Câu 6.52 trang 205 SBT Toán Đại 10 Nâng cao...

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 $\tan \left( {\alpha  + \beta } \right) = 2\tan \alpha$.

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 $\tan \left( {\alpha  + \beta } \right) = 2\tan \alpha .$