因計不出答案,而一直頭痛,請幫忙!

請見以下方法:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Power6.jpg

要留意的是 sin6 x + cos6 x 應等於 (sin2 x + cos2 x)(sin4 x + sin2 x cos2 x + cos4 x), 只需設 a = sin2 x 和 b = cos2 x 灰後代入恆等式 a3 + b3 = (a + b)(a2 + ab + b2) 即可.

2007-02-16 22:53:06 補充：
最後少少打錯:sin^6 x 十 cos^6 x 應該係 (sin^2 x 十 cos^2 x)(sin^4 x - sin^2 x cos^2 x 十 cos^4 x) 至岩.而 a^3 十 b^3 = (a 十 b)(a^2 -ab 十 b^2)

(sinx)^6 + (cosx)^6 + 2asinxcosx >= 0

[ (sinx)^2 + (cosx)^2 ] [ (sinx)^2 - sinx cosx + (cosx)^2 ] + 2a sinx cosx >= 0

1 - (1-2a) sinx cosx >= 0

sinx cosx <= 1/(1-2a) (if 1 - 2a >= 0) ------------------ Case 1
sinx cosx >= 1/(1-2a) (if 1 - 2a <= 0) ------------------ Case 2

Since sin2x = 2 sinx cosx
sinx cosx = sin2x/2

The maximum and minimum value of sinx cox is 1/2 and -1/2.

Case 1:
1/(1-2a) >= 1/2 (the maximum value of sinx cosx)
1 >= 1/2 - a
-a <= 1/2
a >= -1/2

Combine with the condition (if 1 - 2a >= 0), gives a <= 1/2

Therefore, -1/2 <= a <= 1/2.

Case 2:
1/(1-2a) <= -1/2 (the maximum value of sinx cosx)
1 >= a - 1/2
a <= 3/2

Combine with the condition (if 1 - 2a >= 0), gives a >= 1/2

Therefore, it is no solution.

From case 1 and case 2, the answer is -1/2 <= a <= 1/2

我已經証實了我的答案是對的, 你只需要代入 a=1/2 入去, 就能明白了.