Question

In a triangle ABC prove that cosA/2+ cosB/2+cosC/2=pie-A/2*pie-B/2*pie-C/2

Answer 1

Given :- in triangle ABC, prove that :- cosA/2+cosB/2+cosC/2 = 4cos(π-A/4)cos(π-B/4)cos(π-C/4)

Solution :-

Solving LHS,

→ cosA/2 + cosB/2 + cosC/2

→ (cosA/2 + cosB/2) + cosC/2

using :-

• cosP + cosQ = 2•cos(C+D/2)•cos(C-D/2)
• C = π - (A + B) => C/2 = π/2 - (A + B)/2

→ 2[cos(A/2 + B/2)/2•cos(A/2 - B/2)/2] + cos{π/2 - (A+B/2)}

→ 2[cos(A+B/4) • cos(A-B/4)] + sin(A+B/2)

using :-

• sin2P = 2sinP•cosP

→ 2[cos(A+B/4) • cos(A-B/4)] + 2•sin(A+B/4)•cos(A+B/4)

taking common now,

→ 2cos(A+B/4)[cos(A-B/4) + sin(A+B/4)]

using :-

A + B = π - C => (A+B/4) = (π-C/4)

→ 2cos(π - C/4)[cos(A-B/4) + sin(A+B/4)]

using :-

• sin P = cos(π/2 - P)

→ 2cos(π - C/4)[cos(A-B/4) + cos{π/2 - (A+B/4)}]

→ 2cos(π - C/4)[cos(A-B/4) + cos{(2π - (A+B)/4}]

using :-

• cos P + cos Q = 2•cos(C+D/2)•cos(C-D/2)

→ 2cos(π - C/4)[2•cos{(A-B/4)+(2π-(A+B)/4)}/2 • cos{(A-B/4)-(2π-(A+B)/4)}/2]

→ 2cos(π - C/4)[2•cos{(A - B + 2π-A-B)/8}•cos{(A-B-2π+A+B)/8}]

→ 4cos(π - C/4)[cos{(2π-2B)/8}•cos{(2A-2π)/8}]

→ 4cos(π - C/4)[cos(π-B/4)•cos(A-π/4)]

using :-

• cos (-P) = cos P

→ 4cos(π - C/4)[cos(π-B/4)•cos(π-A/4)]

→ 4•cos(π-A/4)•cos(π-B/4)•cos(π-C/4) = RHS (Proved).

{Excellent Question.}

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