Câu 6.38 trang 202 SBT Đại số 10 Nâng cao

Chứng minh rằng, với mọi \(\alpha \), với mọi số nguyên k, ta có:

\(\sin \left( {\alpha  + k\dfrac{\pi }{2}} \right) = \left\{ \begin{array}{l}{\left( { - 1} \right)^l}\sin \alpha \,\,\,\,\,nếu\,\,k - 2l\\{\left( { - 1} \right)^l}\cos \alpha \,\,\,\,nếu\,\,k = 2l + 1;\end{array} \right.\)

\(\cos \left( {\alpha  + k\dfrac{\pi }{2}} \right) = \left\{ \begin{array}{l}{\left( { - 1} \right)^l}\cos \alpha \,\,\,\,\,nếu\,\,k = 2l\\{\left( { - 1} \right)^{l + 1}}\sin \alpha \,\,\,\,nếu\,\,k = 2l + 1;\end{array} \right.\)

\(\tan \left( {\alpha  + k\dfrac{\pi }{2}} \right) = \left\{ \begin{array}{l}\tan \alpha \,\,\,\,\,\,\,\,nếu\,\,k = 2l + 1\\ - \cot \alpha \,\,\,\,nếu\,\,k = 2l + 1\,\end{array} \right.\)

(khi các biểu thức này có nghĩa)

Giải:

• \(\sin \left( {\alpha  + 2l\dfrac{\pi }{2}} \right) = \sin \left( {\alpha  + l\pi } \right) = {\left( { - 1} \right)^l}\sin \alpha \);

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

• \(\begin{array}{l}\cos \left( {\alpha  + 2l\dfrac{\pi }{2}} \right) = \cos \left( {\alpha  + l\pi } \right) = {\left( { - 1} \right)^l}\cos \alpha \\\cos \left[ {\alpha  + \left( {2l + 1} \right)\dfrac{\pi }{2}} \right] = \cos \left( {\alpha  + \dfrac{\pi }{2} + l\pi } \right)\\ = {\left( { - 1} \right)^l}\cos \left( {\alpha  + \dfrac{\pi }{2}} \right) = {\left( { - 1} \right)^l}\left( { - \sin \alpha } \right)\\ = {\left( { - 1} \right)^{l + 1}}\sin \alpha \end{array}\)

• Từ đó

\(\begin{array}{l}\tan \left( {\alpha  + 2l\dfrac{\pi }{2}} \right) = \tan \alpha ;\\\tan \left[ {\alpha  + \left( {2l + 1} \right)\dfrac{\pi }{2}} \right] =  - \cot \alpha .\end{array}\)

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