Giả sử các biểu thức sau có nghĩa, chứng minh rằng:
a) \(\tan \alpha = \dfrac{{\sin \alpha + \sin 2\alpha }}{{1 + \cos \alpha + \cos 2\alpha }};\)
b) \({\tan ^2}\alpha = \dfrac{{2\sin 2\alpha - \sin 4\alpha }}{{2\sin 2\alpha + \sin 4\alpha }}\).
a)
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\(\begin{array}{l}\dfrac{{\sin \alpha + \sin 2\alpha }}{{1 + \cos \alpha + \cos 2\alpha }} = \dfrac{{\sin \alpha \left( {1 + 2\cos \alpha } \right)1}}{{1 + \cos \alpha + 2{{\cos }^2}\alpha - 1}}\\ = \dfrac{{\sin \alpha \left( {1 + 2\cos \alpha } \right)}}{{\cos \alpha \left( {1 + 2\cos \alpha } \right)}} = \tan \alpha \end{array}\)
b)
\(\begin{array}{l}\dfrac{{2\sin 2\alpha - \sin 4\alpha }}{{2\sin 2\alpha + \sin 4\alpha }} = \dfrac{{2\sin 2\alpha \left( {1 - \cos 2\alpha } \right)}}{{2\sin 2\alpha \left( {1 + \cos \alpha } \right)}}\\ = \dfrac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} = {\tan ^2}\alpha .\end{array}\)