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Cho biết \(\sin {15^0} = \dfrac{{\sqrt 6 – \sqrt 2 }}{4}.\)
a) Tính \(\tan15^0\).
b) Chứng minh \(2\sin 15^0\cos 15^0=\sin 30^0\).
Giải
a)
\({\cos ^2}{15^0} = 1 – {\left( {\dfrac{{\sqrt 6 – \sqrt 2 }}{4}} \right)^2}\)
\(= \dfrac{{8 + 2\sqrt {12} }}{{16}}\)
\(= \dfrac{{{{(\sqrt 6 )}^2} + 2\sqrt 6 \sqrt 2 + {{(\sqrt 2 )}^2}}}{{16}}\)
\(= \dfrac{{{{(\sqrt 6 + \sqrt 2 )}^2}}}{{16}}.\)
Do \(15^0<90^0\) nên \(\cos 15^0>0\), suy ra \(\cos {15^0} = \dfrac{{\sqrt 6 + \sqrt 2 }}{4}.\)
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\(\tan {15^0} = \dfrac{{\sin {{15}^0}}}{{\cos {{15}^0}}}\)
\(= \dfrac{{\sqrt 6 – \sqrt 2 }}{{\sqrt 6 + \sqrt 2 }}\)
\(= \dfrac{{{{(\sqrt 6 – \sqrt 2 )}^2}}}{{6 – 2}} = 2 – \sqrt 3 .\)
b)
\(2\sin {15^0}\cos {15^0} \)
\(= 2.\dfrac{{\sqrt 6 – \sqrt 2 }}{4}.\dfrac{{\sqrt 6 + \sqrt 2 }}{{42}} \)
\(= \dfrac{1}{2} = {\mathop{\rm s}\nolimits} {\rm{in3}}{0^0}\).