amath唔識

2007-05-13 3:16 am
sinx+sin2x+sin3x+sin4x+sin5x=0
的通解
更新1:

你知唔知你咁做會冇人黎睇架--

回答 (2)

2007-05-13 6:00 am
✔ 最佳答案
sinx+sin2x+sin3x+sin4x+sin5x=0
(sinx+sin5x)+(sin2x+sin4x)+sin3x=0
  x+5x   x-5x    2x+4x   2x-4x
2sin---cos---+2sin----cos----+sin3x = 0
   2    2      2 2
   6x   -4x     6x   -2x
2sin---cos---+2sin---cos---+sin3x = 0
   2    2     2    2
2sin3xcos2x+2sin3xcosx+sin3x = 0
sin3x(2cos2x+2cosx+1) = 0
sin3x[2(2cos平方x-1)+2cosx+1] = 0
sin3x(4cos平方x-2+2cosx+1) = 0
sin3x(4cos平方x+2cosx-1) = 0
sin3x = 0       or  4cos平方x+2cosx-1 = 0
3x = 180n°          x=360m°±72 or x=360k°±144
x = 60n°          
                          where n, m, k are integer
參考: 自己
2007-05-17 10:27 pm
無乜唔妥喎...


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