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Câu hỏi: Chứng minh rằng
\(\cos \dfrac{\pi }{{32}} = \dfrac{1}{2}\sqrt {2 + \sqrt {2 + \sqrt {2 + \sqrt 2 } } } .\)
\(\cos \dfrac{\pi }{{32}} = \dfrac{1}{2}\sqrt {2 + \sqrt {2 + \sqrt {2 + \sqrt 2 } } } .\)
Lời giải chi tiết
Ta có \(\cos \dfrac{\pi }{4} = \dfrac{1}{2}\sqrt 2 ;\)
\(\cos \dfrac{\pi }{8} = \sqrt {\dfrac{{1 + \cos \dfrac{\pi }{4}}}{2}}\)
\( = \sqrt {\dfrac{{2 + \sqrt 2 }}{4}} = \dfrac{1}{2}\sqrt {2 + \sqrt 2 } .\)
\(\begin{array}{l}\cos \dfrac{\pi }{{16}} = \sqrt {\dfrac{{1 + \cos \dfrac{\pi }{8}}}{2}} \\ = \sqrt {\dfrac{{2 + \sqrt {2 + \sqrt 2 } }}{4}} = \dfrac{1}{2}\sqrt {2 + \sqrt {2 + \sqrt 2 } } ;\end{array}\)
\(\begin{array}{l}\cos \dfrac{\pi }{{32}} = \sqrt {\dfrac{{1 + \cos \dfrac{\pi }{{16}}}}{2}} \\ = \sqrt {\dfrac{{2 + \sqrt {2 + \sqrt {2 + \sqrt 2 } } }}{4}} \\ = \dfrac{1}{2}\sqrt {2 + \sqrt {2 + \sqrt {2 + \sqrt 2 } } } .\end{array}\)
Ta có \(\cos \dfrac{\pi }{4} = \dfrac{1}{2}\sqrt 2 ;\)
\(\cos \dfrac{\pi }{8} = \sqrt {\dfrac{{1 + \cos \dfrac{\pi }{4}}}{2}}\)
\( = \sqrt {\dfrac{{2 + \sqrt 2 }}{4}} = \dfrac{1}{2}\sqrt {2 + \sqrt 2 } .\)
\(\begin{array}{l}\cos \dfrac{\pi }{{16}} = \sqrt {\dfrac{{1 + \cos \dfrac{\pi }{8}}}{2}} \\ = \sqrt {\dfrac{{2 + \sqrt {2 + \sqrt 2 } }}{4}} = \dfrac{1}{2}\sqrt {2 + \sqrt {2 + \sqrt 2 } } ;\end{array}\)
\(\begin{array}{l}\cos \dfrac{\pi }{{32}} = \sqrt {\dfrac{{1 + \cos \dfrac{\pi }{{16}}}}{2}} \\ = \sqrt {\dfrac{{2 + \sqrt {2 + \sqrt {2 + \sqrt 2 } } }}{4}} \\ = \dfrac{1}{2}\sqrt {2 + \sqrt {2 + \sqrt {2 + \sqrt 2 } } } .\end{array}\)