2019-2020学年新一线同步人教A版数学必修一练习：5.5.2 简单的三角恒等变换 .docx
5.5.2简单的三角恒等变换课后篇巩固提升1.cos28-14的值为()A.2-14B.2+14C.24D.22解析cos28-14=1+cos42-14=2+14.答案B2.已知为第一象限角,且tan =43,则sin 2的值为()A.55B.-55C.55D.15解析因为为第一象限角,且tan =43,所以cos =35,而2是第一或第三象限角.当2是第一象限角时,sin 2=1-cos2=55;当2是第三象限角时,sin 2=-1-cos2=-55,故sin 2=55.答案C3.若函数f(x)=(1+3tan x)cos x,则f12=()A.6-22B.-3C.1D.2解析f(x)=1+3sinxcosxcos x=cos x+3sin x=2sinx+6,f12=2sin12+6=2sin4=2.答案D4.已知tan =2,则2sin2+1cos2-4的值是()A.53B.-134C.135D.134解析tan =2,2sin2+1cos2-4=2sin2+sin2+cos2cos2-2=3sin2+cos2sin2=3sin2+cos22sincos=3tan2+12tan=322+122=134.答案D5.已知1-sinxcosx=35,则cosx1+sinx的值等于()A.35B.-35C.53D.-53解析因为1-sinxcosx1+sinxcosx=1-sin2xcos2x=cos2xcos2x=1,而1-sinxcosx=35,所以1+sinxcosx=53,故cosx1+sinx=35.答案A6.已知sin+3+sin =-435,则cos+23=()A.-45B.-35C.35D.45解析sin3+sin =sin3cos +cos3sin +sin =-435,32sin +32cos =-435,即32sin +12cos =-45.sin+6=-45.故cos+23=cos+2+6=-sin+6=45.答案D7.若tan =17,则1+cos2sin2=.解析因为tan =sin21+cos2=17,所以1+cos2sin2=7.答案78.已知f(x)=sin x+3cos x,且锐角满足f()=2,则=.解析因为f(x)=sin x+3cos x=212sinx+32cosx=2sinx+3,又因为f()=2,所以2sin+3=2,解得=6.答案69.已知cosx-6=m,则cos x+cosx-3=.解析因为cos x+cosx-3=cos x+cos xcos 3+sin xsin 3=32cos x+32sin x=3cosx-6,所以cos x+cosx-3=3m.答案3m10.已知sin =1213,sin(+)=45,均为锐角,求cos 2的值.解0<<2,cos =1-sin2=513,0<<2,0<<2,0<+<,若0<+<2,sin(+)<sin ,+<,<0,与已知矛盾,2<+<,cos(+)=-35,cos =cos (+)-=cos(+)cos +sin(+)sin =-35513+451213=3365.0<<2,0<2<4,cos 2=1+cos2=76565.11.已知sin A+sin B+sin C=0,cos A+cos B+cos C=0,求证:cos2A+cos2B+cos2C=32.证明由已知,得sin A+sin B=-sin C,cos A+cos B=-cos C.和差化积,得2sinA+B2cosA-B2=-sin C.2cosA+B2cosA-B2=-cos C.当cosA-B2=0时,sin C=cos C=0,不满足题意,cosA-B20.,得tanA+B2=tan C.cos(A+B)=1-tan2A+B21+tan2A+B2=1-tan2C1+tan2C=cos 2C.2+2,得2+2cos(A-B)=1,即cos(A-B)=-12,cos2A+cos2B+cos2C=12(1+cos 2A+1+cos 2B+1+cos 2C)=32+122cos(A+B)cos(A-B)+cos 2C=32+122cos2C-12+cos2C=32.
