Question

If ∆ABC is a right triangle and if ∠A = `pi/2` then prove that cos B – cos C = `- 1 + 2sqrt(2) cos  "B"/2  sin  "C"/2`

Answer 1

A + B + C = 180°

Given A = 90°

∴ B + C = 90°

⇒ `("B" + "C")/2` = 45°

`"B"/2 + "C"/2` = 45°

R.H.S = `- 1 + 2sqrt(2) cos  "B"/2 sin  "C"/2`

= `1 + sqrt(2) (2cos  "B"/2 sin  "C"/2)`

We know that 2 cosA sinB = sin(A + B) – sin(A – B)

= `- 1 + sqrt(2) (sin  (("B" + "C"))/2 - sin  (("B" - "C"))/2)`

= `- 1 + sqrt(2) (sin 45^circ - sin  (("B" - "C"))/2)`

= `- 1 + sqrt(2) (1/sqrt(2) - sin  (("B" - "C")^2)/2)`

= `- 1 + 1 - sqrt(2) sin  (("B" - "C"))/2`

= `- sqrt(2) sin  (("B" - "C"))/2`  .....(1)

L.H.S = cos B – cos C

= `2 sin  (("B" + "C"))/2 sin  (("C" - "B"))/2`

= `2 sin 45^circ sin  (("C" - "B"))/2`

= `2(1/sqrt(2)) sin  ((-("B" - "C"))/2)`

= `- sqrt(2) sin  (("B" - "C"))/2` .....(2)

From (1) and (2)

⇒ L.H.S = R.H.S