Question

Prove that a^2Sin(B-C)+b^2Sin(C-A)+c^2Sin(A-B)=0

Answer 1

{ correcting the question }

Before we solve this problem, we must know some formulae:

a/sinA = b/sinB = c/sinC ..... (1)

cosA = (b² + c² - a²)/(2bc)

cosB = (c² + a² - b²)/(2ca)

cosC = (a² + b² - c²)/(2ab)

Proof:

From (1), we write

a/sinA = b/sinB = c/sinC = k (≠ 0)

Then sinA = a/k, sinB = b/k, sinC = c/k

L.H.S. = a² sin(B - C)/(sinB + sinC) + b² sin(C - A)/(sinC + sinA) + c² sin(A - B)/(sinA + sinB)

= a² (sinB cosC - cosB sinC)/(sinB + sinC)

+ b² (sinC cosA - cosC sinA)/(sinC + sinA)

+ c² (sinA cosB - cosA sinB)/(sinA + sinB)

= a² [{(b/k) . (a² + b² - c²)/(2ab)} - {(c² + a² - b²)/(2ca) . (c/k)}]/(b/k + c/k)

+ b² [{(c/k) . (b² + c² - a²)/(2bc)} - {(a² + b² - a²)/(2ab) . (a/k)}]/(c/k + a/k)

+ c² [{(a/k) . (c² + a² - b²)/(2ca)} - {(b² + c² - a²)/(2bc) . (b/k)}]/(a/k + b/k)

= a/(2k) * (a² + b² - c² - c² - a² + b²) * k/(b + c)

+ b/(2k) * (b² + c² - a² - a² - b² + c²) * k/(c + a)

+ c/(2k) * (c² + a² - b² - b² - c² + a²) * k/(a + b)

= {a/(b + c) * (b² - c²)} + {b/(c + a) * (c² - a²)} + {c/(a + b) * (a² - b²)}

= a (b - c) + b (c - a) + c (a - b)

= ab - ca + bc - ab + ca - bc

= 0 = R.H.S.

Hence proved.

Answer 2

Answer:

0

Step-by-step explanation:

this answer is way easier than the other answers

∑a^2sin(B-c)/sinB+sinC

∑a^2(2sinB+C/2*cosB-C/2)/2sinB+C/2*cosB-C/2

∑a^sin*(B-C/2)/sin(90-A/2)

∑a^sin*(B-C/2)/sin/cosA/2

by mollweides rule

b-c/a = sin(B-C/2)/cosA/2

∑a^2(b-c/a)

∑a(b-c)

a(b-c)+b(c-a)+c(a-b)

ab-ac+bc-ab+ca-cb=0