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Câu hỏi: a) Chứng minh rằng nếu \(\cos \left( {\alpha + \beta } \right) = 0\) thì \(\sin \left( {\alpha + 2\beta } \right) = \sin \alpha \).
b) Chứng minh rằng nếu \(\sin \left( {2\alpha + \beta } \right) = 3\sin \beta \) và \(\cos \alpha \ne 0,\cos \left( {\alpha + \beta } \right) \ne 0\) thì \(\tan \left( {\alpha + \beta } \right) = 2\tan \alpha \)
b) Chứng minh rằng nếu \(\sin \left( {2\alpha + \beta } \right) = 3\sin \beta \) và \(\cos \alpha \ne 0,\cos \left( {\alpha + \beta } \right) \ne 0\) thì \(\tan \left( {\alpha + \beta } \right) = 2\tan \alpha \)
Lời giải chi tiết
a) Nếu \(\cos \left( {\alpha + \beta } \right) = 0\) thì
\(\begin{array}{l}\sin \left( {\alpha + 2\beta } \right) = \sin \alpha \cos 2\beta + \sin 2\beta \cos \alpha \\ = \sin \alpha \left({1 - 2{{\sin }^2}\beta } \right) + 2\sin \beta \cos \beta \cos \alpha \\ = \sin \alpha + 2\sin \beta \left({ - \sin \alpha \sin \beta + \cos \alpha \cos \beta } \right)\\ = \sin \alpha + 2\sin \beta \cos \left({\alpha + \beta } \right) = \sin \alpha \end{array}\)
Ta có
\(\begin{array}{l}\sin \left( {2\alpha + \beta } \right) = 3\sin \beta \\ \Leftrightarrow 2\sin \alpha \cos \alpha \cos \beta + \left({2{{\cos }^2}\alpha - 1} \right)\sin \beta = 3\sin \beta \\ \Leftrightarrow \cos \alpha \sin \left({\alpha + \beta } \right) = 2\sin \beta \left(1 \right)\end{array}\)
Mặt khác
\(\begin{array}{l}\sin \left( {2\alpha + \beta } \right) = 2\sin \beta \\ \Leftrightarrow 2\sin \alpha \cos \alpha \cos \beta + \left({1 - 2{{\sin }^2}\alpha } \right)\sin \beta = 3sin\beta \\ \Leftrightarrow \sin \alpha \cos \left({\alpha + \beta } \right) = \sin \beta \left(2 \right)\end{array}\)
Từ (1) và (2) suy ra \(\cot \alpha \tan \left( {\alpha + \beta } \right) = 2.\) Do đó \(\tan \left( {\alpha + \beta } \right) = 2\tan \alpha .\)
a) Nếu \(\cos \left( {\alpha + \beta } \right) = 0\) thì
\(\begin{array}{l}\sin \left( {\alpha + 2\beta } \right) = \sin \alpha \cos 2\beta + \sin 2\beta \cos \alpha \\ = \sin \alpha \left({1 - 2{{\sin }^2}\beta } \right) + 2\sin \beta \cos \beta \cos \alpha \\ = \sin \alpha + 2\sin \beta \left({ - \sin \alpha \sin \beta + \cos \alpha \cos \beta } \right)\\ = \sin \alpha + 2\sin \beta \cos \left({\alpha + \beta } \right) = \sin \alpha \end{array}\)
Ta có
\(\begin{array}{l}\sin \left( {2\alpha + \beta } \right) = 3\sin \beta \\ \Leftrightarrow 2\sin \alpha \cos \alpha \cos \beta + \left({2{{\cos }^2}\alpha - 1} \right)\sin \beta = 3\sin \beta \\ \Leftrightarrow \cos \alpha \sin \left({\alpha + \beta } \right) = 2\sin \beta \left(1 \right)\end{array}\)
Mặt khác
\(\begin{array}{l}\sin \left( {2\alpha + \beta } \right) = 2\sin \beta \\ \Leftrightarrow 2\sin \alpha \cos \alpha \cos \beta + \left({1 - 2{{\sin }^2}\alpha } \right)\sin \beta = 3sin\beta \\ \Leftrightarrow \sin \alpha \cos \left({\alpha + \beta } \right) = \sin \beta \left(2 \right)\end{array}\)
Từ (1) và (2) suy ra \(\cot \alpha \tan \left( {\alpha + \beta } \right) = 2.\) Do đó \(\tan \left( {\alpha + \beta } \right) = 2\tan \alpha .\)