Question

With the usual notations prove that `2{asin^2  "C"/(2) + "c"sin^2  "A"/(2)}` = a – b + c.

Answer 1

L.H.S. = `2{a sin^2  "C"/2 + "c"sin^2  "A"/2}`

= `2 {a ((1 – cos "C")/2) + c(1 – cos A/2)}`

= `2{(a - a cos "C" + c - c cos"A")}`

= `a + c - (c cos "A" + a cos "C")`  `...[∵ "By Projection Rule b" = c cos"A" + a cos "C"]`

= a + c − b

= a - b + c

L. H. S. = R.H.S.

Answer 2

L.H.S. = `2{a sin^2  "C"/2 + "c"sin^2  "A"/2}`

= `a(2sin^2  "C"/2)+c(2sin^2  "A"/2)`

= a (1 - cos C) + c (1 - cos A)

= `a[1 - (a^2+b^2-c^2)/(2ab)]+c[(b^2+c^2-a^2)/(2bc)]`     ...[By cosine rule]

= `a[(2ab-a^2-b^2+c^2)/(2ab)]+c[(2bc-b^2-c^2+a^2)/(2bc)]`

= `(2ab-a^2-b^2+c^2)/(2b)+(2bc-b^2-c^2+a^2)/(2b)`

= `(2ab-a^2-b^2+c^2+2bc-b^2-c^2+a^2)/(2b)`

= `(2ab-2b^2+2bc)/(2b)`

= a - b + c

= RHS