三角學 3條 15分 好易(2)

1.cos3x/2．cosx/2+sin3x/2．sinx/2=√2/2
1/2(cos2x+cosx)+1/2(cos2x-cosx)=√2/2
cos2x+cosx+cos2x-cosx=√2
2cos2x=√2
cos2x=√2/2
2x=45˚,315˚,405˚or 675˚
x=22.5˚,157.5˚,202.5˚ or 337.5˚

2.tan2x+tanx=√3-√3．tan2xtanx
tan2x+tanx=√3(1-tan2xtanx)
(tan2x+tanx)/(1-tan2xtanx)=√3
tan3x=√3
3x=60˚,240˚,420˚,600˚,780˚or 960˚
x=20˚,80˚,140˚,200˚,260˚or320˚

3. tan(A+B)
=(tanA+tanB)/(1-tanAtanB)
=(1/5+1/3)/[1-(1/5)(1/3)]
=4/7
tan(C+D)
=(tanC+tanD)/(1-tanCtanD)
=(1/8+1/7)/[(1-(1/8)(1/7)]
=3/11
tan(A+B+C+D)
=[tan(A+B)+tan(C+D)]/[1-tan(A+B)tan(C+D)]
=(4/7+3/11)/[1-(4/7)(3/11)]
=1

∵tan(A+B+C+D)=1
∴A+B+C+D=π/4

1.cos3x/2 ．cosx/2+sin3x/2．sinx/2=開方2/2
cos(3x/2-x/2)=開方2/2
cos2x/2=開方2/2
cosx=開方2/2
x=0 or x=360

2.tan2x+tanx=開方3-開方3．tan2x．tanx
tan2x+tanx= 開方3(1-tan2xtanx)
(tan2x+tanx)/(1-tan2xtanx)=開方3
tan(2x+x)=開方3
tan3x=開方3
3x=60,240,420,600,780,960
x=20,80,140,200,260,320

3.設ABCD為銳角 己知tanA=1/5 tanB=1/3 tanC=1/8
tanD=1/7 試明A+B+C+D=兀/4

tanA=1/5 A=11.30993247
計哂A,B,C,D的數值再加埋就等如兀/4

2008-03-02 09:34:08 補充：
超凡學生..cos3x/2 是不等於 1/2(cos3x)的..所以你的答案錯了....