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Question
For triangle ABC, show that: `sin (A + B)/2 = cos C/2`
Sum
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Solution
We know that for a triangle ΔABC
∠A + ∠B + ∠C = 180°
`(angleB + angleA)/2 = 90^circ - (angleC)/2`
`sin((A + B)/2) = sin(90^circ - C/2)`
= `cos(C/2)`
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Chapter 21: Trigonometrical Ratios [Sine, Consine, Tangent of an Angle and their Reciprocals] - EXERCISE 21 (F) [Page 334]
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