問:已知 A +
B + C = π,sin
A = 1/2 (sin B + sin C),證明 cot (B/2) cot (C/2) = 3
解:
sin
A = 1/2 (sin B + sin C)
sin
A = 1/2 [cos (π/2 - B) + cos (π/2 - C)]
sin
A = 1/2 [cos (1/2) (π - 2B) + cos (1/2) (π - 2C)]
sin
A = 1/2 [cos (1/2) (A - B + C) + cos (1/2) (A + B - C)]
2
sin (A/2) cos (A/2) = cos (A/2) cos (1/2) (B - C)
2
sin (A/2) = cos (1/2) (B - C)
2
cos (1/2) (π - A) = cos (1/2) (B - C)
2
cos (1/2) (B + C) = cos (1/2) (B - C)
1/2 [cos (1/2) (B + C) + cos (1/2) (B - C)] = -3/2
[cos (1/2) (B +C) - cos (1/2) (B - C)]
cos
(B/2) cos (C/2) = 3 sin (B/2) sin (C/2)
cos
(B/2) ÷ sin (B/2) = 3 sin (C/2) ÷ cos (C/2)
cot
(B/2) = 3 tan (C/2)
cot
(B/2) cot (C/2) = 3
~ * ~ * ~ * ~ * ~ * ~ * ~ * ~ * ~
各位同學:千萬別以為可以由解題第一步想到中間那些步驟,栗是由最後一步往回做,然後將演算步驟倒轉抄一遍的。如果連這個技倆也不會,M2的考試成績便凍過水了。
咁我結冰喇...
回覆刪除[版主回覆12/30/2010 08:40:00]您又唔使考, 結唔結冰都冇所謂啦
不過我幾鍾意呢條, 幾好玩.
這是“倒行逆施”的最好範例....
回覆刪除[版主回覆12/30/2010 21:02:00]我對學生一向倒行逆施,終於給慧行先生發現了