Question

A+B+C =2S , then sin S- sin( S-A) - sin (S-B) - sin (S-C) =​

Answer 1

Step-by-step explanation:

We have,

$\purple{ \bold{ \mathbb{ A + B + C = 2S}}}$

Now,

$\sin(S) - \sin(S -A) - \sin(S - B) - \sin(S - C)$

$= 2\cos \bigg( \frac{S +S - A}{2} \bigg)\sin \bigg ( \frac{ S - S + A}{2} \bigg) - 2\sin \bigg( \frac{2S - B - C}{2} \bigg)\cos\bigg( \frac{ C - B }{2} \bigg)$

$= 2\cos \bigg( \frac{2S - A}{2} \bigg)\sin \bigg ( \frac{ A}{2} \bigg) - 2\sin \bigg( \frac{2S - B - C}{2} \bigg)\cos\bigg( \frac{ C - B }{2} \bigg)$

$= 2\cos \bigg( \frac{B + C}{2} \bigg)\sin \bigg ( \frac{ A}{2} \bigg) - 2\sin \bigg( \frac{ A }{2} \bigg)\cos\bigg( \frac{ C - B }{2} \bigg)$

$= 2 \sin \bigg ( \frac{ A}{2} \bigg) \bigg \{ \cos \bigg( \frac{B + C}{2} \bigg)-\cos\bigg( \frac{ C - B }{2} \bigg) \bigg \}$

$= 2 \sin \bigg ( \frac{ A}{2} \bigg) \bigg \{ - 2\sin \frac{1}{2} \bigg( \frac{B + C}{2} + \frac{ C - B }{2} \bigg) \sin \frac{1}{2} \bigg( \frac{B + C}{2} - \frac{ C - B }{2} \bigg) \bigg \}$

$= 2 \sin \bigg ( \frac{ A}{2} \bigg) \bigg \{ - 2\sin \frac{1}{2} \bigg( \frac{B + C + C - B }{2} \bigg) \sin \frac{1}{2} \bigg( \frac{B + C - C + B }{2} \bigg) \bigg \}$

$= 2 \sin \bigg ( \frac{ A}{2} \bigg) \bigg \{ - 2\sin \frac{1}{2} \bigg( \frac{ C + C }{2} \bigg) \sin \frac{1}{2} \bigg( \frac{B + B }{2} \bigg) \bigg \}$

$= 2 \sin \bigg ( \frac{ A}{2} \bigg) \bigg \{ - 2\sin \frac{1}{2} \bigg( \frac{2 C }{2} \bigg) \sin \frac{1}{2} \bigg( \frac{2B }{2} \bigg) \bigg \}$

$= 2 \sin \bigg ( \frac{ A}{2} \bigg) \bigg \{ - 2\sin \bigg( \frac{ C }{2} \bigg) \sin \bigg( \frac{B }{2} \bigg) \bigg \}$

$= - 4 \sin \bigg ( \frac{ A}{2} \bigg) \sin \bigg( \frac{B }{2} \bigg) \sin \bigg( \frac{ C }{2} \bigg)$