Let y = sec^2 [x - (π/6)] sec^2 [x - (π/2)].
(a) Show that y = 4 / {cos[2x - (2π / 3)] + 1/2} ^2
(b) Hence, find the minimum value of y.
F.4 trigonometry~~~
2010-02-25 4:28 am
回答 (1)
2010-02-25 7:16 am
✔ 最佳答案
(a) Consider cos2 (x - π/6) cos2 (x - π/2)= [cos (x - π/6) cos (x - π/2)]2
= (1/2)2 [cos (2x - 2π/3) + cos 2π/3]2
= (1/4) [cos (2x - 2π/3) - 1/2]2
Thus 1/y = 4/[cos (2x - 2π/3) - 1/2]2
(b) -3/2 <= cos (2x - 2π/3) - 1/2 <= 1/2
0 <= [cos (2x - 2π/3) - 1/2]2 <= 9/4
4/9 <= 1/[cos (2x - 2π/3) - 1/2]2
Hence min. of y = 16/9
參考: Myself
收錄日期: 2021-04-13 17:06:36
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