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TS EAMCET 04 May 2019 Shift 1 Solved Paper

Section: Mathematics

Question : 22 of 160

Marks: +1, -0

If A + B + C = π, then sin A - sin B + sin C =

4 cos $\left(\frac{A}{2}\right)$ cos $\left(\frac{B}{2}\right)$ cos $\left(\frac{C}{2}\right)$
4 sin $\left(\frac{A}{2}\right)$ sin $\left(\frac{B}{2}\right)$ cos $\left(\frac{C}{2}\right)$
sin $\left(\frac{A}{2}\right)$ cos $\left(\frac{B}{2}\right)$ sin $\left(\frac{C}{2}\right)$
4 sin $\left(\frac{A}{2}\right)$ cos $\left(\frac{B}{2}\right)$ sin $\left(\frac{C}{2}\right)$

Solution:

Consider the expression,

A+B+C=\pi

\text{So}

\sin A-\sin B+\sin C \;\; =2 \cos \frac{C}{2}\left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)+2 \sin \frac{C}{2} \cos \frac{C}{2}

\;\; =2 \sin \frac{C}{2}\left(\frac{A-B}{2}\right)+2 \sin \frac{C}{2} \cos \frac{C}{2}

\;\; =2 \sin \frac{C}{2}\left[\sin \left(\frac{A-B}{2}\right)+\cos \frac{C}{2}\right]

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

Further simplify the above

\sin A-\sin B+\sin C \;\; =2 \sin \frac{C}{2}\left(2 \sin \frac{A}{2} \cos \frac{B}{2}\right)

\;\; =4 \sin \left(\frac{A}{2}\right) \cos \left(\frac{B}{2}\right) \sin \left(\frac{C}{2}\right)

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