a, b and c are the 3 sides of a triangle and the corresponding angles are A, B and C. Prove that (a - b) / (a + b) = tan[(A - B)/2] / tan[(A + B)/2].
Trig. Question
2009-08-01 5:08 am
回答 (2)
2009-08-01 5:26 am
✔ 最佳答案
(a - b) / (a + b) = tan[(A - B)/2] / tan[(A + B)/2]. By sine law:
L.H.S. = (sinA - sinB) / (sinA + sinB)
= 2cos[(A+B)/2] * sin[(A-B)/2 ] / {2sin[(A+B)/2] * cos[(A-B)/2] }
= sin[(A-B)/2] / cos[(A-B)/2] * cos[(A+B)/2] / sin[(A+B)/2]
= sin[(A-B)/2] / cos[(A-B)/2] / {sin[(A+B)/2] / cos[(A+B)/2]}
= tan[(A - B)/2] / tan[(A + B)/2]
= R.H.S.
2009-08-01 5:27 am
正切定律
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