For triangle ABC, show that. SIN A+B/2=COS C/2

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Solution

In a ΔABC, ∠A + ∠B + ∠C = 180° [Angle sum property]

⇒ ∠A + ∠B = 180° - ∠C

⇒ (∠A + ∠B) / 2 = (180° - ∠C) / 2

⇒ (∠A + ∠B) / 2 = (180° - ∠C) / 2

⇒ (∠A + ∠B) / 2 = (90° - ∠C/ 2)

⇒ sin (∠A + ∠B) / 2 = sin (90° - ∠C/ 2)

⇒ sin (∠A + ∠B) / 2 = cos ∠C/ 2. [sin (90 - θ) = cos θ)]

Therefore, sin (∠A + ∠B) / 2 in terms of angle c is cos ∠C/ 2.

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