Question:
In triangle ABC, show that sin²A /2 sin²B C/2=1
Answer:
Complete question is:
In triangle ABC, show that sin2 A/2 * sin2B/2 * sin2 C/2 = 1 - 2 * sin A/2 * sin
B/2 * sin C/2
Solution:
In a triangle ABC,
A + B + C = 180
LHS
sin2 A/2 + sin2 B/2 + sin2 C/2
= (1/2) * (2sin2 A/2 + 2sin2 B/2 + 2sin2 C/2)
= (1/2) * (1 - cos A + 1 - cos B + 1 - cos C)
= (1/2) * {3 -(cos A + cos B + cos C)}
= (1/2) * [3 - {2cos (A + B)/2 * cos(A - B)/2 + cos C)}]
= (1/2) * [3 - {2cos(90 - C/2) * cos(A - B)/2 + cos C}]
= (1/2) * [3 - {2sin C/2 * cos(A - B)/2 + 1 - 2sin2 C/2}]
= (1/2) * [3 - 1 - 2sin C/2 * {cos(A + B)/2 -sin C/2)}]
= (1/2) * [2 - 2sin C/2 * {cos(A - B)/2 - sin(90- (A+B)/2}] {since sin(90 - θ) = cos θ)
= (1/2) * {2 - sin C/2 * cos(A - B)/2 - cos(A + B)/2}
= (1/2) * {2 - 2sin C/2* (2sin A/2 * sin B/2)}
= (1/2) * (2 - 4sin A/2 * sin B/2 * sin C/2)
= 1 - 2*sin A/2 * sin B/2 * sin C/2
= RHS
So, sin2 A/2 * sin2B/2 * sin2 C/2 = 1 - 2 * sin A/2 * sin B/2 * sin C/2
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