Câu 6.62 trang 207 SBT Đại số 10 Nâng cao

Chứng minh rằng với mọi \(\alpha \) mà \(\sin 2\alpha  \ne 0\), ta có

\(\sin \left( {\cot \alpha } \right) + \sin \left( {\tan \alpha } \right) = 2\sin \left( {\dfrac{1}{{\sin 2\alpha }}} \right)\cos \left( {\cot 2\alpha } \right)\)

Giải:

Đặt \(u = \dfrac{1}{2}\left( {\tan \alpha  + \cot \alpha } \right),\) \(v = \dfrac{1}{2}\left( {\tan \alpha  - \cot \alpha } \right)\) thì \(u + v = \tan \alpha ,u - v = \cot \alpha \). Khi đó ta có

\(\begin{array}{l}\sin \left( {\tan \alpha } \right) + \sin \left( {\cot \alpha } \right)\\ = \sin \left( {u + v} \right) + \sin \left( {u - v} \right)\\ = 2\sin u\cos v\\ = 2\sin \left[ {\dfrac{1}{2}\left( {\dfrac{{\sin \alpha }}{{\cos \alpha }} + \dfrac{{\cos \alpha }}{{\sin \alpha }}} \right)} \right].\cos \left[ {\dfrac{1}{2}\left( {\dfrac{{\sin \alpha }}{{\cos \alpha }} - \dfrac{{\cos \alpha }}{{\sin \alpha }}} \right)} \right]\\ = 2\sin \left( {\dfrac{1}{{2\sin \alpha \cos \alpha }}} \right).\cos \left( {\dfrac{{{{\sin }^2}\alpha  - {{\cos }^2}\alpha }}{{2\sin \alpha \cos \alpha }}} \right)\\ = 2\sin \left( {\dfrac{1}{{\sin 2\alpha }}} \right).\cos \left( {\cot 2\alpha } \right).\end{array}\)

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