793 8.2 The Law of Cosines Now we obtain an expression for 1 - cos A. 1 - cos A = 1 - b2 + c2 - a2 2bc (1111)1111* cos A, from (1) = 2bc + a2 - b2 - c2 2bc = a2 - 1b2 - 2bc + c22 2bc Regroup. = a2 - 1b - c22 2bc = 3a - 1b - c243a + 1b - c24 2bc = 1a - b + c21a + b - c2 2bc Distributive property = 21s - b2 # 21s - c2 2bc Use equations (5) and (6). 1 - cos A = 21s - b21s - c2 bc Lowest terms (7) Similarly, it can be shown that 1 + cos A = 2s1s - a2 bc . (8) Recall the double-angle identities for cos 2u. Find a common denominator, and distribute the - sign. Factor the perfect square trinomial. Factor the difference of squares. Pay attention to signs. cos 2u = 2 cos2 u - 1 cos A = 2 cos2 a A 2b - 1 Let u = A 2 . 1 + cos A = 2 cos2 a A 2b Add 1. 2s1s - a2 bc = 2 cos2 a A 2b Substitute. (1+)11* From (8) s1s - a2 bc = cos2 a A 2b Divide by 2. cos a A 2b = Bs1s - a2 bc (9) cos 2u = 1 - 2 sin2 u cos A = 1 - 2 sin2 a A 2b Let u = A 2 . 1 - cos A = 2 sin2 a A 2b 21s - b21s - c2 bc = 2 sin2 a A 2b Substitute. (111111)111111* From (7) 1s - b21s - c2 bc = sin2 a A 2b Divide by 2. sin a A 2b = B1s - b21s - c2 bc (10) Subtract 1. Multiply by -1. The area of triangle ABC can be expressed as follows. = 1 2 bc sin A Area formula 2 = bc sin A Multiply by 2. 2 bc = sin A Divide by bc. (11)
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