Difficult General solution

Consider cos2 x + cos2 2x:
If we want cos2 x + cos2 2x = 0, then both cos x and cos 2x should be zero, thus:
cos x = 0 → x = π/2, 3π/2, 5π/2, ...... also with -π/2, -3π/2, -5π/2, ......
cos 2x = 0 → x = π/4, 3π/4, 5π/4, ...... also with -π/4, -3π/4, -5π/4, ......
Therefore, none of their roots match with each other and hence cos2 x + cos2 2x = 0 has no solution.
Instead, cos2 x + cos2 2x > 0
Hence cos2 x + cos2 2x + cos2 3x > 0 since cos2 3x is non-negative. Eventually:
cos2 x + cos2 2x + cos2 3x + ... + cosn nx > 0
Thus, for (a) and (b), there is no general solution.
Also, for the same reason, when n is even in (c), there is also no general solution since the sum must be positive.
When n is odd in (c), taking n = 3 as an example:
cos3 x + cos3 2x + cos3 3x = (cos3 x + cos3 3x) + cos3 2x
= (cos x + cos 3x)(cos2 x - cos x cos 3x + cos2 3x) + cos3 2x
= 2 cos 2x cos x (cos2 x - cos x cos 3x + cos2 3x) + cos3 2x
= cos 2x [2 cos x (cos2 x - cos x cos 3x + cos2 3x) + cos2 2x]
Therefore, one of the general solutions will be that of cos 2x = 0.
And in general case for n being odd:
cosn x + cosn 2x + cosn 3x + ... + cosn nx
After grouping:
cosn x+ cosn nx = (cos x + cos nx) [cosn-1 x - ... + cosn-1 nx]
= 2 cos [(n + 1)x/2] cos [(n - 1)x/2] [cosn-1 x - ... + cosn-1 nx]
cosn 2x+ cosn (n - 1)x = [cos 2x + cos (n - 1)x] [cosn-1 2x - ... + cosn-1 (n - 1)x]
= 2 cos [(n + 1)x/2] cos [(n - 3)x/2] [cosn-1 2x - ... + cosn-1 (n - 1)x]
Continuing to the last pair:
cosn [(n - 1)x/2] + cosn [(n + 3)x/2] = [cos [(n - 1)x/2] + cos [(n + 3)x/2]] [cosn-1 [(n - 1)x/2] - ... + cosn-1 [(n + 3)x/2]]
= 2 cos [(n + 1)x/2] cos x [cosn-1 [(n - 1)x/2] - ... + cosn-1 [(n + 3)x/2]]
Including the last term: cosn [(n + 1)x/2], we can see that cos [(n + 1)x/2] is a factor of cosn x + cosn 2x + cosn 3x + ... + cosn nx.
Hence one of the general solutions of cosn x + cosn 2x + cosn 3x + ... + cosn nx = 0 is cos [(n + 1)x/2] = 0.

我在Pure Maths書中找到一條公式：
(cos 2θ + cos 4θ + ...... + cos 2nθ)sin θ = sin nθ cos (n + 1)θ
這對解答part (b)是非常有用的。

2009-08-10 02:58:04 補充：
(a)
沒有其他捷徑，只有把其餘的cos² 2x和cos² 3x這兩項都轉換成以cos x來表示，這樣會得出一條以cos x來表示的bicubic equation。

(b)
想辦法消除每項的二次方，然後用(cos 2x + cos 4x + ...... + cos 2nx)sin x = sin nx cos (n + 1)x化簡。雖然我不敢保證這樣做能否可以100%找到analytical solution，但至少萬一找不到analytical solution，都可以用電腦來計算。

(c)
完全no idea，因為實在too general。

2009-08-14 16:28:44 補充：

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參考資料：
my wisdom of maths + formula from http://mathworld.wolfram.com/SquareRoot.html + http://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Logarithmic_forms

想問你用邊本參考書架??有冇名字提供?

2009-08-02 21:16:19 補充：
哦...因為每次看你的發問問題...都是十分奇怪...>

part a 已經黎玩complex root? 咁激-,-

2009-08-02 11:27:00 補充：
呢D題目係我搵書果陣見到...但依家我搵5番本書T^TSOR

同complex root 有關 =]