2018-2019學(xué)年高中數(shù)學(xué) 第三章 三角恒等變換 3.1 兩角和與差的正弦、余弦和正切公式3課后習(xí)題 新人教A版必修4.doc

3.1.3二倍角的正弦、余弦、正切公式課后篇鞏固探究A組基礎(chǔ)鞏固1.cos 12-sin 12cos 12+sin 12=()A.-32B.-12C.12D.32解析原式=cos212-sin212=cos6=32,故選D.答案D2.若tan =3,則sin2cos2的值等于()A.2B.3C.4D.6解析sin2cos2=2sincoscos2=2tan =23=6.答案D3.已知sin4-x=35,則cos 2-2x的值為()A.1925B.1625C.1425D.725解析cos2-2x=cos 24-x=1-2sin24-x=1-2352=725.答案D4.若為銳角,3sin =tan =2tan ,則tan 2等于()A.34B.43C.-34D.-43解析因?yàn)闉殇J角,3sin =tan ,所以cos =13,則tan =22,即tan =2,所以tan 2=2tan1-tan2=-43.答案D5.若sin+cossin-cos=12,則tan 2=()A.-34B.34C.-43D.43解析等式sin+cossin-cos=12左邊分子、分母同時(shí)除以cos (顯然cos 0),得tan+1tan-1=12,解得tan =-3,tan 2=2tan1-tan2=34.答案B6.已知2,sin =55,則tan 2=.解析由2,sin =55,得cos =-255,tan =sincos=-12,tan 2=2tan1-tan2=-43.答案-437.化簡(jiǎn):2sin21+cos2cos2cos2=.解析原式=2sin22cos2cos2cos2=tan 2.答案tan 28.若cos(75+)=13,則sin(60+2)=.解析依題意,cos(75+)=13,則cos(150+2)=2cos2(+75)-1=2132-1=-79,sin(60+2)=-cos(90+60+2)=-cos(150+2)=79.答案799.求下列各式的值:(1)2cos2-12tan4-sin24+;(2)23tan 15+tan215;(3)sin 10sin 30sin 50sin 70.解(1)原式=cos22tan4-cos22-4-=cos22tan4-cos24-=cos22sin4-cos4-=cos2sin24-2=cos2cos2=1.(2)原式=3tan 30(1-tan215)+tan215=333(1-tan215)+tan215=1.(3)(方法一)sin 10sin 30sin 50sin 70=12cos 20cos 40cos 80=2sin20cos20cos40cos804sin20=sin40cos40cos804sin20=sin80cos808sin20=116sin160sin20=116.(方法二)令x=sin 10sin 50sin 70,y=cos 10cos 50cos 70.則xy=sin 10cos 10sin 50cos 50sin 70cos 70=12sin 2012sin 10012sin 140=18sin 20sin 80sin 40=18cos 10cos 50cos 70=18y.y0,x=18.從而有sin 10sin 30sin 50sin 70=116.10.已知sin +cos =355,0,4,sin-4=35,4,2.(1)求sin 2和tan 2的值;(2)求cos(+2)的值.解(1)由題意得(sin +cos )2=95,即1+sin 2=95,sin 2=45,又易知20,2,cos 2=1-sin22=35,tan 2=sin2cos2=43.(2)4,2,-40,4,sin-4=35,cos-4=45,sin 2-4=2sin-4cos-4=2425.又sin 2-4=-cos 2,cos 2=-2425.又易知22,sin 2=725.又cos2=1+cos22=45,cos =255,sin =55,cos(+2)=cos cos 2-sin sin 2=255-2425-55725=-11525.B組能力提升1.4sin 80-cos10sin10=()A.3B.-3C.2D.22-3解析4sin 80-cos10sin10=4cos10sin10-cos10sin10=2sin20-cos10sin10=2sin(30-10)-cos10sin10=2(sin30cos10-cos30sin10)-cos10sin10=-3.答案B2.若0,2,且cos2+cos2+2=310,則tan =()A.12B.14C.13D.13或-7解析cos2+cos2+2=cos2-sin 2=cos2-2sin cos =cos2-2sincossin2+cos2=1-2tantan2+1=310,整理得3tan2+20tan -7=0,解得tan =13或tan =-7.又0,2,所以tan =13,故選C.答案C3.若tan =12,則tan2+4=.解析tan =12,tan 2=2tan1-tan2=2121-14=43.tan2+4=tan2+tan41-tan2tan4=43+11-43=-7.答案-74.已知角,為銳角,且1-cos 2=sin cos ,tan(-)=13,則=.解析由1-cos 2=sin cos ,得1-(1-2sin2)=sin cos ,即2sin2=sin cos .為銳角,sin 0,2sin =cos ,即tan =12.(方法一)由tan(-)=tan-tan1+tantan=tan-121+12tan=13,得tan =1.為銳角,=4.(方法二)tan =tan(-+)=tan(-)+tan1-tan(-)tan=13+121-1312=1.為銳角,=4.答案45.已知函數(shù)f(x)=4cos4x-2cos2x-1sin4+xsin4-x.(1)求f-1112的值;(2)當(dāng)x0,4時(shí),求函數(shù)g(x)=12f(x)+sin 2x的最大值和最小值.解(1)f(x)=(1+cos2x)2-2cos2x-1sin4+xsin4-x=cos22xsin4+xcos4+x=2cos22xsin2+2x=2cos22xcos2x=2cos 2x,所以f-1112=2cos-116=2cos 6=3.(2)g(x)=cos 2x+sin 2x=2sin2x+4.因?yàn)閤0,4,所以2x+44,34,所以當(dāng)x=8時(shí),g(x)max=2,當(dāng)x=0時(shí),g(x)min=1.6.已知函數(shù)f(x)=4tan xsin2-xcosx-3-3.(1)求f(x)的定義域與最小正周期;(2)討論f(x)在區(qū)間-4,4上的單調(diào)性.解(1)f(x)的定義域?yàn)閤x2+k,kZ.f(x)=4tan xcos xcosx-3-3=4sin xcosx-3-3=4sin x12cosx+32sinx-3=2sin xcos x+23sin2x-3=sin 2x+3(1-cos 2x)-3=sin 2x-3cos 2x=2sin2x-3.所以f(x)的最小正周期T=22=.(2)令z=2x-3,函數(shù)y=2sin z的單調(diào)遞增區(qū)間是-2+2k,2+2k,kZ.由-2+2k2x-32+2k,得-12+kx512+k,kZ.設(shè)A=-4,4,B=x-12+kx512+k,kZ,易知AB=-12,4.所以,當(dāng)x-4,4時(shí),f(x)在區(qū)間-12,4上單調(diào)遞增,在區(qū)間-4,-12上單調(diào)遞減.