引用第21樓卡卡....於2007-09-03 21:06發表的“”:
有冇人同我講我錯係邊..thanks
其實我都計到你個answer, 不過無咁煩
將sin 3a 當 sin(2a+a) 咁expand 再simplify 就同你一樣
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而ckl個sol 中, cos2A=cos5A 是有問題
因我based on cos2A=cos5A
2A = 2πn±5A ,where n is an integer
-3A=2πn or 7A=2πn
A=-3/2 πn or 2/7 πn , where n is an integer
試 a = 2/7 π (n=1) 時, LHS=/=RHS