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Question
In ΔABC, A + B + C = π show that
sin A + sin B + sin C = `4cos "A"/2 cos "B"/2 cos "C"/2 `
Solution
L.H.S. = sin A + sin B + sin C
= `2sin(("A" + "B")/2)*cos(("A" - "B")/2) + 2sin "C"/2*cos "C"/2`
In ΔABC, A + B + C = π
∴ A + B = π – C
∴ `sin(("A" + "B")/2) = sin((pi - "C")/2) = sin(pi/2 - "C"/2) = cos "C"/2` ...(i)
and `cos(("A" + "B")/2) = cos((pi - "C")/2) = cos(pi/2 - "C"/2) = sin "C"/2` ...(ii)
∴ L.H.S. = `2*cos "C"/2*cos(("A" - "B")/2) + 2cos(("A" + "B")/2)*cos "C"/2` ...[From (i) and (ii)]
= `2*cos "C"/2*[cos(("A" - "B")/2) + cos(("A" + "B")/2)]`
= `2*cos "C"/2*2cos[(("A" + "B")/2 + ("A" - "B")/2)/2]*cos[(("A" + "B")/2 - ("A" - "B")/2)/2]`
= `2*cos "C"/2*(2cos "A"/2 *cos "B"/2)`
= `4*cos "A"/2* cos "B"/2*cos "C"/2`
= R.H.S.
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