# Can you prove that?

## $\sin \left(S - A\right) \cdot \sin \left(S - B\right) + \sin \left(S - C\right) \cdot \sin \left(S\right) = \sin A \cdot \sin B \mathmr{if} A + B + C = 2 S$

Jul 15, 2018

#### Explanation:

Things to know:-$2 \sin A \sin B = \cos \left(A - B\right) - \cos \left(A + B\right) \mathmr{and} \cos \left(- A\right) = \cos A$

$L H S = \sin \left(S - A\right) \sin \left(S - B\right) + \sin \left(S - C\right) \sin S$

$= \frac{1}{2} \left[2 \sin \left(S - A\right) \sin \left(S - B\right) + 2 \sin \left(S - C\right) \sin S\right]$

$= \frac{1}{2} \left[\cos \left(\cancel{S} - A \cancel{- S} + B\right) - \cos \left(S - A + S - B\right) + \cos \left(\cancel{S} - C \cancel{- S}\right) - \cos \left(S - C + S\right)\right]$

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

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

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

$= \frac{1}{2} \left[2 \sin A \sin B\right] = \sin A \sin B = R H S$