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```
sin (B+ C -A)+sin(C+ A- B )+sin(A+B-C)=4sinAsinBsinC

```
3 years ago RAHUL ROHILLA
873 Points
```							sin( B+C-A ) + sin( C+A-B ) + sin( A+B-C )=> sin( π - 2A ) + sin( π - 2B ) + sin( π - 2C )=> sin (2A ) + sin( 2B ) + sin( 2C )=> 2sin( A + B)*cos(A - B) + 2sin( C)*Cos(C) ; use sinC + sinD = 2sin(C+D)/2 * cos(C -D)/2,=> 2sin( π - C)*cos(A - B) + 2sin( C)* cos(C)=> 2sin(C)*cos(A - B) + 2sin( C)*cos{π - (A+B)} ; use A + B + C = π=> 2sin(C)[cos(A - B) - Cos(A+B)] ; use cos (π - θ) = -cosθ=> 2sin(C)[2sin(A) *sin(B)] ; use cos C - CosD = 2sin(C+D)/2 * sin(D - C)/2=> 4sin(A) sin(B) sin(C) ----- RHS hence Proved
```
2 years ago
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