Re: [中學] 兩題三角
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: Sent from JPTT on my Samsung SM-T320.
2.
i. 0 = sin(A+B+C)
= cos(A+B)sin(C) + sin(A+B)cos(C)
= cosAcos(B)sin(C) - sin(A)sin(B)sin(C)
+ sin(A)cos(B)cos(C) + cos(A)sin(B)cos(C)
因題設說明三角形ABC非為直角三角形，
=> 0 = tan(C) - tan(A)tan(B)tan(C) + tan(A) + tan(B)
所以 tan(A) + tan(B) + tan(C) = tan(A)tan(B)tan(C)
ii. tan(A/2)tan(B/2) = 1 - [tan(A/2) + tan(B/2)]/cot(C/2)
= 1 - tan(C/2)tan(A/2) - tan(C/2)tan(B/2)
類推可得
tan(A/2)tan(B/2) + tan(B/2)tan(C/2) + tan(C/2)tan(A/2)
= 3 - 2 * [tan(A/2)tan(B/2) + tan(B/2)tan(C/2) + tan(C/2)tan(A/2)]
=> tan(A/2)tan(B/2) + tan(B/2)tan(C/2) + tan(C/2)tan(A/2) = 1
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