四川省成都市新都一中数学必修4同步检测：第一章 第6课时 诱导公式 WORD版含答案.docx

1、第6课时诱导公式(二)基础达标(水平一)1.若cos+2=23,则sin 等于().A.23 B.-23 C.53 D.-53【解析】cos+2=-sin =23,sin =-23.【答案】B2.已知sin+4=13,则cos4-的值为().A.223B.-223C.13D.-13【解析】cos4-=cos2-4+=sin+4=13.【答案】C3.已知角A,B,C为ABC的内角,则下列表达式为常数的是().A.sin(A+B)+sin CB.cos(B+C)-cos AC.sinA+B2cosC2D.cosB+C2cosA2【解析】A+B+C=,A+B2=2-C2,sinA+B2=sin2-C

2、2=cosC2,sinA+B2cosC2=cosC2cosC2=1,故选C.【答案】C4.已知sin2+cos2-=15,(0,),则1tan的值为().A.43B.34C.-43D.-34【解析】sin2+cos2-=cos +sin =15,(cos +sin )2=1+2sin cos =125,sin cos =-1225,又(0,),故sin =45,cos =-35,因此1tan=-34.【答案】D5.化简sin(-1071)cos 9+sin(-171)sin(-261)的结果为.【解析】原式=sin(9-1080)cos 9+sin(9-180)sin(99-360)=sin

3、9cos 9-sin 9sin 99=sin 9cos 9-sin 9sin(90+9)=sin 9cos 9-sin 9cos 9=0.【答案】06.已知tan(3+)=2,则sin(-3)+cos(-)+sin2-2cos2+-sin(-)+cos(+)=.【解析】由tan(3+)=2,得tan =2.原式=sin(-)-cos+cos+2sinsin-cos=-sin+2sinsin-cos=sinsin-cos=tantan-1=22-1=2.【答案】27.求证: sin(540-x)tan(900-x)1tan(450-x)tan(810-x)cos(360-x)sin(-x)=si

4、n x.【解析】左边=sin(180-x)tan(-x)1tan(90-x)tan(90-x)cosxsin(-x)=sinx-tanxcos(90-x)sin(90-x)cos(90-x)sin(90-x)-1tanx=-cos xsinxcosxsinxcosx-1tanx=sin x=右边,故原等式得证.拓展提升(水平二)8.若sin 是2x2-5x+2=0的一个根,则sin-32sin32-tan2(2-)cos2-cos2+sin(+)的值为().A.2B.-2C.12D.-12【解析】由2x2-5x+2=0,得x=12或x=2.所以sin =12.原式=cos(-cos)tan2s

5、in(-sin)(-sin)=-1sin=-2.【答案】B9.A,B,C为ABC的三个内角,下列关系式中不成立的是().cos(A+B)=cos C;cosB+C2=sinA2;tan(A+B)=-tan C;sin(2A+B+C)=sin A.A.B.C.D.【解析】cos(A+B)=cos(-C)=-cos C,错,排除B、D;cosB+C2=cos-A2=cos2-A2=sinA2,正确,排除A,选C.【答案】C10.若2,32,tan(-5)=-34,则sin2+cos2-的值为.【解析】tan(-5)=tan =-34,即sincos=-34.又sin2+cos2=1,2,32,且tan =-34,cos =-45,sin =35,sin2+cos2-=cos +sin =-15.【答案】-1511.化简:sin4k-14-+cos4k+14-(kZ).【解析】当k为奇数时,原式=sin-4-+cos+4-=sin4+-cos4-=sin2-4-cos4-=cos4-cos4-=0;当k为偶数时,原式=sin-4-+cos4-=-sin4+cos4-=-sin2-4-+cos4-=-cos4-+cos4-=0.综上,原式=0.