Đáp án:
Giải thích các bước giải:
Câu 248:
\(\begin{array}{l}
\cos A + \cos B + \cos C - 1\\
= 2\cos \frac{{A + B}}{2}\cos \frac{{A - B}}{2} - 2{\sin ^2}\frac{C}{2}\\
= 2\sin \frac{C}{2}\cos \frac{{A - B}}{2} - 2{\sin ^2}\frac{C}{2}\\
= 2\sin \frac{C}{2}\left( {\cos \frac{{A - B}}{2} - \sin \frac{C}{2}} \right)\\
= 2\sin \frac{C}{2}\left( {\cos \frac{{A - B}}{2} - \cos \frac{{A + B}}{2}} \right)\\
= 2\sin \frac{C}{2}\left[ { - 2\sin \frac{A}{2}\sin \left( { - \frac{B}{2}} \right)} \right]\\
= 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}
\end{array}\)
Suy ra \(\cos A + \cos B + \cos C = 1 + 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\) hay \(a=1,b=4\) hay \(a.b=4\)
