Câu 6.64 trang 207 SBT Toán Đại 10 Nâng cao: Chứng minh rằng...

Chứng minh rằng

$\cos \dfrac{\pi }{{32}} = \dfrac{1}{2}\sqrt {2 + \sqrt {2 + \sqrt {2 + \sqrt 2 } } } .$

Ta có $\cos \dfrac{\pi }{4} = \dfrac{1}{2}\sqrt 2 ;$

$\cos \dfrac{\pi }{8} = \sqrt {\dfrac{{1 + \cos \dfrac{\pi }{4}}}{2}}$

$= \sqrt {\dfrac{{2 + \sqrt 2 }}{4}}  = \dfrac{1}{2}\sqrt {2 + \sqrt 2 } .$

$\begin{array}{l}\cos \dfrac{\pi }{{16}} = \sqrt {\dfrac{{1 + \cos \dfrac{\pi }{8}}}{2}} \\ = \sqrt {\dfrac{{2 + \sqrt {2 + \sqrt 2 } }}{4}}  = \dfrac{1}{2}\sqrt {2 + \sqrt {2 + \sqrt 2 } } ;\end{array}$

$\begin{array}{l}\cos \dfrac{\pi }{{32}} = \sqrt {\dfrac{{1 + \cos \dfrac{\pi }{{16}}}}{2}} \\ = \sqrt {\dfrac{{2 + \sqrt {2 + \sqrt {2 + \sqrt 2 } } }}{4}} \\ = \dfrac{1}{2}\sqrt {2 + \sqrt {2 + \sqrt {2 + \sqrt 2 } } } .\end{array}$