Đề bài

Cho \(\sin {15^ \circ } = \frac{{\sqrt 6  - \sqrt 2 }}{4}.\) a) Tính \(\sin {75^ \circ },\,\,\cos {105^ \circ },\,\,\tan {165^ \circ }.\) b) Tính giá trị của biểu thức \(A = \sin {75^ \circ }.\cos {165^ \circ } + \cos {105^ \circ }.\sin {165^ \circ }.\)

Lời giải chi tiết

Ta có: \({\sin ^2}{15^ \circ } + {\cos ^2}{15^ \circ } = 1\)\(\begin{array}{l} \Rightarrow \,\,{\cos ^2}{15^ \circ } = 1 - {\sin ^2}{15^ \circ } = \frac{{2 + \sqrt 3 }}{4}\\ \Rightarrow \,\,\cos {15^ \circ } = \sqrt {\frac{{2 + \sqrt 3 }}{4}}  = \sqrt {\frac{{8 + 4\sqrt 3 }}{{16}}}  = \sqrt {\frac{{{{\left( {\sqrt 6 } \right)}^2} + 2.\sqrt 6 .\sqrt 2  + {{\left( {\sqrt 2 } \right)}^2}}}{{16}}} \\ \Rightarrow \,\,\cos {15^ \circ } = \sqrt {\frac{{{{\left( {\sqrt 6  + \sqrt 2 } \right)}^2}}}{{16}}}  = \frac{{\sqrt 6  + \sqrt 2 }}{4}.\end{array}\)Ta có: \(\tan {15^ \circ } = \frac{{\sin {{15}^ \circ }}}{{\cos {{15}^ \circ }}} = \frac{{\sqrt 6  - \sqrt 2 }}{4}:\frac{{\sqrt 6  + \sqrt 2 }}{4} = 2 - \sqrt 3 \)a) \(\sin {75^ \circ } = \sin \left( {{{90}^ \circ } - {{15}^ \circ }} \right) = \cos {15^ \circ } = \frac{{\sqrt 6  + \sqrt 2 }}{4}.\)\(\cos {105^ \circ } = \cos \left( {{{180}^ \circ } - {{75}^ \circ }} \right) =  - \cos {75^ \circ } =  - \cos \left( {{{90}^ \circ } - {{15}^ \circ }} \right) =  - \sin {15^ \circ } = \frac{{\sqrt 2  - \sqrt 6 }}{4}.\) \(\tan {165^ \circ } = \tan \left( {{{180}^ \circ } - {{15}^ \circ }} \right) =  - \tan {15^ \circ } = \sqrt 3  - 2.\)b) \(A = \sin {75^ \circ }.\cos {165^ \circ } + \cos {105^ \circ }.\sin {165^ \circ }.\)Ta có: \(\cos {165^ \circ } = \cos \left( {{{180}^ \circ } - {{15}^ \circ }} \right) =  - \cos {15^ \circ } =  - \frac{{\sqrt 2  + \sqrt 6 }}{4}.\)Ta có: \(\sin {165^ \circ } = \tan {165^ \circ }.\cos {165^ \circ } =  - \left( {\sqrt 3  - 2} \right).\frac{{\sqrt 2  + \sqrt 6 }}{4} = \frac{{\sqrt 6  - \sqrt 2 }}{4}.\)\(\begin{array}{l}A = \sin {75^ \circ }.\cos {165^ \circ } + \cos {105^ \circ }.\sin {165^ \circ }\\A = \frac{{\sqrt 6  + \sqrt 2 }}{4}.\left( { - \frac{{\sqrt 2  + \sqrt 6 }}{4}} \right) + \frac{{\sqrt 2  - \sqrt 6 }}{4}.\frac{{\sqrt 6  - \sqrt 2 }}{4}\\A = \frac{{ - 4\sqrt 3  - 8}}{{16}} + \frac{{ - 8 + 16\sqrt 3 }}{{16}} = \frac{{ - 16}}{{16}} =  - 1.\end{array}\)