Question

sin(a)+sin(b)+sin(c) transformat in produs
nu ma intereseaza neaparat rezultatul, mai degraba algoritmul de calcul

Answer 1

Răspuns:

$sinA+sinB+sinC=2sin\frac{A+B}{2}*cos\frac{A-B}{2}+sinC$

Stim ca $A+B+C=180^*=>A+B=180-C$

$sin\frac{A+B}{2}=sin\frac{180^*-C}{2}=sin(90^*-\frac{C}{2})=cos \frac{C}{2}$

$sinC=sin(2*\frac{C}{2})=2sin\frac{C}{2}*cos\frac{C}{2}$

Inlocuim:

$2cos\frac{C}{2}*cos\frac{A-B}{2}+2sin\frac{C}{2}*cos\frac{C}{2}$

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