# If A+B+C=pi then prove that sin(B+2C)+sin(C+2A)+sin(A+2B)=4sin((B-C)/2)*sin((C-A)/2)*sin((A-B)/2)?

Nov 8, 2017

Kindly refer to a Proof in the Explanation.

#### Explanation:

$A + B + C = \pi \Rightarrow A + B = \pi - C .$

$\therefore A + 2 B = \left(A + B\right) + B = \left(\pi - C\right) + B = \pi - \left(C - B\right) .$

$\therefore \sin \left(A + 2 B\right) = \sin \left(\pi - \left(C - B\right)\right) = \sin \left(C - B\right) .$

Similarly,

$\sin \left(B + 2 C\right) = \sin \left(A - C\right) , \mathmr{and} , \sin \left(C + 2 A\right) = \sin \left(B - A\right) .$

$\therefore \sin \left(A + 2 B\right) + \sin \left(B + 2 C\right) + \sin \left(C + 2 A\right) ,$

$= \sin \left(C - B\right) + \sin \left(A - C\right) + \sin \left(B - A\right) .$

Let, $C - B = x , A - C = y \mathmr{and} B - A = z .$

Then, $x + y + z = 0.$

$\therefore x + y = - z \ldots \ldots \left(1\right) \mathmr{and} z = - \left(x + y\right) \ldots \ldots \left(2\right) .$

$\text{The L.H.S.=} \sin x + \sin y + \sin z ,$

$= 2 \sin \left(\frac{x + y}{2}\right) \cos \left(\frac{x - y}{2}\right) + \sin z ,$

$= 2 \sin \left(- \frac{z}{2}\right) \cos \left(\frac{x - y}{2}\right) + \sin z , \ldots \ldots \ldots \ldots \left[\because \left(1\right)\right] ,$

$= - 2 \sin \left(\frac{z}{2}\right) \cos \left(\frac{x - y}{2}\right) + 2 \sin \left(\frac{z}{2}\right) \cos \left(\frac{z}{2}\right) , \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left[\because \sin 2 a = 2 \sin a \cos a\right] ,$

$= 2 \sin \left(\frac{z}{2}\right) \left\{\cos \left(\frac{z}{2}\right) - \cos \left(\frac{x - y}{2}\right)\right\} ,$

$= 2 \sin \left(\frac{z}{2}\right) \left\{\cos \left(- \frac{x + y}{2}\right) - \cos \left(\frac{x - y}{2}\right)\right\} , \ldots \ldots . \left[\because \left(2\right)\right] ,$

$= 2 \sin \left(\frac{z}{2}\right) \left\{\cos \left(\frac{x + y}{2}\right) - \cos \left(\frac{x - y}{2}\right)\right\} ,$

$= 2 \sin \left(\frac{z}{2}\right) \left\{- 2 \sin \left(\frac{x}{2}\right) \sin \left(\frac{y}{2}\right)\right\} ,$

$= - 4 \sin \left(\frac{x}{2}\right) \sin \left(\frac{y}{2}\right) \sin \left(\frac{z}{2}\right) ,$

$= - 4 \sin \left(\frac{C - B}{2}\right) \sin \left(\frac{A - C}{2}\right) \sin \left(\frac{B - A}{2}\right) ,$

$= - 4 \sin \left(- \frac{B - C}{2}\right) \sin \left(- \frac{C - A}{2}\right) \sin \left(- \frac{A - B}{2}\right) ,$

$= + 4 \sin \left(\frac{B - C}{2}\right) \sin \left(\frac{C - A}{2}\right) \sin \left(\frac{A - B}{2}\right) ,$

$\text{=The R.H.S.}$

Enjoy Maths.!