1、10.1.2两角和与差的正弦必备知识基础练1.化简sin 16cos 44+sin 74sin 44的值为()A.32B.-32C.12D.-12答案A解析sin 16cos 44+sin 74sin 44=sin 16cos 44+cos 16sin 44=sin(16+44)=sin 60=32,故选A.2.化简:sinx+3+sinx-3=()A.-sin xB.sin xC.-cos xD.cos x答案B解析sinx+3+sinx-3=12sin x+32cos x+12sin x-32cos x=sin x.3.若sin6-=cos6+,则tan =()A.-1B.0C.12D.1
2、答案A解析由已知得12cos -32sin =32cos -12sin ,因此1-32sin =3-12cos ,于是tan =-1.4.已知,32,sin =-14,32,2,cos =45,则+为()A.第一象限角B.第二象限角C.第三象限角D.第四象限角答案B解析由已知得cos =-154,sin =-35,+52,72.所以sin(+)=sin cos +cos sin =315-4200,故+为第二象限角.5.(2021天津和平高一期末)已知tan A=2tan B,sin(A+B)=14,则sin(A-B)=()A.13B.14C.112D.-112答案C解析由tan A=2tan
3、 B得sinAcosA=2sinBcosB,即sin Acos B=2cos Asin B.sin(A+B)=14,sin Acos B+cos Asin B=14.sin Acos B=16,cos Asin B=112.则sin(A-B)=sin Acos B-cos Asin B=16-112=112.故选C.6.已知a=(2sin 35,2cos 35),b=(cos 5,-sin 5),则ab=.答案1解析ab=2sin 35cos 5-2cos 35sin 5=2sin 30=1.7.化简:sin(-150)+cos(-120)cos=.答案-1解析原式=sincos150-cos
4、sin150+coscos120+sinsin120cos=-32sin-12cos-12cos+32sincos=-1.8.化简求值:(1)sin(+)cos(-)+cos(+)sin(-);(2)cos(70+)sin(170-)-sin(70+)cos(10+).解(1)原式=sin(+-)=sin 2.(2)原式=cos(70+)sin(10+)-sin(70+)cos(10+)=sin (10+)-(70+)=sin(-60)=-32.关键能力提升练9.(2021江苏苏州昆山校级月考)已知234,若cos(-)=1213,sin(+)=-35,则sin 2=()A.13B.-13C.
5、5665D.-1665答案D解析2sin ,cos -sin =22.sin-54=12.13.若cos =-13,sin =-33,2,32,2,则sin(+)的值为.答案539解析cos =-13,2,sin =1-cos2=223.sin =-33,32,2,cos =1-sin2=63.sin(+)=sin cos +cos sin =22363+-13-33=539.14.(2020辽宁沈阳和平期末)已知,为锐角,cos =17,cos(+)=-1114.(1)求sin(+)的值;(2)求sin 的值.解(1),为锐角,cos(+)=-1114,2+,sin(+)=1-cos2(+)
6、=5314.(2)为锐角,cos =17,sin =1-cos2=1-(17)2=437.sin =sin(+)-=cos sin(+)-sin cos(+)=175314-437-1114=32.学科素养创新练15.已知函数f(x)=sin2x+6+sin2x-6+cos 2x+a(aR,a为常数).(1)求函数f(x)的最小正周期及增区间;(2)当x0,2时,f(x)的最小值为-2,求a的值.解(1)f(x)=2sin 2xcos6+cos 2x+a=3sin 2x+cos 2x+a=2sin2x+6+a,f(x)的最小正周期T=22=.当2k-22x+62k+2,kZ,即k-3xk+6(kZ)时,函数f(x)是增函数,故该函数的增区间为k-3,k+6(kZ).(2)当x0,2时,2x+66,76,当x=2时,f(x)取得最小值.2sin22+6+a=-2,a=-1.5
