2019版高中人版B版数学必修四练习：1-2-4 诱导公式 WORD版含答案.docx

1、1.2.4诱导公式课后篇巩固探究一、A组基础巩固1.sin-3+2sin43+3sin23等于()A.1B.12C.0D.-1答案:C2.函数f(x)=cosx3(xZ)的值域为()A.-1,-12,0,12,1B.-1,-12,12,1C.-1,-32,0,32,1D.-1,-32,32,1答案:B3.已知a=tan-76,b=cos234,c=sin-334,则a,b,c的大小关系是()A.bacB.abcC.bcaD.acb解析:因为a=-33,b=22,c=-22,所以bac.答案:A4.化简1+2sin(-2)cos(-2)的结果是()A.sin 2-cos 2B.cos 2-sin

2、 2C.(cos 2-sin 2)D.无法确定解析:原式=|sin(-2)+cos(-2)|=|sin2-cos2|=sin2-cos2.答案:A5.已知sin(-)=log814,且-2,0,则tan(-)的值为()A.-255B.255C.255D.52解析:由已知得sin(-)=sin=log814=-23.-2,0,cos=1-sin2=53,tan=-255,tan(-)=-tan=255.答案:B6.设tan(5+)=m,则sin(-3)+cos(-)sin(-)-cos(+)的值为.解析:由题意知tan=m,原式=-sin-cos-sin+cos=-tan-1-tan+1=m+1

3、m-1.答案:m+1m-17.已知sin3-=12,则cos6+=.解析:3-+6+=2,cos6+=sin3-=12.答案:128.化简:(1)1+cos2+sin2-tan(+);(2)sin(2-)cos(+)cos2+cos112-cos(-)sin(3-)sin(-)sin92+.解:(1)原式=1+(-sin)costan=1-sin2=cos2.(2)原式=(-sin)(-cos)(-sin)cos5+2-(-cos)sin(-)-sin(+)sin4+2+=-sin2cos-cos2-(-cos)sin-(-sin)sin2+=sin2cossin-cossin2cos=-si

4、ncos=-tan.9.已知sin(5+)=lg1310,求cos(2+)的值.解:sin(5+)=sin(+)=-sin,lg1310=lg10-13=-13,sin =13,cos(2+)=cos =1-sin2=1-132=223.二、B组能力提升1.已知函数f(x)=cosx2,则下列等式成立的是()A.f(2-x)=f(x)B.f(2+x)=f(x)C.f(-x)=f(x)D.f(-x)=-f(x)解析:f(-x)=cos-x2=cosx2=f(x).答案:C2.已知f(cos x)=sin x,设x是第一象限角,则f(sin x)为()A.sec xB.cos xC.sin xD.

5、1-sin x解析:f(sinx)=fcos2-x=sin2-x=cosx.答案:B3.已知为锐角,2tan(-)-3cos2+5=0,tan(+)+6sin(+)-1=0,则sin 的值是()A.355B.377C.31010D.13解析:由已知可知-2tan+3sin+5=0,tan-6sin-1=0.所以tan=3.又tan=sincos,所以9=sin2cos2=sin21-sin2.所以sin2=910.因为为锐角,所以sin=31010.答案:C4.已知sin(2+)tan(+)tan(3-)cos2-tan(-)=1,则3sin2+3sincos+2cos2的值是()A.1B.2

6、C.3D.6解析:因为原式=sintantan(-)-sintan=tan=1,所以3sin2+3cos2sin2+3sincos+2cos2=3tan2+3tan2+3tan+2=1.答案:A5.sin21+sin22+sin23+sin288+sin289+sin290的值等于()A.45B.4512C.45+22D.90+22答案:B6.tan 1tan 2tan 3tan 89=.解析:因为tan1tan89=1,所以tan1tan2tan89=11144个tan 45=1.答案:17.已知f(x)=sinx,x0,则f-116+f116=.解析:f-116=sin-116=sin6=

7、12,f116=f56-1=f-16-2=sin-6-2=-52,所以f-116+f116=12-52=-2.答案:-28.导学号73764015已知是第三象限的角,且f()=sin(-)cos(2-)tan-+32cot(-)sin(-).(1)化简f();(2)若cos-32=15,求f()的值.解:(1)f()=sin(-)cos(2-)tan32-cot(-)sin(-)=sincoscot-cotsin=-cos.(2)cos-32=-sin,sin=-15.又是第三象限的角,cos=-1-sin2=-256,f()=256.9.导学号73764016已知tan ,1tan是关于x的方程3x2-3kx+3k2-13=0的两实根,且30.因为30,sin0,cos0,故取k=433.于是tan+1tan=sincos+cossin=1sincos=433,即sincos=34.所以(sin+cos)2=1+2sincos=2+32.因为sin+cos0,所以sin+cos=-3+12.于是cos(3+)+sin(+)=cos(+)+sin(+)=-(cos+sin)=3+12.