ute40zf07t的部落格

標題:

A.maths trigo

發問:

1.Show that for any ΔABC, cos (A/2) = √ ((a+b+c)(b+c-a)/4bc) 2.Given ΔABC, a)show that (i) sin(A+C/2)-sin(B+C/2)=0 (<--- no need to do this part) (ii)sin(A-C/2)-sin(B-C/2) =2sinC sin((A-B)/2) b) Using the results of (a), show that c sin((A-B)/2) =(a-b)cos (C/2) THX very much!!! 更新: Is there any other method to do these questions? Coz I have not learnt compound angles yet....

• Intel(R) 82945G Express Chipset Family
• [10分] 呂明才排球女仔C GRADE
• Probability_9@1@

 

此文章來自奇摩知識+如有不便請留言告知

最佳解答:

(1) By the cosine law: a2 = b2 + c2 - 2bc cos A cos A = (b2 + c2 - a2)/2bc 2cos2 (A/2) - 1 = (b2 + c2 - a2)/2bc 2cos2 (A/2) = (b2 + c2 - a2)/2bc + 1 2cos2 (A/2) = (b2 + 2bc + c2 - a2)/2bc cos2 (A/2) = [(b + c)2 - a2]/4bc cos2 (A/2) = (b + c + a)(b + c - a)/4bc cos (A/2) = √[(b + c + a)(b + c - a)/4bc] (2) a ii) sin (A - C/2) - sin (B - C/2) = 2 sin {[(A - C/2) - (B - C/2)]/2} cos {[(A - C/2) + (B - C/2)]/2} = 2 sin [(A - B)/2] cos [(A + B - C)/2] = 2 sin [(A - B)/2] cos [(180° - C - C)/2] = 2 sin [(A - B)/2] cos (90° - C) = 2 sin [(A - B)/2] sin C (b) Adding (a)(i) and (ii) together: sin (A + C/2) - sin (B + C/2) + sin (A - C/2) - sin (B - C/2) = 2 sin [(A - B)/2] sin C [sin (A + C/2) + sin (A - C/2)] - [sin (B + C/2) + sin (B - C/2)] = 2 sin [(A - B)/2] sin C 2 sin A cos (C/2) - 2 sin B cos (C/2) = 2 sin C sin [(A - B)/2] cos (C/2) (sin A - sin B) = sin C sin [(A - B)/2] cos C/sin [(A - B)/2] = sin C/(sin A - sin B) cos C/sin [(A - B)/2] = c/(a - b) (By sine law) (a - b) cos C = c sin [(A - B)/2]

其他解答: