Question

Chứng minh:

\(\dfrac{1-cos\alpha-cos2\alpha+cos3\alpha}{1-2cos\alpha}=2sin^2\alpha\)

Answer 1

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.