Răspuns:
[tex]sinA+sinB+sinC=2sin\frac{A+B}{2}*cos\frac{A-B}{2}+sinC[/tex]
Stim ca [tex]A+B+C=180^*=>A+B=180-C[/tex]
[tex]sin\frac{A+B}{2}=sin\frac{180^*-C}{2}=sin(90^*-\frac{C}{2})=cos \frac{C}{2}[/tex]
[tex]sinC=sin(2*\frac{C}{2})=2sin\frac{C}{2}*cos\frac{C}{2}[/tex]
Inlocuim:
[tex]2cos\frac{C}{2}*cos\frac{A-B}{2}+2sin\frac{C}{2}*cos\frac{C}{2}[/tex]
[tex]2cos\frac{C}{2}(cos\frac{A-B}{2}+sin\frac{C}{2})[/tex]
