Phương pháp giải:

Sử dụng linh hoạt các công thức biến đổi tổng thành tích và tích thành tổng.

Lời giải chi tiết:

* Xét đáp án A:

$\begin{array}{l}\sin A + \sin B + \sin C = 2\sin \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2} + \sin C\\ = 2\sin \dfrac{{\pi  - C}}{2}\cos \dfrac{{A - B}}{2} + \sin C = 2\sin \left( {\dfrac{\pi }{2} - \dfrac{C}{2}} \right)\cos \dfrac{{A - B}}{2} + 2\sin \dfrac{C}{2}\cos \dfrac{C}{2}\\ = 2\cos \dfrac{C}{2}\cos \dfrac{{A - B}}{2} + 2\sin \dfrac{C}{2}\cos \dfrac{C}{2} = 2\cos \dfrac{C}{2}\left( {\cos \dfrac{{A - B}}{2} + \sin \dfrac{C}{2}} \right)\\ = 2\cos \dfrac{C}{2}\left( {\cos \dfrac{{A - B}}{2} + \cos \left( {\dfrac{\pi }{2} - \dfrac{C}{2}} \right)} \right) = 2\cos \dfrac{C}{2}\left( {\cos \dfrac{{A - B}}{2} + \cos \dfrac{{A + B}}{2}} \right)\\ = 2\cos \dfrac{C}{2}.2\cos \dfrac{A}{2}\cos \dfrac{B}{2} = 4\cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2}\end{array}$

$\Rightarrow$ Đáp án A đúng.

* Xét đáp án B:

$\begin{array}{l}\cos A + \cos B + \cos C = 1 + 4\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}\\ \Leftrightarrow \cos A + \cos B + \cos C - 1 = 4\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}\\ \Leftrightarrow 2\cos \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2} - 2{\sin ^2}\dfrac{C}{2} = 4\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}\\ \Leftrightarrow 2\cos \left( {\dfrac{\pi }{2} - \dfrac{C}{2}} \right)\cos \dfrac{{A - B}}{2} - 2{\sin ^2}\dfrac{C}{2} = 4\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}\\ \Leftrightarrow 2\sin \dfrac{C}{2}\cos \dfrac{{A - B}}{2} - 2{\sin ^2}\dfrac{C}{2} = 4\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}\\ \Leftrightarrow 2\sin \dfrac{C}{2}\left( {\cos \dfrac{{A - B}}{2} - \sin \dfrac{C}{2}} \right) = 4\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}\\ \Leftrightarrow \cos \dfrac{{A - B}}{2} - \cos \left( {\dfrac{\pi }{2} - \dfrac{C}{2}} \right) = 2\sin \dfrac{A}{2}\sin \dfrac{B}{2}\\ \Leftrightarrow \cos \dfrac{{A - B}}{2} - \cos \dfrac{{A + B}}{2} = 2\sin \dfrac{A}{2}\sin \dfrac{B}{2}\\ \Leftrightarrow  - 2\sin \dfrac{{A - B + A + B}}{4}\sin \dfrac{{A - B - A - B}}{4} = 2\sin \dfrac{A}{2}\sin \dfrac{B}{2}\\ \Leftrightarrow  - \sin \dfrac{A}{2}\sin \dfrac{{ - B}}{2} = \sin \dfrac{A}{2}\sin \dfrac{B}{2}\,\,\left( {luon\,\,dung} \right)\end{array}$

$\Rightarrow$ Đáp án B đúng.

* Xét đáp án C:

$\begin{array}{l}\,\,\,\,\sin 2A + \sin 2B + \sin 2C = 2\sin \left( {A + B} \right)\cos \left( {A - B} \right) + \sin 2C\\ = 2\sin \left( {\pi  - C} \right)\cos \left( {A - B} \right) + \sin 2C = 2\sin C\cos \left( {A - B} \right) + 2\sin C\cos C\\ = 2\sin C\left( {\cos \left( {A - B} \right) + \cos C} \right) = 2\sin C.2\cos \dfrac{{C + A - B}}{2}\cos \dfrac{{C - A + B}}{2}\\ = 4\sin Ccos\dfrac{{\pi  - 2B}}{2}\cos \dfrac{{\pi  - 2A}}{2} = 4\sin C\cos \left( {\dfrac{\pi }{2} - B} \right)\cos \left( {\dfrac{\pi }{2} - A} \right)\\ = 4\sin C\sin B\sin A = 4\sin A\sin B\sin C\end{array}$

$\Rightarrow$ Đáp án C đúng.

Chọn D.