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Question
In , ΔABC prove that
`"sin"(("B" - "C")/2) = (("b" - "c")/"a") "cos"("A"/2)`
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Solution
RHS = `(("b"-"c")/"a") "cos""A"/2` ...(by sine rule)
=`(("k" "sin" "B" - "k" "sin" "C")/("k" "sin""A")) . "cos""A"/2`
= `["k"["sin""B" - "sin""C"]]/("k""sin""A") ."cos""A"/2`
=`[[2"cos" ("B"+"C")/2 . "sin" ("B" -"C")/2]]/(2 "sin" "A"/2 "cos""A"/2) . "cos""A"/2`
= `(2"cos" (pi/2-"A"/2) "sin"
(("B"-"C")/2))/(2 "sin" "A"/2)`
=` "sin"(("B"-"C")/2) = "LHS"`
Hence, the required result is proved.
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