3.2　簡單的三角恒等變換

3.2簡單的三角恒等變換【選題明細表】 知識點、方法題號化簡求值1,2,3,4,13恒等式證明8簡單應(yīng)用5,6,7綜合應(yīng)用9,10,11,121.下列各式中,值為12的是(B)(A)sin 15cos 15(B)cos26-sin26(C)tan301-tan230 (D)1+cos602解析:選項A中,原式=12sin 30=14;選項B中,原式=cos 3=12;選項C中,原式=122tan301-tan230=12tan 60=32;選項D中,原式=cos 30=32.故選B.2.(2019泰安高一期末)已知cos =-15,523,那么sin 2等于(D)(A)105(B)-105(C)155(D)-155解析:因為523,所以54232,所以sin 20.由cos =1-2sin22,得sin 2=-1-cos2=-(1+15)12=-155.故選D.3.已知2sin =1+cos ,則tan 2等于(B)(A)12(B)12或不存在(C)2(D)2或不存在解析:2sin =1+cos ,即4sin 2cos 2=2cos22,當(dāng)cos 2=0時,tan 2不存在,當(dāng)cos 20時,tan 2=12.故選B.4.化簡(sin 2+cos 2)2+2sin2(4-2)得(C)(A)2+sin (B)2+2sin(-4)(C)2 (D)2+2sin(+4)解析:原式=1+2sin 2cos 2+1-cos2(4-2)=2+sin -cos(2-)= 2+sin -sin =2.故選C.5.使函數(shù)f(x)=sin(2x+)+3cos(2x+)為奇函數(shù)的的一個值是(D)(A)6(B)3(C)2(D)23解析:f(x)=sin(2x+)+3cos(2x+)=2sin(2x+3+).當(dāng)=23時,f(x)=2sin(2x+)=-2sin 2x是奇函數(shù).故選D.6.函數(shù)f(x)=sin(2x-4)-22sin2x的最小正周期是.解析:f(x)=22sin 2x-22cos 2x-2(1-cos 2x)=22sin 2x+22cos 2x- 2=sin(2x+4)-2,所以T=22=.答案:7.若向量a=(2sin ,-1),b=(cos ,2sin2+m)(R),且ab,則m的最小值為.解析:因為a=(2sin ,-1),b=(cos ,2sin2+m)(R),且ab,所以2sin cos =2sin2+m,所以m=-2sin2 +2sin cos =cos 2+sin 2-1=2sin(2+4)-1,因為R,所以2sin(2+4)-2,2,所以m的最小值為-2-1.答案:-2-18.求證:1+sin4-cos42tan=1+sin4+cos41-tan2.證明:原式等價于1+sin 4-cos 4=2tan1-tan2(1+sin 4+cos 4),即1+sin 4-cos 4=tan 2(1+sin 4+cos 4)(*)而(*)式右邊=tan 2(1+cos 4+sin 4)=sin2cos2(2cos22+2sin 2cos 2)=2sin 2cos 2+2sin22=sin 4+1-cos 4=左邊,所以(*)式成立,原式得證.9.函數(shù)y=cos2(x-12)+sin2(x+12)-1(A)(A)是奇函數(shù)(B)是偶函數(shù)(C)既是奇函數(shù)又是偶函數(shù)(D)既不是奇函數(shù)又不是偶函數(shù)解析:y=1+cos(2x-6)2+1-cos(2x+6)2-1=12cos(2x-6)-cos(2x+6)=12sin 2x,是奇函數(shù).故選A.10.(2019全國卷)函數(shù)y=sin x-3cos x的圖象可由函數(shù)y= sin x+3cos x的圖象至少向右平移個單位長度得到.解析:y=sin x-3cos x=2sin(x-3),y=sin x+3cos x=2sin(x+3),y=2sin(x+3)的圖象至少向右平移23個單位長度得到y(tǒng)=2sin(x+3-23) =2sin(x-3)的圖象.答案:2311.關(guān)于函數(shù)f(x)=sin xcos x-cos2x,給出下列命題:f(x)的最小正周期為2;f(x)在區(qū)間(0,8)上為增函數(shù);直線x=38是函數(shù)f(x)圖象的一條對稱軸;函數(shù)f(x)的圖象可由函數(shù)f(x)=22sin 2x的圖象向右平移8個單位得到;對任意xR,恒有f(4+x)+f(-x)=-1.其中正確命題的序號是.解析:f(x)=12sin 2x-1+cos2x2=22sin(2x-4)-12,顯然錯;x(0,8)時,2x-4(-4,0),函數(shù)f(x)為增函數(shù),故正確;令2x-4=2+k,kZ得x=38+k2,kZ,顯然x=38是函數(shù)f(x)圖象的一條對稱軸,故正確;f(x)=22sin 2x的圖象向右平移8個單位得到y(tǒng)=22sin 2(x-8)=22sin(2x-4),故錯;f(4+x)+f(-x)=22sin(2x+4)-12+22sin(-2x-4)-12=22sin(2x+4)-22sin (2x+4)-1=-1,故正確.答案:12.已知函數(shù)f(x)=sin(2-x)sin x-3cos2x.(1)求f(x)的最小正周期和最大值;(2)討論f(x)在6,23上的單調(diào)性.解:(1)f(x)=sin(2-x) sin x-3cos2x=cos x sin x-32(1+cos 2x)=12sin 2x-32(1+cos 2x)=12sin 2x-32cos 2x-32=sin(2x-3)-32,因此f(x)的最小正周期為,最大值為2-32.(2)當(dāng)x6,23時,有02x-3,從而當(dāng)02x-32,即6x512時,f(x)單調(diào)遞增,當(dāng)22x-3,即512x23時,f(x)單調(diào)遞減.綜上可知,f(x)在6,512上單調(diào)遞增;在512,23上單調(diào)遞減.13.(2019日照高一期末)若4,2,sin 2=378,則sin 等于(D)(A)35(B)45(