數學難題
Given sinA + sinB + sinC = 0 and cosA + cosB + cosC = 0,show that sin2A + sin2B + sin2C = 0
請各位大大指點指點。。。。0吾該。。。:icon081: Originally posted by bazookay at 2005-9-10 01:55 PM:
Given sinA + sinB + sinC = 0 and cosA + cosB + cosC = 0,
show that sin2A + sin2B + sin2C = 0
請各位大大指點指點。。。。0吾該。。。:icon081:
mmmm.........my method is somewhat long...
let e(iA) = cosA+isinA
then e(iA)+e(iB)+e(iC)=0
Since |e(iA)|=|e(iB)|=|e(iC)|=1, it represents an equilateral triangle thats one vertex is in the origin in the Argand plane.
Let A = @, B=@+2pi/3, C=@-2pi/3
Consider e(2iA)+e(2iB)+e(2iC), Since the magnitude of each complex number is 1 also, and only the angel is doubled, the resulting complex numbers are of arguments 2@, 2@+4pi/3, 2@ - 4pi/3.
But 2@+4pi/3 = 2@ - 2pi/3 and 2@-4pi/3 = 2@+2pi/3, So the 3 complex numbers will form an equil. triangle too. So e(2iA)+e(2iB)+e(2iC)=0, then
sin2A+sin2B+sin2C=0 and cos2A+cos2B+cos2C=0...
:icon064:諗唔到d簡單d既方法...:icon068: Originally posted by Erlkoing上的KK at 2005-9-11 04:30:
mmmm.........my method is somewhat long...
let e(iA) = cosA+isinA
then e(iA)+e(iB)+e(iC)=0
Since |e(iA)|=|e(iB)|=|e(iC)|=1, it represents an equilateral triangle thats one vertex is in the ori ...
除左Complex，我記得好似仲有個方法但比較長 Originally posted by Erlkoing上的KK at 11-9-2005 04:30 AM:
mmmm.........my method is somewhat long...
let e(iA) = cosA+isinA
then e(iA)+e(iB)+e(iC)=0
Since |e(iA)|=|e(iB)|=|e(iC)|=1, it represents an equilateral triangle thats one vertex is in the ori ...
thanks very much for your posting.....it is because i never look and think of the relationship between trigo and complex field....you have given me a new idea of solving trigo problems with complex methods....but you said that your method is long, and 天生射手 mentioned that he got a another (longer) method to tackle it, i have been thinking of playing around with the trigo identities but to no avail....:(
by the way thanks x 1000 to your suggested solution to this problem....:icon8::) you are good man... Originally posted by 天生射手 at 11-9-2005 11:50 AM:
除左Complex，我記得好似仲有個方法但比較長
長0吾緊要。。。儘管post 出來睇睇。。。無話邊個方法好定0吾好。。。重要0既個solution係link 緊去邊一個新 topic。。。。:)
學0吾同0既solution。。。可以同時接觸0吾同0既數學topics。。。。:) Originally posted by bazookay at 2005-9-11 12:03 PM:
thanks very much for your posting.....it is because i never look and think of the relationship between trigo and complex field....you have given me a new idea of solving trigo problems with com ...
做trigo problem呢...因為成日有sin a同cos a的關係...所以好煩的....所以用complex係最好的...(gwa)....
其實好似之後如果讀signal....個wave係Acos(wt+@)...佢地都會轉做complex...因為好方便的....
and it seems that sin nA + sin nB + sin nC = cos nA + cos nB + cos nC = 0
for n != multiples of 3...唔知岩唔岩...
[ Last edited by Erlkoing上的KK on 11-9-2005 at 05:24 PM ] 在網上搵到一個solution
但係佢係prove cos2A+cos2B+cos2C = 0
http://bbs.ee.ntu.edu.tw/boards/Math/7/28.html
: : cosA + cosB + cosC = 0 ---(1)
: : sinA + sinB + sinC = 0 ---(2)
: : (1)式平方得
: : (cosA)^2 + (cosB)^2 + (cosC)^2 + 2(cosAcosB +cosBcosC+cosCcosA) = 0 --(3)
: : (2)式平方得
: : (sinA)^2 + (sinB)^2 + (sinC)^2 + 2(sinAsinB +sinBsinC+sinCsinA) = 0 --(4)
: : (3) -(4)得
: : ( (cosA)^2 - (sinA)^2) + ( (cosB)^2 - (sinB)^2) + ( (cosC)^2 - (sinC)^2) +
: : 2((cosAcosB-sinAsinB+cosBcosC-sinBsinC+cosCcosA-sinCsinA) = 0
: : 化簡
: : cos2A+cos2B+cos2C+(cos(A+B)+cos(B+C)+cos(C+A)) = 0 --(5)
: : 在(5)式中我有A,B,C為三角形三個內角之懷疑,若我懷疑成真則繼續化簡得
: 這是錯的 三角形三個內角不可能有 sinA+sinB+sinC=0 這樣的條件
: : cos2A+cos2B+cos2C+(cos(PI-C)+cos(PI-A)+cos(PI-B)) = 0
: : 化簡
: : cos2A+cos2B+cos2C+(cosC+cosA+cosB)=0
: : 由(1)式已知故
: : cos2A+cos2B+cos2C + 0 = 0
: : 所以
: : cos2A+cos2B+cos2C = 0得證
但係我睇唔明... Originally posted by lidouble at 2005-9-11 07:20 PM:
在網上搵到一個solution
但係佢係prove cos2A+cos2B+cos2C = 0
http://bbs.ee.ntu.edu.tw/boards/Math/7/28.html
: : cosA + cosB + cosC = 0 ---(1)
: : sinA + sinB + sinC = 0 ---(2)
: : ...
我直頭唔明佢既思路
所以我話數學唔係畀一般人讀o既:icon032: Originally posted by 2076 at 2005-9-11 07:21 PM:
我直頭唔明佢既思路
所以我話數學唔係畀一般人讀o既:icon032:
冇錯
唔岩我地讀的
岩岩諗左可唔可以將佢地乘埋
(sinA+sinB+sinC)(cosA+cosB+cosC)
=sinAcosA+sinAcosB+sinAcosC+sinBcosA+sinBcosB+sinBcosC+sinCcosA+sinCcosB+sinCcosC
=1/2(sin(A+A)+sin(A-A)+sin(A+B)+sin(A-B)+sin(A+C)+sin(A-C)
+sin(B+A)+sin(B-A)+sin(B+B)+sin(B-B)+sin(B+C)+sin(B-C)
+sin(C+A)+sin(C-A)+sin(C+B)+sin(C-B)+sin(C+C)+sin(C-C))
=1/2(sin(A+A)+sin(A+B)+sin(A-B)+sin(A+C)+sin(A-C)
+sin(B+A)+sin(B-A)+sin(B+B)+sin(B+C)+sin(B-C)
+sin(C+A)+sin(C-A)+sin(C+B)+sin(C-B)+sin(C+C))
=1/2(sin(2A)+sin(2B)+sin(2C)+2(sin(A+B)+sin(A+C)+sin(B+C)))
sin(A+B)+sin(A+C)+sin(B+C)
=sinAcosB+sinBcosA+sinAcosC+sinCcosA+sinCcosB+sinBcosC
=sinA(cosB+cosC)+sinB(cosA+cosC)+sinC(cosA+cosB)
=-sinAcosA-sinBcosB-sinCcosC
=-1/2(sin2A+sin2B+sin2C)
(sinA+sinB+sinC)(cosA+cosB+cosC)
=1/2(sin(2A)+sin(2B)+sin(2C)-(sin2A+sin2B+sin2C))
=0....:(:(:(
結果係晒左d時間
計返條題目出黎:icon116: Originally posted by lidouble at 2005-9-11 07:30 PM:
冇錯
唔岩我地讀的
岩岩諗左可唔可以將佢地乘埋
(sinA+sinB+sinC)(cosA+cosB+cosC)
=sinAcosA+sinAcosB+sinAcosC+sinBcosA+sinBcosB+sinBcosC+sinCcosA+sinCcosB+sinCcosC
=1/2(sin(A+A)+sin(A-A)+sin ...
e個forum有幾個讀數學系既人
試下問下佢地:icon032: Originally posted by 2076 at 2005-9-11 07:31 PM:
e個forum有幾個讀數學系既人
試下問下佢地:icon032:
我記得我上pmath堂教過
唔記得左唔甘心:icon110::icon110: Originally posted by lidouble at 2005-9-11 07:35 PM:
我記得我上pmath堂教過
唔記得左唔甘心:icon110::icon110:
:eek2:
原來係pure maths數 難怪我唔識:icon032: Originally posted by 2076 at 2005-9-11 07:37 PM:
:eek2:
原來係pure maths數 難怪我唔識:icon032:
但係我記得好似係用返o個10條式
代下拆下就記到 Originally posted by lidouble at 11-9-2005 07:30 PM:
冇錯
唔岩我地讀的
岩岩諗左可唔可以將佢地乘埋
(sinA+sinB+sinC)(cosA+cosB+cosC)
=sinAcosA+sinAcosB+sinAcosC+sinBcosA+sinBcosB+sinBcosC+sinCcosA+sinCcosB+sinCcosC
=1/2(sin(A+A)+sin(A-A)+sin ...
0吾好話晒左時間。。。你0既構思亦都彼左大家一D解決問題0既0吾同思路。。。做數學係0甘架喇。。。solve難題0既時候。。你好似0係一間黑mongmong0既大屋入面。。。伸手不見五指。。後尾摸下摸下。。大概知到D傢私﹐臺凳響邊。。跟住繼續去摸。。突然摸到燈掣。。。啪。。。間屋著晒燈。。睇到野。。就係你搵到solution 0既時候喇。。。
所以千祈0吾好話晒時間。。。。:icon118:
thanks v much for your reply。。。:) Originally posted by lidouble at 2005-9-11 07:30 PM:
冇錯
唔岩我地讀的
岩岩諗左可唔可以將佢地乘埋
(sinA+sinB+sinC)(cosA+cosB+cosC)
=sinAcosA+sinAcosB+sinAcosC+sinBcosA+sinBcosB+sinBcosC+sinCcosA+sinCcosB+sinCcosC
=1/2(sin(A+A)+sin(A-A)+sin ...
其實一開始乘埋已經唔好lu....因為你知a=0 and b=0, 但係ab=0只係imply a=0 or b=0, 係condition度已經蝕左.. Originally posted by lidouble at 2005-9-11 07:20 PM:
在網上搵到一個solution
但係佢係prove cos2A+cos2B+cos2C = 0
http://bbs.ee.ntu.edu.tw/boards/Math/7/28.html
: : cosA + cosB + cosC = 0 ---(1)
: : sinA + sinB + sinC = 0 ---(2)
: : ...
唔係啦...哩個唔係proof啦.....哩個係一個人整左個好似proof的物體, 然後有人話佢其中一步錯左ja...
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