1、第3课时二倍角的正弦、余弦、正切公式分层演练 综合提升A级基础巩固1.函数y=1-2cos2x的最小正周期是()A.4 B.2C.D.2答案:C2.若为第二象限角,sin =35,则sin 2=()A.-2425 B.-1225 C.1225D.2425答案:A3.若tan =-13,则cos 2=()A.-45B.-15 C.15 D.45答案:D4.若x-2,0,cos x=45,则tan 2x等于()A.724B.-724 C.247 D.-247答案:D5.已知2,cos =-45.(1)求tan 的值;(2)求sin 2+cos 2的值.解:(1)因为cos =-45,2,所以sin
2、 =35,所以tan =sincos=-34.(2)由(1)易得sin 2=2sin cos =-2425,cos 2=2cos2-1=725,所以sin 2+cos 2=-2425+725=-1725.B级能力提升6.化简sin2x2cosx1+tan xtan x2=()A.cos x B.tan xC.sin x D.12sin 2x解析:原式=2sinxcosx2cosx1+sinxcosxsinx2cos x2=sin xcosxcos x2+sinxsinx2cosxcos x2=sin xcos x2cosxcos x2=sinxcosx=tan x.答案:B7.若为锐角,cos
3、(+15)=35,则cos(2-15)=17250.解析:因为为锐角,cos(+15)=35,所以sin(+15)=45.所以sin(2+30)=2sin(+15)cos(+15)=2425,cos(2+30)=2cos2(+15)-1=2925-1=-725.所以cos(2-15)=cos(2+30-45)=cos(2+30)cos 45+sin(2+30)sin 45=-72522+242522=17250.8.已知函数f(x)=cos2x+sin xcos x,xR.(1)求f6的值;(2)若sin =35,且2,求f2+24.解:(1)f(6)=cos26+sin 6cos 6=(32
4、)2+1232=3+34.(2)因为f(x)=cos2x+sin xcos x=1+cos2x2+12sin 2x=12+12(sin 2x+cos 2x)=12+22sin(2x+4),所以f(2+24)=12+22sin(+12+4)=12+22sin(+3)=12+22(12sin +32cos ).又因为sin =35,且(2,),所以cos =-45.所以f(2+24)=12+22(1235-3245)=10+32-4620.C级挑战创新9.多空题已知sin =35,2,则tan 2=-247,cos 2=725.解析:因为sin =35,(2,),所以cos =-1-sin2=-4
5、5,所以tan =sincos=-34,tan 2=2tan1-tan 2=-247,cos 2=2cos2-1=725.10.开放探究题某同学在一次研究性学习中发现,以下五个式子的值都等于同一个常数:sin213+cos217-sin 13cos 17;sin215+cos215-sin 15cos 15;sin218+cos212-sin 18cos 12;sin2(-18)+cos248-sin(-18)cos 48;sin2(-25)+cos255-sin(-25)cos 55.(1)请根据式求出这个常数.(2)根据(1)的计算结果,将该同学的发现推广为三角恒等式,并证明你的结论.解:
6、(1)sin215+cos215-sin 15cos 15=1-12sin 30=1-14=34.(2)三角恒等式为sin2+cos2(30-)-sin cos(30-)=34.证法1:sin2+cos2(30-)-sin cos(30-)=sin2+(cos 30cos +sin 30sin )2-sin (cos 30cos +sin 30sin )=sin2+34cos2+32sin cos +14sin2-32sin cos -12sin2=34sin2+34cos2=34.证法2:sin2+cos2(30-)-sin cos(30-)=1-cos22+1+cos(60-2)2-sin (cos 30cos +sin 30sin )=12-12cos 2+12+12(cos 60cos 2+sin 60sin 2)-32sin cos -12sin2=12-12cos 2+12+14cos 2+34sin 2-34sin 2-14(1-cos 2)=1-14cos 2-14+14cos 2=34.
