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Question
For ΔABC , prove that :
`sin((A + B)/2) = cos"C/2`
Solution
`sin((A + B)/2) = cos"C/2`
We know that for a triangle ΔABC
`<A + <B + <C = 180^circ`
`(<B + <A)/2 = 90^circ - (<C)/2`
`sin((A+B)/2) = sin(90^circ - C/2)`
= `cos(C/2)`
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