Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos 2x = 2{\cos ^2}x - 1\\
\sin 2x = 2\sin x.\cos x\\
2\sin x.\sin y = - \left( {\cos \left( {x + y} \right) - \cos \left( {x - y} \right)} \right)\\
\cot \frac{C}{2} = \frac{{2\sin A.\sin B}}{{\sin C}}\\
\Leftrightarrow \frac{{\cos \frac{C}{2}}}{{\sin \frac{C}{2}}} = \frac{{2\sin A.\sin B}}{{2.\sin \frac{C}{2}.\cos \frac{C}{2}}}\\
\Leftrightarrow \frac{{\cos \frac{C}{2}}}{{\sin \frac{C}{2}}}.2.\sin \frac{C}{2}.\cos \frac{C}{2} = 2\sin A.\sin B\\
\Leftrightarrow 2{\cos ^2}\frac{C}{2} = 2\sin A.\sin B\\
\Leftrightarrow \cos C + 1 = - \left( {\cos \left( {A + B} \right) - \cos \left( {A - B} \right)} \right)\\
\Leftrightarrow \cos C + 1 = - \left( { - \cos \left( {180^\circ - A - B} \right) - \cos \left( {A - B} \right)} \right)\\
\Leftrightarrow \cos C + 1 = - \left( { - \cos C - \cos \left( {A - B} \right)} \right)\\
\Leftrightarrow \cos C + 1 = \cos C + \cos \left( {A - B} \right)\\
\Leftrightarrow \cos \left( {A - B} \right) = 1\\
\Leftrightarrow \widehat A - \widehat B = 0^\circ \\
\Leftrightarrow \widehat A = \widehat B
\end{array}\)
Vậy tam giác ABC cân tại C.
