1、考点规范练18同角三角函数的基本关系及诱导公式考点规范练B册第11页基础巩固组1.若cos(3-x)-3cosx+2=0,则tan x等于() A.-12B.-2C.12D.13答案:D解析:cos(3-x)-3cosx+2=0,-cos x+3sin x=0,tan x=13,故选D.2.sin296+cos-293-tan254=()A.0B.12C.1D.-12答案:A解析:原式=sin4+56+cos-10+3-tan6+4=sin56+cos3-tan4=12+12-1=0.3.已知f()=sin(-)cos(2-)cos(-)tan,则f-313的值为()A.12B.-13C.-1
2、2D.13答案:C解析:f()=sincos-costan=-cos ,f-313=-cos-313=-cos10+3=-cos3=-12.4.(2015福建泉州期末)若tan =2,则2sin2+1sin2的值为()A.53B.-134C.135D.134答案:D解析:因为2sin2+1sin2=3sin2+cos22sincos=3tan2+12tan=322+122=134.故选D.5.(2015成都外国语学校月考)已知tan(-)=34,且2,32,则sin+2=()A.45B.-45C.35D.-35导学号92950779答案:B解析:由tan(-)=34,得tan =34,即sin
3、 =34cos ,又因为2,32,所以为第三象限的角,将sin =34cos ,代入sin2+cos2=1,得cos =-45,所以sin+2=cos =-45.6.(2015河北衡水模拟)已知cos512+=13,且-2,则cos12-等于()A.223B.-13C.13D.-223答案:D解析:cos512+=sin12-=13,又-2,71212-1312,cos12-=-1-sin212-=-223.7.若sin 是5x2-7x-6=0的根,则sin-32sin32-tan2(2-)cos2-cos2+sin(+)=()A.35B.53C.45D.54答案:B解析:由5x2-7x-6=
4、0,解得x=-35或2.则sin =-35.原式=cos(-cos)tan2sin(-sin)(-sin)=1-sin=53.8.已知sin+3cos3cos-sin=5,则sin2-sin cos 的值是()A.25B.-25C.-2D.2答案:A解析:由sin+3cos3cos-sin=5,得tan+33-tan=5,解得tan =2.则sin2-sin cos =sin2-sincossin2+cos2=tan2-tantan2+1=25.9.(2015浙江绍兴二模)若f(cos x)=cos 2x,则f(sin 15)=.答案:-32解析:f(sin 15)=f(cos 75)=cos 150=cos(180-30)=-cos 30=-32.10.已知sin cos =18,且42,则cos -sin 的值为.答案:-32解析:当4cos ,cos -sin log2x的解集为.答案:(0,2)解析:由f(31)=asin531+btan531=asin5+btan5=f(1)=1,则f(31)log2x,即1log2x,解得0x0,cos 0,所以cos 1|cos|+sin 1|sin|=-1+1=0,即原式等于0.
