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Question
If A, B, C are the interior angles of a ΔABC, show that `cos[(B+C)/2] = sin A/2`
Solution
We have to prove: `cos[(B+C)/2] = sin A/2`
Since we know that in triangle ABC
`A + B + C = 180^@`
`=> B + C = 180^@ - A`
Dividing by 2 on both sides, we get
`=> (B +C)/2 = 90^@ - A/2`
`=> cos (B + C)/2 = cos (90^@ - A/2)`
`=> cos (B + C)/2 = sin A/2`
Proved
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