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Question
In an ∆ABC, prove the following, `"a"sin ("A"/2 + "B") = ("b" + "c") sin "A"/2`
Solution
`"a"sin ("A"/2 + "B") = ("b" + "c") sin "A"/2`
`(sin("A"/2 + "B"))/(sin "A"/2) = ("b" + "c")/"a"` ......(1)
`("b" + "c")/"a" = (2"R"sin "B" + 2"R" sin "C")/(2"R" sin"A")`
= `(sin "B" + sin "C")/sin"A"`
= `(sin(("B" + "C")/2) * sin (("B" - "C")/2))/(2 sin "A"/2 * cos "A"/2)`
= `((sin pi/2 - "A"/2) * cos ("B"/2 - "C"/2))/(sin "A"/2 * cos "A"/2)`
= `(cos "A"/2 * cos ["B"/2 - (pi/2 - ("A"/2 + "B"/2))])/(sin "A"/2 * cos "A"/2)`
= `(cos["B"/2 - pi/2 + "A"/2 + "B"/2])/(sin "A"/2)`
= `(cos["B" +"A"/2 - pi/2])/(sin "A"/2)`
= `(cos[pi/2 - ("B" + "A"/2)])/(sin "A"/2) cos(- theta)` = cos θ
`("b" + "c")/"a" = (sin ("A"/2 + "B"))/(sin "A"/2)`
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