Question

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

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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Answer 2

Step-by-step explanation:

Given In triangle ABC,prove that cos A/2+ cos B/2 + cos C/2 = 4 cos(π-A/4)cos(π-B/4)cos(π-C/4)

• Now we have the l. h.s
• So cos (A/2) + cos (B/2) + cos (C/2)
• So we have cos C + cos D = 2 cos (C + D / 2) . cos (C – D / 2)
• 2 cos (A/2 + B/2) / 2 . cos (A/ 2 – B/2) / 2 + cos (C/2)
• 2 cos (A + B / 4). cos (A – B / 4) + cos (π – (A + B) / 2 )
• (Now A + B + C = π
•   C = π – (A + B) )
• 2 cos (A + B / 4) cos (A – B / 4) + sin (A + B / 2)
• (sin x = s sin x/ 2 cos x/ 2)
• 2 cos (A + B / 4) cos (A – B / 4) + 2 sin (A + B / 4 cos (A + B / 4)
• Taking common we get
• 2 cos (A + B / 4) [cos (A - B / 4 +  sin (A + B / 4) ]
• 2 cos (A + B / 4 ) [ cos (A – B / 4) + cos (π / 2 – A + B / 4) ]
• 2 cos (A + B / 4) [ cos (A – B / 4) + cos (2 π – (A + B) /4 ]
• using cos C and cos D formula we get
• 2 cos (A + B / 4) [ 2 cos (A – B / 4 + 2π – (A + B ) / 4) / 2). cos (A – B / 4 - 2π – (A + B) / 4) / 2]
• 2 cos (A + B / 4) [2 cos (A – B + 2π – A – B / 8 ) ] cos (A – B - 2π + A + B / 8)
• 2 cos (A + B / 4) [ 2 cos ( 2 (π – B /8) . cos (2 (- π + B /8) ]
• 2 cos (A + B / 4) [2 cos (π – B / 4). cos (π – A / 4) ] (cos (- theta) = cos theta)
• 2 cos (π – C / 4)  2 cos(π – B / 4).cos (π – A / 4)
• 4 cos (π – C / 4). cos (π – B / 4)..cos (π – A / 4)

Reference link will be

https://brainly.in/question/26132182