Question

If A+B+C=180, sinA-sinB+sinC=

Answer 1

$\textbf{Given:}$

$\displaystyle\,A+B+C=180^{\circ}$

$\displaystyle\,sin\,A-sin\,B+sin\,C$

$\text{Using}$

$\boxed{\bf\sin\,C-sin\,D=2\,cos(\frac{C+D}{2})\,sin(\frac{C-D}{2})}$

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

$=\displaystyle2\,cos(\frac{180^{\circ}-C}{2})\,sin(\frac{A-B}{2})+sin\,C$

$\text{using}$

$\boxed{\bf\sin\,A=2\,sin(\frac{A}{2})\,cos(\frac{A}{2})}$

$=\displaystyle\,2\,cos(90^{\circ}-(\frac{C}{2}))\,sin(\frac{A-B}{2})+2\,sin\,(\frac{C}{2})\,cos(\frac{c}{2})$

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

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

$=\displaystyle\,2\,sin(\frac{C}{2})[sin(\frac{A-B}{2})+cos(\frac{180^{\circ}-(A+B)}{2})]$

$=\displaystyle\,2\,sin(\frac{C}{2})\,[sin(\frac{A-B}{2})+cos\,(90^{\circ}-(\frac{A+B}{2}))]$

$\boxed{\bf\,sin(A+B)+sin(A-B)=2\,sin\,A\,cos\,B}$

$=\displaystyle\,2\,sin(\frac{C}{2})\,[sin(\frac{A-B}{2})+sin(\frac{A+B}{2})]$

$=\displaystyle\,2\,sin(\frac{C}{2})[2\,sin(\frac{A}{2})\,cos(\frac{B}{2})]$

$=\displaystyle\,4\,sin(\frac{A}{2})\,cos(\frac{B}{2})\,sin(\frac{C}{2})$

$\implies\boxed{\bf\,sin\,A-sin\,B+sin\,C=4\,sin(\frac{A}{2})\,cos(\frac{B}{2})\,sin(\frac{C}{2})}$