Question

23.) In a AABC, prove that
sin3A sin(B-C) + sin3B sin (C - A) + sin3C sin(A-B) = 0
​

Answer 1

Step-by-step explanation:

sin^3a sin(b-c) +sin^3b sin(c-a) +sin^3c sin(a-b) +sin (a+b+c) sin(b-c) sin(c-a) sin(a-b) =0

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

sin (a+b+c)[sin(b-c) sin(c-a)] sin(a-b) =

sin (a+b+c)[(-1/2)(cos(b-a)-cos((b+a)-2c)] sin(a-b) =

(-1/2)[sin (a+b+c)sin(a-b)](cos(b-a)-cos((b+a)-2c) =

+(-1/2)[(-1/2)(cos(c+2a)-cos(c+2b)(cos(b-a)-cos((b+a)-2c) =

+(1/4)(cos(c+2a)-cos(c+2b))(cos(b-a)-cos((b+a)-2c) =

+(1/4)(cos(c+2a)-cos(c+2b))(cos(b-a)-cos((b+a)-2c) =

+(1/4)[cos(c+2a)cos(b-a)-cos(c+2b)cos(b-a)-cos(c+2a)cos((b+a)-2c)+cos(c+2b)cos((b+a)-2c)] =

+(1/4)(1/2)[cos(a+b+c)+cos(c-b+3a)-cos(c-a+3b)-cos(c+a+b)-cos(3a+b-c)-cos(3c-b+a)+cos(3b+a-c)+cos(b-a+3c)] =

+(1/8)[cos(a+b+c)+cos(c-b+3a)-cos(c-a+3b)-cos(c+a+b)-cos(3a+b-c)-cos(3c-b+a)

+cos(3b+a-c)+cos(b-a+3c)] =

+(1/8)[cos(c-b+3a)-cos(c-a+3b)-cos(3a+b-c)-cos(3c-b+a)+cos(3b+a-c)+cos(b-a+3c)] =

+(1/8)[cos(3a+(c-b))-cos(3b+(c-a))-cos(3a-(c-b))-cos(3c-(b-a))+cos(3b-(c-a))+cos(3c+(b-a))] =

+(1/8)[cos(3a+(c-b))-cos(3a-(c-b))+cos(3b-(c-a))-cos(3b+(c-a))+cos(3c+(b-a))-cos(3c-(b-a)] =

+(1/8)(-2sin(3a)sin(c-b)+2sin(3b)sin(c-a)-2sin(3c)sin(b-a)) =

But Sin(-x)=-sinx

+(1/4)(sin(3a)sin(b-c)+sin(3b)sin(c-a)+sin(3c)sin(a-b)) =

Your question becomes:

sin^3a sin(b-c) +sin^3b sin(c-a) +sin^3c sin(a-b)+sin (a+b+c) sin(b-c) sin(c-a) sin(a-b) =0

sin^3a sin(b-c) +sin^3b sin(c-a) +sin^3c sin(a-b)+(1/4)(sin(3a)sin(b-c)+sin(3b)sin(c-a)+sin(3c)sin(a-b))=0

sin(b-c)[sin^3a +(1/4)sin(3a)] + sin(c-a)[sin^3b+(1/4)sin(3b)] + sin(a-b)[sin^3c+ (1/4)sin(3c)]=0

sin(b-c)[(sin a)^3 +(1/4)sin(3a)] + sin(c-a)[(sin b)^3+(1/4)sin(3b)] + sin(a-b)[(sin c)^3+ (1/4)sin(3c)]=0

(sinx)^3 = 3/4 sinx- 1/4sin3x --> (sinx)^3 + (1/4)sin3x = 3/4sinxsin(b-c)[(3/4)sin(a)] + sin(c-a)[(3/4)sin(b)] + sin(a-b)[(3/4)sin(c)]=0

divide 3/4

sin(b-c)sin(a) + sin(c-a)sin(b) +sin(a-b)sin(c)=0

2[Cos(b-c+a)-cos(b-c-a)]+ 2[cos(c-a+b)-cos(c-a-b)]+ 2[cos(a-b+c)-cos(a-b-c)]

=0

Divide 2

Cos(b-c+a)-cos(b-c-a)+cos(c-a+b)-cos(c-a-b)+cos(a-b+c)-cos(a-b-c)=0

But cos(-x)= cosx so cos(b-c+a)= cos(-b+c-a)

Cos(b-c+a)-cos(b-c-a)+cos(c-a+b)-cos(c-a-b)+cos(a-b+c)-cos(a-b-c)=0

-cos(b-c-a)+cos(c-a+b)+cos(a-b+c)-cos(a-b-c)=0

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

-cos(b-c-a)+cos(c-a-b)+cos(a-b+c)-cos(a-b-c)=0

so also

-cos(b-c-a)+cos(a-b+c)=0

The result , wherever a,b,c the value is zero , so the right question is “proof that sin^3a sin(b-c) +sin^3b sin(c-a) +sin^3c sin(a-b) +sin (a+b+c) sin(b-c) sin(c-a) sin(a-b) =0 for any value a,b,c”.

Answer 2

Since ABC is a triangle

A + B + C = π -------(1)

LHS

$= ∑ [ sin(3A)sin(B-C) ]$

(The given expression is written in this manner for simplicity)

$=\frac{1}{2}∑ [ 2sin(3A)sin(B-C) ]$

$=\frac{1}{2}∑ [cos(3A-B+C) - cos(3A+B- C)]$

$=\frac{1}{2}∑ [cos(2A+A+C-B) - cos(2A+A+B- C)]$

$=\frac{1}{2}∑ [cos(2A+π-B-B) - cos(2A+π-C- C)]$

$=\frac{1}{2}∑ [cos(2A+π-2B) - cos(2A+π-2C)]$

$=\frac{1}{2}∑ [-cos(2A-2B) + cos(2A-2C)]$

$=\frac{1}{2} [( cos(2A-2C) -cos(2A-2B) )$

$+ ( cos(2B-2A)-cos(2B-2C))$

$+( cos(2C-2B)-cos(2C-2A))]$

$=\frac{1}{2} [( cos(2C-2A)-cos(2A-2B))$

$+ ( cos(2A-2B)-cos(2B-2C))$

$+( cos(2B-2C) -cos(2C-2A))]$

$= \frac{1}{2}\times 0$

$= 0$

Hence Proved