Lời giải:
Áp dụng công thức: \(\cos a-\cos b=-2\sin \frac{a+b}{2}\sin \frac{a-b}{2}\) ta có:
\(\text{VT}=\frac{1-\cos 2\alpha-(\cos 3\alpha-\cos \alpha)}{1-2\cos \alpha}=\frac{1-\cos 2\alpha+(-2)\sin 2\alpha\sin \alpha}{1-2\cos \alpha}(*)\)
Lại có:
\(\cos 2\alpha=\cos ^2\alpha-\sin ^2\alpha=(\cos ^2\alpha+\sin ^2\alpha)-2\sin ^2\alpha\)
\(=1-2\sin ^2\alpha\)
\(\Rightarrow 1-\cos 2\alpha=2\sin ^2\alpha(**)\)
Từ \((*); (**)\Rightarrow \text{VT}=\frac{2\sin ^2\alpha-2\sin 2\alpha\sin \alpha}{1-2\cos \alpha}\)
\(=\frac{2\sin ^2\alpha-4\sin \alpha\cos \alpha\sin \alpha}{1-2\cos \alpha}=\frac{2\sin ^2\alpha(1-2\cos \alpha)}{1-2\cos \alpha}=2\sin ^2\alpha\)
Ta có đpcm.
