+sin,求tan(+)的值.解tan =cos-sincos+sin=1-tan1+tan=tan4-,因?yàn)?均為銳

10、角,所以-44-4,02,又y=tan x在-2,2上是單調(diào)函數(shù),所以=4-,即+=4,tan(+)=1.7.已知向量a=(cos ,sin ),b=(cos ,sin ),|a-b|=255.(1)求cos(-)的值;(2)若-202,且sin =-513,求sin 的值.解(1)a=(cos ,sin ),b=(cos ,sin ),|a|=|b|=1,|a-b|2=a2-2ab+b2=1+1-2(cos cos +sin sin )=2-2cos(-).又|a-b|=255,|a-b|2=2-2cos(-)=45,cos(-)=35.(2)-202,0-0)有最大值1和最小值-4,求a,b的值.解f(x)=22(cos x-sin x)sin4+x-2asin x+b=12(cos2x-sin2x)-2asin x+b=12(1-2sin2x)-2as