Question

In a ∆ABC, prove the following, a(cos B + cos C) = `2("b" + "c") sin^2  "A"/2`

Answer 1

We have `"a"/sin"A" = "b"/sin"B" ="c"/sin"C"` = 2R

`"a"/sin"A"` = 2R ⇒ a = 2R sin A

`"b"/sin"B"` = 2R ⇒ b = 2R sin B

`"c"/sin"C"` = 2R ⇒ c = 2R sin C

a(cos B + cos C) = `2("b" + "c") sin^2  "A"/2`

`(cos "B" + cos "C")/(2sin^2  "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"`

= `(2sin (("B" + "C")/2) * cos  (("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"/ 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 - ("A"/2 + "B")])/(sin "A"/2) cos(- theta)` = cos θ

= `(sin ("A"/2 + "B"))/(sin  "A"/2)`

= `(2 sin  "A"/2 * sin ("A"/2 + "B"))/(2sin  "A"/2 * sin  "A"/2)`

= `(2sin  "A"/2 (sin  "A"/2 cos "B" + cos  "A"/2 sin "B"))/(2 sin  "A"/2 sin  "A"/2)`

= `(2sin^2  "A"/2 cos "B" + 2sin  "A"/2 cos  "A"/2 sin "B")/(2 sin^2  "A"/2)`

= `((1 - cos "A") cos "B" + sin "A" sin "B")/(2sin^2  "A"/2)`

= `(cos "B" - cos"A" cos"B" + sin"A" sin"B")/(2sin^2  "A"/2)`

= `(cos "B" - (cos "A" cos "B" + sin "A" sin "B"))/(2sin^2  "A"/2)`

= `(cos "B" - cos("A" + "B"))/(2 sin^2  "A"/2)`

= `(cos"B" - cos(180 - "C"))/(2sin^2  "A"/2)`

`("b"+ "c")/"a" = (cos"B" + cos"C")/(2sin^2  "A"/2)` .....(2)

From equations (1) and (2), result follows