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Question
In ΔABC, A + B + C = π show that
`sin^2 "A"/2 + sin^2 "B"/2 - sin^2 "C"/2 = 1 - 2cos "A"/2 cos "B"/2 sin "C"/2`
Solution
L.H.S. = `sin^2 ("A"/2) + sin^2 ("B"/2) - sin^2 ("C"/2)`
= `(1 - cos"A")/2 + (1 - cos"B")/2 -sin^2("C"/2)`
= `1/2 - 1/2cos "A" + 1/2 - 1/2 cos"B" - sin^2("C"/2)`
= `1 - 1/2(cos"A" + cos"B") - sin^2("C"/2)`
= `1 - 1/2 xx 2cos(("A" + "B")/2)*cos(("A" - "B")/2) - sin^2("C"/2)`
= `1 - cos(pi/2 - "C"/2)*cos(("A" - "B")/2) - sin^2("C"/2)` ...[∵ A + B + C = π]
= `1 - sin("C"/2)*cos(("A" - "B")/2) -sin^2("C"/2)`
= `1 - sin("C"/2)[cos(("A" - "B")/2) + sin("C"/2)]`
= `1 - sin("C"/2)[cos(("A" - "B")/2) + sin{pi/2 - (("A" + "B")/2)}]` ...[∵ A + B + C = π]
= `1 - sin("C"/2)[cos(("A" - "B")/2) + cos(("A" + "B")/2)]`
= `1 - sin("C"/2) xx 2cos((("A" + "B")/2 + ("A" - "B")/2)/2)*cos((("A" + "B")/2 - ("A" - "B")/2)/2)`
= `1 - sin("C"/2) xx 2cos("A"/2)*cos("B"/2)`
= `1 - 2("A"/2)*cos("B"/2)*sin("C"/2)`
= R.H.S.
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