In triangle ABC show that (sin)^2 A/2 = (sin)^ 2 B+C/2 = 1
Asked by Navya Benny | 20th Sep, 2013, 09:35: PM
Expert Answer:
In a triangle sum of all angles is 180 degrees.
A + B + C = 180o
B + C = 180 - A
B + C/2 = 90 - A/2
sin2 (B + C/2) = sin2 (90 - A/2)
sin2 (B + C/2) = cos2 A/2
Hence,
LHS = sin2 A/2 + sin2 (B + C/2) = sin2 A/2 + cos2 A/2 = 1
In a triangle sum of all angles is 180 degrees.
A + B + C = 180o
B + C = 180 - A
B + C/2 = 90 - A/2
sin2 (B + C/2) = sin2 (90 - A/2)
sin2 (B + C/2) = cos2 A/2
Hence,
LHS = sin2 A/2 + sin2 (B + C/2) = sin2 A/2 + cos2 A/2 = 1
Answered by | 22nd Sep, 2013, 02:31: PM
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