Question

prove that in triangle ABC
a^3cos(B-C)+b^3cos(C-A)+c^3cos(A-B)=3abc
​

Answer 1

Question :- Prove that in ∆ABC, a^3cos(B-C)+b^ 3cos(C-A)+c^3cos(A-B) = 3abc. ?

Solution :-

in ∆ABC, sine rule says that,

• a/sinA = b/sinB = c/sinC = k (let).
• a = ksinA .
• b = ksinB .
• c = ksinC .

Solving first term a³cos(B - C) :-

→ a³cos(B - C)

→ a² * a * cos(B - C)

putting value of a from sine rule,

→ a² * (ksinA) * cos(B - C)

using angle sum property of a ∆, A = 180° - (B + C)

→ a² * k*sin{180° - (B + C)} * cos(B - C)

as sin(180° - x) = sin x ,

→ a² * k * sin(B + C) * cos(B - C)

Multiply and divide by 2,

→ (1/2)a²•k • [2sin(B + C) * cos(B - C)]

→ (1/2)a²•k • [sin2B + cos2C]

→ (1/2)a²•k • [2•sinB•cosB + 2•sinC•cosC]

→ (1/2)a²•k•2•[sinB•cosB + sinC•cosC]

→ a²[k•sinB•cosB + k•sinC•cosC]

again, Putting value from sine rule,

→ a²[b•cosB + c•cosC]

Similarly , we get :-

• b³cos(C - A) = b²[c•cosC + a•cosA]
• c³cos(A - B) = c²[a•cosA + b•cosB]

Putting all values in LHS now, we get,

→ a²[b•cosB + c•cosC] + b²[c•cosC + a•cosA] + c²[a•cosA + b•cosB] = 3abc

→ a²b•cosB + a²c•cosC + b²c•cosC + b²a•cosA + c²a•cosA + c²b•cosB = 3abc

→ (a²b•cosB + b²a•cosA) + (b²c•cosC + c²b•cosB) + (a²c•cosC + c²a•cosA) = 3abc

→ ab(a•cosB + b•cosA) + bc(b•cosC + c•cosB) + ac(a•cosC + c•cosA) = 3abc .

Now, we know that, when we drop a perpendicular from C to angle c . It divides the triangle into two right triangles, and angle c into two pieces .

So,

• c = a•cos⁡B + b•cos⁡A

Similarly now,

• a = b•cosC + c•cosB
• b = a•cosC + c•cosA

therefore,

→ ab•c + bc•a + ac•b = 3abc

→ abc + abc + abc = 3abc

→ 3abc = 3abc (Proved).

Learn more :-

It sino + tano = m

tano - sino an

Then express the

values of m²-n² in terms

of M and N

brainly.in/question/13926306

tanA/(1-cotA) + cotA/(1-tanA)

brainly.in/question/16775946

Answer 2

Answer:

hope it's help full for u