Câu 6.57 trang 206 SBT Đại số 10 Nâng cao

Xét các biểu thức

\(\begin{array}{l}S = \sin \alpha  + sin2\alpha  + sin3\alpha  +  \ldots  + \sin n\alpha ,\\T = 1 + \cos \alpha  + \cos 2\alpha  + \cos 3\alpha  +  \ldots cosn\alpha \end{array}\)

(\(n\) là một số nguyên dương)

Chứng minh

a) \(S\sin \dfrac{\alpha }{2} = \sin \dfrac{{n\alpha }}{2}\sin \dfrac{{\left( {n + 1} \right)\alpha }}{2}\)

b) \(T\sin \dfrac{\alpha }{2} = \cos \dfrac{{n\alpha }}{2}\sin \dfrac{{\left( {n + 1} \right)\alpha }}{2}\)

Giải:

a) Với \(k = 1,2,3, \ldots ,n,\) ta có:

\(\sin k\alpha \sin \dfrac{\alpha }{2} = \dfrac{1}{2}\left[ {\cos \dfrac{{\left( {2k - 1} \right)\alpha }}{2} - \cos \dfrac{{\left( {2k + 1} \right)\alpha }}{2}} \right]\)

Nên

\(\begin{array}{l}S.\sin \dfrac{\alpha }{2} = \dfrac{1}{2}\left[ {\left( {\cos \dfrac{\alpha }{2} - \cos \dfrac{{3\alpha }}{2}} \right) + \left( {\cos \dfrac{{3\alpha }}{2} - \cos \dfrac{{5\alpha }}{2}} \right)} \right.\\\left. { +  \ldots  + \left( {\cos \dfrac{{\left( {2n - 1} \right)\alpha }}{2} - \cos \dfrac{{\left( {2n + 1} \right)\alpha }}{2}} \right)} \right]\\ = \dfrac{1}{2}\left[ {\left( {\cos \dfrac{\alpha }{2} - \cos \dfrac{{\left( {2n + 1} \right)\alpha }}{2}} \right)} \right]\\ = \sin \dfrac{{n\alpha }}{2}\sin \dfrac{{\left( {n + 1} \right)\alpha }}{2}\end{array}\)

b) Với \(k = 1,2,3 \ldots ,n,\) ta có:

\(\cos k\alpha \sin \dfrac{\alpha }{2} = \dfrac{1}{2}\left[ {\sin \dfrac{{\left( {2k + 1} \right)\alpha }}{2} - \sin \dfrac{{\left( {2k - 1} \right)\alpha }}{2}} \right]\)

nên

\(\begin{array}{l}T\sin \dfrac{\alpha }{2} = \sin \dfrac{\alpha }{2} + \dfrac{1}{2}\left[ {\left( {\sin \dfrac{{3\alpha }}{2} - \sin \dfrac{\alpha }{2}} \right) + \left( {\sin \dfrac{{5\alpha }}{2} - \sin \dfrac{{3\alpha }}{2}} \right)} \right.\\\left. { +  \ldots  + \left( {\sin \dfrac{{\left( {2n + 1} \right)\alpha }}{2} - \sin \dfrac{{\left( {2n - 1} \right)\alpha }}{2}} \right)} \right]\\ = \dfrac{1}{2}\left[ {\sin \dfrac{{\left( {2n + 1} \right)\alpha }}{2} + \sin \dfrac{\alpha }{2}} \right]\\ = \cos \dfrac{{n\alpha }}{2}\sin \dfrac{{\left( {n + 2} \right)\alpha }}{2}\end{array}\)

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