1、 T湖北黄冈师院数学系 f bY o 438000 t T,“ ZE , V F %. + T, ( H, / .1 ? Tcos( A+ B)= cosAcosB- sinAsinB.m1m2 : m1, !OD= 1. vAOF VvECF V vEDC._ NEDC =NA OF = A,OA = OB- ABaOB = OC #cosA=OD#cosB#cosA=cosA#cosBaA B= EC= CD#sinA= OD#sinB#sinA= sinA#sinB, _ OA = cosA#cosB- sinA#sinB. OA = OD#cos( A+ B) = cos( A+ B)
2、,_ cos( A+ B)= cosA#cosB- sinA#sinB.2 ? Tsin2A= 2sinA#cosA. : m2, !A O= OC= 1. BC=A C # sinA, A C =2AD, AD = AO #cosA= cosA, _ BC= 2sinA#cosA. BC=OC#sin2A= sin2A,_ sin2A= 2sinA#cosA.3 M Ttan A2 = sinA1+ cosA= 1- cosAsinA .m3 : m3, !A O= OC = 1. vACB V vBDC,_ NCAB = NDCB= A2 .tan A2 =DCAD =DBDC ,7DC=
3、 OC#sinA= sinA, AD = A O+ OD = AO +OC#cosA= 1+ cosA, DB= OB- OD= OB-OC#cosA= 1- cosA, _ tan A2 = sinA1+ cosA=1- cosAsinA .4 ? TsinA=2tan A21+ tan2 A2m4 : m4, !A O= OC= 1. ODAD =tan A2 , OD2+ AD2= 1, _ AD 2tan2 A2+ AD2 = 1, _ AD#21#2005年第2期 河北理科教学研究 问题讨论= 11+ tan2 A2, OD = A D # tan A2 =tan A21+ tan2
4、 A2.vA OD V vA CB,BCOD =ACA O ,BC =OD#ACAO = OD#AC,7A C=2AD, _ BC = OD#2AD =2tan A21+ tan2 A2.BC= OC#sinA= sinA, _ sinA=2tan A21+ tan2 A2.5 TsinA#cosB= 12 sin( A+ B) + sin( A- B)m5 : m5, !OA= OB = 1. A BCD0, OEF0L( EAB),_ EF = 12 ( DA +CB).7DA = OA#sin( A+ B)= sin( A+ B), CB= OB#sin( A-B)= sin( A- B
5、), EF= 12 sin( A+ B)+ sin( A- B) _ EF= OE#sinA= OA #cosB#sinA=sinA#cosB. _ sinA#cosB= 12 sin( A+ B) + sin( A- B) .(上接第20页)!sinA= 2- k2 ( a+ b), cosA= 2+ k2 ( a- b), sinB= 2- k2 ( c + d), cosB= 2+ k2( c - d) ,5| sin( A+ B) | = | sinAcosB+cosAsinB| = 4- k24 #| ( a+ b) ( c- d) + ( a- b) ( c+ d) | = 4-
6、k22 | ac - bd| 1, _ |ac- bd| 24- k2.6 45Q, I_ 6 ( 1979 ML 8 5)X=Qf f ( x ) = x2+ bx + c ( b, c I R), p: | f (1)| , | f (2) | , | f (3) | Bl12 . :L !| f ( 1)| , | f ( 2)| , | f (3)|l12 , f (1)= 1+ a+ b, f (2)= 4+ 2a+ b,f (3) = 9+ 3a+ b, _ f (1) - 2f ( 2) + f (3)=2, _ | f ( 1)| + 2| f (2)| + | f (3)| | f ( 1)- 2f (2)+ f (3)| = 2, 6BZ ,| f (1) | , | f (2)| , | f ( 3)|l12 ,_ | f (1) | + 2| f (2)| + | f ( 3) | 2,A B ,#L !, 5 .#22#2005年第2期 河北理科教学研究 问题讨论
