考點規(guī)范練19兩角和與差的正弦、余弦與正切公式基礎(chǔ)鞏固組1.計算cos 42cos 18-cos 48sin 18的結(jié)果等于()A.12B.33C.22D.32答案A解析原式=sin48cos18-cos48sin18=sin(48-18)=sin30=12.2.已知sin =55,則sin4-cos4的值為()A.-15B.-35C.15D.35答案B解析因為sin=55,所以sin4-cos4=(sin2-cos2)(sin2+cos2)=sin2-cos2=-cos2=2sin2-1=-35.3.(2018全國1)已知角的頂點為坐標(biāo)原點,始邊與x軸的非負(fù)半軸重合,終邊上有兩點A(1,a),B(2,b),且cos 2=23,則|a-b|=()A.15B.55C.255D.1答案B解析cos2=2cos2-1=23,cos2=56,sin2=16,tan2=15,即tan=55.由于a,b的正負(fù)性相同,不妨設(shè)tan>0,即tan=55,由三角函數(shù)定義得a=55,b=255,故|a-b|=55.故選B.4.已知為銳角,且7sin =2cos 2,則sin+3=()A.1+358B.1+538C.1-358D.1-538答案A解析由7sin=2cos2得7sin=2(1-2sin2),即4sin2+7sin-2=0,sin=-2(舍去)或sin=14.為銳角,cos=154,sin+3=1412+15432=1+358,故選A.5.已知0<<2,cos+6=-45,則sin(-)=()A.33+410B.-33+410C.-33-410D.33-410答案B解析0<<2,6<+6<23,又cos+6=-45,sin+6=35.sin(-)=-sin=-sin+6-6=-sin(+6)cos6-cos(+6)sin6=-3532-4512=-33+410.故選B.6.(2017課標(biāo)高考)函數(shù)f(x)=2cos x+sin x的最大值為.答案5解析f(x)22+1=5.7.(2018全國2)已知sin +cos =1,cos +sin =0,則sin(+)=.答案-12解析由題意sin+cos=1,cos+sin=0,得(sin+cos)2+(cos+sin)2=sin2+cos2+2sincos+cos2+sin2+2cossin=2+2sin(+).則2+2sin(+)=1,即sin(+)=-12.8.若sin(+x)+cos(+x)=12,則sin 2x=,1+tanxsinxcosx-4=.答案-34-823解析sin(+x)+cos(+x)=-sinx-cosx=12,即sinx+cosx=-12,兩邊平方得sin2x+2sinxcosx+cos2x=14,即1+sin2x=14,則sin2x=-34,由1+tanxsinxcosx-4=1+sinxcosx22sinx(cosx+sinx)=2sinxcosx=22sin2x=22-34=-823,故答案為-34,-823.能力提升組9.若sin 2=55,sin(-)=1010,且4,32,則+的值是()A.74B.94C.54或74D.54或94答案A解析因為4,故22,2,又sin2=55,故22,4,2,cos2=-255.,32,故-2,54,于是cos(-)=-31010,cos(+)=cos2+(-)=cos2cos(-)-sin2sin(-)=-255-31010-551010=22,且+54,2,故+=74.10.將函數(shù)f(x)=sin32+x(cos x-2sin x)+sin2x的圖象向左平移8個單位長度后得到函數(shù)g(x)的圖象,則g(x)具有性質(zhì)()A.在0,4上單調(diào)遞增,為奇函數(shù)B.周期為,圖象關(guān)于4,0對稱C.最大值為2,圖象關(guān)于直線x=2對稱D.在-2,0上單調(diào)遞增,為偶函數(shù)答案A解析f(x)=sin32+x(cosx-2sinx)+sin2x=-cos2x+sin2x+sin2x=sin2x-cos2x=2sin2x-4,g(x)=2sin2x+8-4=2sin2x,g(x)為奇函數(shù),且在0,4上是增函數(shù).故選A.11.設(shè)f(x)=1+cos2x+sin2x2sin2+x+asin (x+4)的最大值為3,則常數(shù)a=()A.1B.1或-5C.-2或4D.7答案B解析f(x)=2cos2x+2sinxcosx2cosx+asinx+4=2cosx+2sinx+asinx+4=2sinx+4+asinx+4=(a+2)sinx+4,則|a+2|=3,a=1或a=-5.故選B.12.已知函數(shù)f(x)=asin x+bcos x(a0)在x=4處取得最小值,則函數(shù)f34-x是()A.偶函數(shù)且它的圖象關(guān)于點(,0)對稱B.偶函數(shù)且它的圖象關(guān)于點32,0對稱C.奇函數(shù)且它的圖象關(guān)于點(,0)對稱D.奇函數(shù)且它的圖象關(guān)于點32,0對稱答案C解析函數(shù)f(x)=asinx+bcosx=a2+b2sin(x+)(a0)的周期為2,在x=4處取得最小值,故有22(a+b)=-a2+b2,即有b=a,f(x)=2asinx+4.則f34-x=2asin(-x)=2asinx.則函數(shù)y=f34-x為奇函數(shù),對稱中心為(k,0),kZ,故選C.13.已知函數(shù)f(x)=cos2x2+32sin x-12(>0,xR),若函數(shù)f(x)在區(qū)間(,2)內(nèi)沒有零點,則的取值范圍是()A.0,512B.0,51256,1112C.0,56D.0,51256,1112答案D解析f(x)=122cos2x2-1+32sinx=12cosx+32sinx=sinx+6,當(dāng)x(,2)時,x+6+6,2+6,依題意,+6k2+6(k+1)k-16k2+512,kZ,由k2+512>k-16,可得k<76,k=0時,0,512,當(dāng)k=1時,56,1112,所以的取值范圍是0,51256,1112,故選D.14.(2018浙江紹興5月模擬)已知函數(shù)f(x)=cos2x-sin2x+6,則f6=,該函數(shù)的最小正周期為.答案0解析由題意可得,f(x)=1+cos2x2-1-cos(2x+3)2=12cos2x+12cos2xcos3-sin2xsin3=-1232sin2x-32cos2x=-32sin2x-3.則f6=-32sin26-3=0,函數(shù)的最小正周期為T=22=.15.已知,0,2,且sinsin=cos(+),(1)若=6,則tan =;(2)tan 的最大值為.答案(1)35(2)24解析由sinsin=cos(+),化簡可得:sin(1+sin2)=12sin2cos,則tan=12sin21+sin2.(1)若=6,則tan=12sin31+122=32121+14=35.(2)tan=12sin21+sin2=sin23-cos2=-sin23-cos2,看成是圓心為(0,0),半徑r=1的圓上的點與點(3,0)的連線的斜率問題,直線過(3,0),設(shè)方程為y=k(x-3),d=r=1,即1=|3k|k2+1,解得k=24.tan的最大值為24.故答案為:35,24.16.(2018浙江慈溪中學(xué)模擬)若sin +3cos =255,-3,6,tan+3=4,則tan(-)=.答案-76解析由題意可得,sin+3cos=2sin+3=255,sin+3=55,-3,6,tan+3=12,tan(-)=tan(+3)-(+3)=12-41+2=-76.17.(2018浙江金華十校調(diào)研)已知函數(shù)f(x)=sin2x-6+2cos2x-1.(1)求函數(shù)f(x)的最大值及其相應(yīng)x的取值集合;(2)若4<<2且f()=45,求cos 2的值.解(1)f(x)=sin2x-6+2cos2x-1=sin2xcos6-cos2xsin6+cos2x=32sin2x+12cos2x=sin2x+6.即當(dāng)2x+6=2k+2(kZ),解得當(dāng)x=k+6(kZ)時,f(x)max=1.其相應(yīng)x的取值集合為x|x=k+6,kZ.(2)由題意,f()=sin2+6=45.由4<<2,得23<2+6<76,根據(jù)同角三角函數(shù)基本關(guān)系可知cos2+6=-35.因此cos2=cos(2+6)-6=cos2+6cos6+sin2+6sin6=-3532+4512=-33+410.18.設(shè)函數(shù)f(x)=sin2x-cos2x+23sin xcos x+的圖象關(guān)于直線x=對稱,其中,為常數(shù),且12,1.(1)求函數(shù)f(x)的最小正周期;(2)若y=f(x)的圖象經(jīng)過點4,0,求函數(shù)f(x)在區(qū)間0,35上的取值范圍.解(1)f(x)=sin2x+23sinxcosx-cos2x+=3sin2x-cos2x+=2sin2x-6+,圖象關(guān)于直線x=對稱,2-6=2+k,kZ.=k2+13,又12,1,令k=1時,=56符合要求,函數(shù)f(x)的最小正周期為2256=65;(2)f4=0,2sin2564-6+=0,=-2,f(x)=2sin53x-6-2,f(x)-1-2,2-2.