Q62 / 160 · TS EAMCET 2018 (05 May Shift 2) · Mathematics

If $A+B+C=2 S$, then $\sin (S-A)+\sin (S-B)-\sin C=$

  • A
    $-4 \sin \frac{S-A}{2} \sin \frac{S-B}{2} \sin \frac{C}{2}$
  • Correct answer
    B
    $4 \sin \frac{S-A}{2} \sin \frac{S-B}{2} \sin \frac{C}{2}$
  • C
    $-4 \sin \frac{S-A}{2} \sin \frac{S-B}{2} \cos \frac{C}{2}$
  • D
    $4 \sin \frac{S-A}{2} \sin \frac{S-B}{2} \cos \frac{C}{2}$

Solution

Consider, $\sin (S-A)+\sin (S-B)-\sin C$ $$ \begin{aligned} & =2 \sin \left(\frac{2 S-(A+B)}{2}\right) \cos \left(\frac{B-A}{2}\right)-\sin C \\ & =2 \sin \left(\frac{C}{2}\right) \cos \left(\frac{B-A}{2}\right)-2 \sin \frac{C}{2} \cos \frac{C}{2} \\ & =2 \sin \frac{C}{2}\left[\cos \left(\frac{B-A}{2}\right)-\cos \frac{C}{2}\right] \\ & =2 \sin \frac{C}{2}\left[-2 \sin \left(\frac{B-A+C}{2}\right) \sin \left(\frac{B-A-C}{2}\right)\right] \\ & =4 \sin \frac{C}{2} \sin \left(\frac{B+C-A}{4}\right) \sin \left(\frac{A+C-B}{4}\right) \\ & =4 \sin \frac{C}{2} \sin \left(\frac{2 S-2 A}{4}\right) \sin \left(\frac{2 S-2 B}{4}\right) \\ & =4 \sin \frac{C}{2} \sin \left(\frac{S-A}{2}\right) \sin \left(\frac{S-B}{2}\right) \end{aligned} $$

Hence, the correct answer is option: 'B'.

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