1、诱导公式第一课时 练习1.tan 690=().A.-33 B.33 C.12 D.-122.已知角的顶点在坐标原点,始边与x轴的非负半轴重合,其终边过点(3,-4),则cos(-)=().A.35 B.-35 C.45 D.-453.已知cos(-)=-513,且是第四象限角,则sin(-2+)等于().A.-1213 B.1213 C.1213 D.5124.设tan(5+)=m,则sin(-3)+cos(-)sin(-)-cos(+)的值为().A.m+1m-1 B.m-1m+1 C.-1 D.15.(多选题)已知tan7+=5,则下列结论正确的是().A.tan67-=5 B.tan6
2、7-=-5C.tan87+=5 D.tan87+=-56.已知cos(-75)=-13,且角为第四象限角,则sin(105+)=.7.若cos =23,是第四象限角,求sin(-2)+sin(-3)cos(-3)cos(-)-cos(-)cos(-4)的值.8.(多选题)已知A=sin(k+)sin+cos(k+)cos+tan(k+)tan(kZ),则A的值可以是().A.3 B.-3 C.1 D.-19.当(0,)时,若cos56-=-513,则sin+6的值为().A.1213 B.-1213 C.1213 D.51210.求值:cos7+cos27+cos37+cos47+cos57+
3、cos67=.11.在点P(1,3)在角的终边上;sin =3cos ;sin+cossin-cos=2这三个条件中任选一个,完成下列问题.已知.(1)计算2cos(-)-3sin(+)4cos(-)+sin(2-)的值;(2)计算4sin cos +2cos2的值.12.已知f(n)=sinn3,求f(1)+f(2)+f(3)+f(2020)+f(2021).参考答案1.A2.B3.A4.A5.BC6.2237.【解析】由题意得sin =-53,故sin(-2)+sin(-3)cos(-3)cos(-)-cos(-)cos(-4)=sin-sincos-cos+cos2=52.8.AD9.A
4、10.011.【解析】若选择,则tan =yx=3.若选择,则tan =sincos=3.若选择,因为sin+cossin-cos=2,所以sin +cos =2(sin -cos ),整理可得sin =3cos ,即tan =sincos=3.(1)原式=-2cos+3sin4cos-sin=-2+3tan4-tan=7.(2)原式=4sincos+2cos2sin2+cos2=4tan+2tan2+1=75. 12.【解析】将单位圆平均分为6份,则依次可得到3,23,33,43,53,63各角的终边.通过观察发现:sin3与sin53互为相反数,sin23与sin43互为相反数,且sin33=sin63=0,于是f(1)+f(2)+f(3)+f(6)=0.当n=7,8,9,10,11,12时,角的终边与上述6个角的终边重合,故f(7)+f(8)+f(12)=0,同理f(6n+1)+f(6n+2)+f(6n+6)=0,nN+,而2021=3376-1,所以f(1)+f(2)+f(3)+f(2020)+f(2021)=3370-f(6)=-f(6)=0.
