Question

in triangle abc prove that a^3 sin(b-c)+b^3 sin (c-a)+c^3 sin (a-b)=0

Answer 1

Answer:

Step-by-step explanation:

Hi

Consider a given triangle ABC

Using la of sines, we know that

a/sinA = b/sinB = c/sinC = 2R

a = 2RsinA, b = 2RsinB, c = 2RsinC

But, in  ΔABC, A + B + C = 180,

so A = 180 - B - C

sin A = sin (180 - B - C)

= sin(B + C)

So, a = 2Rsin(B + C)

Similarly, we get b = 2Rsin(A+C)

c = 2Rsin(A + B)

Consider

a³sin(B - C) + b³sin(C - A) + c³sin(A - B)

= a².2Rsin(B + C) sin(B - C)+ b²2rsin(C+A)sin (C - A)

+ c².2Rsin(A + B) sin(A - B)

Using the identity

sin(A + B)*sin(A - B) = sin²A - sin²B, we get

L.H.S

= 2Ra²(sin²B - sin²C) + 2Rb²(sin²C - sin²A) + 2Rc²(sin²A - sin²B)

=But Using sine rule,

asinB = bsinA ,

asinC = c sinA and

csinB = bsinC,

so L.H.S = 0 (Since all terms get cancelled out)

Hope, it helps !