Câu 6.53 trang 205 SBT Đại số 10 Nâng cao

Chứng minh

a) \(4\cos {15^0}\cos {21^0}\cos {24^0} - \cos {12^0} - \cos {18^0}\)

\(= \dfrac{{1 + \sqrt 3 }}{2}\);

b) \(\tan {30^0} + \tan {40^0} + \tan {50^0} + \tan {60^0}\)

\(= \dfrac{{8\sqrt 3 }}{3}\cos {20^0}\);

c) \(\dfrac{1}{{\sin {{18}^0}}} - \dfrac{1}{{\sin {{54}^0}}} = 2;\)

d) \(\tan {9^0} - \tan {27^0} - \tan {63^0} + \tan {81^0} = 4\).

Giải:

a)

\(\begin{array}{l}4\cos {15^0}\cos {21^0}\cos {24^0} - \cos {12^0} - \cos {18^0}\\ = 2\cos {15^0}\left( {\cos {{45}^0} + \cos {3^0}} \right) - 2\cos {15^0}\cos {3^0}\\ = 2\cos {15^0}\cos {45^0}\\ = \cos {60^0} + \cos {30^0} = \dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2}.\end{array}\) 

b)

\(\begin{array}{l}\tan {30^0} + \tan {40^0} + \tan {50^0} + \tan {60^0}\\ = \dfrac{{\sin {{90}^0}}}{{\cos {{30}^0}\cos {{60}^0}}} + \dfrac{{\sin {{90}^0}}}{{\cos {{40}^0}\cos {{50}^0}}}\\ = \dfrac{{\cos {{90}^0} + \cos {{10}^0} + \cos {{90}^0} + \cos {{30}^0}}}{{\dfrac{1}{2}\cos {{10}^0}\cos {{30}^0}}}\\ = \dfrac{{4\cos {{20}^0}\cos {{10}^0}}}{{\cos {{10}^0}\cos {{30}^0}}}\\ = \dfrac{8}{{\sqrt 3 }}\cos {20^0} = \dfrac{{8\sqrt 3 }}{3}\cos {20^0}.\end{array}\)

c)

\(\begin{array}{l}\dfrac{1}{{\sin {{18}^0}}} - \dfrac{1}{{\sin {{54}^0}}} = \dfrac{{\sin {{54}^0} - \sin {{18}^0}}}{{\sin {{18}^0}\sin {{54}^0}}}\\ = \dfrac{{2\cos {{36}^0}\sin {{18}^0}}}{{\sin {{18}^0}\sin {{54}^0}}} = \dfrac{{2\cos {{36}^0}}}{{\sin {{54}^0}}}\\ = \dfrac{{2\cos {{36}^0}}}{{\cos {{36}^0}}} = 2.\end{array}\)

d)

\(\begin{array}{l}\tan {9^0} - \tan {27^0} - \tan {63^0} + \tan {81^0}\\ = \tan {9^0} + \tan {81^0} - \left( {\tan {{27}^0} + \tan {{63}^0}} \right)\\ = \left( {\dfrac{{\sin {9^0}}}{{\cos {9^0}}} + \dfrac{{\sin {{81}^0}}}{{\cos {{81}^0}}}} \right) - \left( {\dfrac{{\sin {{27}^0}}}{{\cos {{27}^0}}} + \dfrac{{\sin {{63}^0}}}{{\cos {{63}^0}}}} \right)\\ = \dfrac{1}{{\sin {9^0}\cos {9^0}}} - \dfrac{1}{{\sin {{27}^0}\cos {{27}^0}}}\\ = \dfrac{2}{{\sin {{18}^0}}} - \dfrac{2}{{\sin {{54}^0}}} = 2.2 = 4\end{array}\)

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