高中数学 第三章 三角恒等变换 3.1.3 二倍角的正弦、余弦、正切公式课后习题 新人教A版必修4.doc

3.1.3二倍角的正弦、余弦、正切公式一、a组1.(2016陕西渭南阶段性测试)cos 12-sin 12cos 12+sin 12=()a.-32b.-12c.12d.32解析:原式=cos212-sin212=cos6=32,故选d.答案:d2.若sin 2=14,且4,2,则cos -sin 的值是()a.32b.34c.-32d.-34解析:(cos -sin )2=1-sin 2=1-14=34.4,2,cos -sin 0,cos -sin =-32.答案:c3.已知向量a=(3,-2),b=(cos ,sin ),若ab,则tan 2的值为()a.125b.-125c.1213d.-1213解析:由已知可得3sin -(-2)cos =0,tan =-23.tan 2=2tan1-tan2=-125.答案:b4.若f(x)=2tan x-2sin2x2-1sinx2cosx2,则f12的值为()a.-43b.-433c.8d.43解析:f(x)=2sinxcosx+2cosxsinx=2(sin2x+cos2x)sinxcosx=4sin2x,f12=4sin6=8.答案:c5.设sin =352,tan(-)=12,则tan(-2)=()a.-247b.-724c.247d.724解析:sin =35,2,cos =-45,tan =-34.又tan(-)=12,tan =-12,tan 2=2tan1-tan2=-43.tan(-2)=tan-tan21+tantan2=-34-431+-34-43=724.答案:d6.若sin6-=13,则cos23+2的值是.解析:sin6-=cos3+=13,cos23+2=cos 23+=2cos23+-1=2132-1=-79.答案:-797.(2016广东深圳南山区期末)已知sin(+)=13,则cos 2=.解析:sin(+)=-sin =13,sin =-13,cos 2=1-2sin2=1-2-132=79.答案:798.化简:2sin21+cos2cos2cos2=.解析:原式=2sin22cos2cos2cos2=tan 2.答案:tan 29.已知函数f(x)=cos2x2-sin x2cos x2-12.(1)求函数f(x)的最小正周期和值域;(2)若f()=3210,求sin 2的值.解:(1)因为f(x)=cos2x2-sin x2cos x2-12=12(1+cos x)-12sin x-12=22cosx+4,所以函数f(x)的最小正周期为2,值域为-22,22.(2)由(1)知,f()=22cos+4=3210,所以cos+4=35.所以sin 2=-cos2+2=-cos 2+4=1-2cos2+4=1-1825=725.10.(2016北京朝阳区高一期末)已知函数f(x)=sin2x+sin xcos x-2.(1)求f(x)的最小正周期;(2)求f(x)的单调增区间.解:(1)f(x)=sin2x+sin xcos x-2=1-cos2x2+12sin 2x-2=22sin2x-4-32,f(x)的最小正周期t=22=.(2)由2k-22x-42k+2,kz可解得f(x)的单调增区间是k-8,k+38(kz).二、b组1.已知sin 2=45,cos 2=-35,则角的终边所在的象限是()a.第一象限b.第二象限c.第三象限d.第四象限解析:sin =2sin 2cos 2=-24250,cos =cos22-sin22=-352-452=-7250,角的终边在第三象限.答案:c2.若向量a=(2,sin x),b=(cos2x,2cos x),则函数f(x)=ab的最小正周期是()a.2b.c.2d.4解析:f(x)=ab=2cos2x+2sin xcos x=1+cos 2x+sin 2x=1+2sin 2x+4,f(x)=ab的最小正周期是.答案:b3.化简1+sin4-cos41+sin4+cos4等于()a.1tan2b.tan 2c.1tand.tan 解析:原式=(1-cos4)+sin4(1+cos4)+sin4=2sin22+2sin2cos22cos22+2sin2cos2=2sin2(sin2+cos2)2cos2(cos2+sin2)=tan 2.答案:b4.若4,2,sin 2=378,则sin =.解析:因为4,2,所以22,所以cos 20,cos 2=-1-sin22=-18.又cos 2=1-2sin2=-18,所以sin2=916,所以sin =34.答案:345.(tan 10-3)sin 40的值为.解析:原式=sin10cos10-3sin 40=sin10-3cos10cos10sin 40=-2cos(10+30)cos10sin 40=-sin80cos10=-1.答案:-16.导学号08720090已知sin +4+sin -4=23,则sin -41-cos2-sin2的值为.解析:sin +4+sin -4=23,sin cos 4+cos sin 4+sin cos 4-cos sin 4=23,即2sin =23,sin =13.sin -41-cos2-sin2=sincos 4-cossin 4(1-cos2)-sin2=22(sin-cos)2sin2-2sincos=2(sin-cos)4sin(sin-cos)=24sin=2413=324.答案:3247.已知cos 4+x=35,1712x74,求sin2x+2sin2x1-tanx的值.解:sin2x+2sin2x1-tanx=2sinxcosx+2sin2x1-sinxcosx=2sinxcosx(cosx+sinx)cosx-sinx.易知cos x+sin x=2sin 4+x,cos x-sin x=2cos4+x.1712x74,534+x2,又cos 4+x=35,sin 4+x=-45.sin 2x=-cos 2+2x=-cos 24+x,原式=-43sin 2x=43cos 2x+2=43cos 2x+4=432cos2x+4-1=-2875.8.导学号08720091已知函数f(x)=53cos2x+3sin2x-4sin xcos x.(1)求f512;(2)若f()=53,2,求角.解:f(x)=53cos2x+3sin2x-4sin xcos x=53cos2x+53sin2x-2sin 2x-43sin2x=53-2sin 2x-23(1-cos 2x)=33-2sin 2x+23cos 2x=