高中數(shù)學(xué)必修4配北師版-課后習(xí)題Word版第3章 第3節(jié)

1、3二倍角的三角函數(shù)課后篇鞏固探究A組基礎(chǔ)鞏固1.若tan =3,則sin2cos2的值等于()A.2B.3C.4D.6解析sin2cos2=2sincoscos2=2sincos=2tan =6.答案D2.1+cos1001-cos100等于()A.-2cos 5B.2cos 5C.-2sin 5D.2sin 5解析原式=2cos2502sin250=2(cos 50-sin 50)=222cos50-22sin50=2sin(45-50)=-2sin 5.答案C3.cos17cos217cos417cos817的值為()A.12B.14C.18D.116解析乘以sin17sin17,利用倍角

2、公式化簡(jiǎn)得116.答案D4.已知322,化簡(jiǎn)12+1212+12cos2的結(jié)果為()A.sin2B.-sin2C.cos2D.-cos2解析322,3420,cos20,原式=12+121+cos2212+12cos2=1+cos2=cos22=-cos2.答案D5.若sin 2=2425,02,則2cos4-的值為()A.15B.-15C.15D.75解析(sin +cos )2=1+sin 2=4925,因?yàn)?2,所以sin +cos =75,則2cos4-=222(cos +sin )=75.答案D6.若sin+cossin-cos=12,則tan 2=()A.-34B.34C.-43D

3、.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B7.已知sin4-x=35,則sin 2x=.答案7258.定義運(yùn)算ab=a2-ab-b2,則sin6cos6=.解析原式=sin26-sin6cos6-cos26=-cos312sin3=-1234.答案-12349.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=cos22tan

4、4-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

5、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.導(dǎo)學(xué)號(hào)93774097已知函數(shù)f(x)=2cos x(sin x-cos x),xR.(1)求函數(shù)f(x)圖像的對(duì)稱中心;(2)求函數(shù)f(x)在區(qū)間8,34上的最小值和最大值.解(1)f(x)=2cos x(sin x-cos x)=sin 2x-cos 2x-1=2

6、sin2x-4-1.令2x-4=k,kZ,得x=k2+8,kZ,因此,函數(shù)f(x)的圖像的對(duì)稱中心為k2+8,-1,kZ.(2)因?yàn)閒(x)=2sin2x-4-1在區(qū)間8,38上為增函數(shù),在區(qū)間38,34上為減函數(shù),又f8=-1,f38=2-1,f34=2sin32-4-1=-2cos4-1=-2,故函數(shù)f(x)在區(qū)間8,34上的最大值為2-1,最小值為-2.B組能力提升1.2cos2x-12tan4-xsin24+x可化簡(jiǎn)為()A.1B.-1C.cos xD.-sin x解析原式=cos2x2tan4-xcos24-x=cos2x2sin4-xcos4-xcos24-x=cos2x2sin4

7、-xcos4-x=cos2xsin2-2x=1.答案A2.若cos =-45,是第三象限的角,則1-tan21+tan2=()A.12B.-12C.35D.-2解析1-tan21+tan2=cos2-sin2cos2+sin2=cos2-sin2cos2+sin2cos2+sin22=cos1+sin,因?yàn)閏os =-45,且是第三象限的角,所以sin =-35,故1-tan21+tan2=-451-35=-2.答案D3.若cos2sin-4=-22,則cos +sin 的值為.解析cos2sin-4=cos2-sin2sincos4-cossin4=2(cos+sin)(cos-sin)si

8、n-cos=-2(cos +sin )=-22,cos +sin =12.答案124.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.若23

9、4,cos(-)=1213,sin(+)=-35,則sin +cos 的值為.解析由題意得0-4,+0,sin +cos =36565.答案365656.若f(x)=cos 2x-2a(1+cos x)的最小值為-12,則a=.解析f(x)=cos 2x-2acos x-2a=2cos2x-2acos x-2a-1,令t=cos x,則-1t1,函數(shù)f(x)可轉(zhuǎn)化為y=2t2-2at-2a-1=2t-a22a22-2a-1,當(dāng)a21,即a2時(shí),當(dāng)t=1時(shí),ymin=2-2a-2a-1=-12,解得a=38,不符合a2,舍去;當(dāng)a2-1,即a-2時(shí),當(dāng)t=-1時(shí),ymin=2+2a-2a-1=1

10、-12,不符合題意,舍去;當(dāng)-1a21,即-2a2時(shí),當(dāng)t=a2時(shí),ymin=-a22-2a-1=-12,解得a=-23,因?yàn)?2a2,所以a=-2+3.綜上所述,a=-2+3.答案-2+37.導(dǎo)學(xué)號(hào)93774098已知sin =1213,sin(+)=45,均為銳角,求cos2的值.解02,cos =1-sin2=513,02,02,0+.若0+2,sin(+)sin ,+,0,與已知矛盾,2+,cos(+)=-35,cos =cos(+)-=cos(+)cos +sin(+)sin =-35513+451213=3365.又20,4,cos2=1+cos2=76565.8.已知函數(shù)f(x)=4tan xsin2-xcosx-33.(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-33=4sin xcosx-33=4sin x12cosx+32sinx3=2sin xcos x+23sin2x-3=sin 2x+3(1-cos 2x)-3=sin 2x-3cos 2x=2sin2x-3.所以f(