\(\cos {\rm{ }}{44^o},{\rm{ sin }}{50^o},{\rm{ sin }}{70^o},{\rm{ cos }}{55^o}\) A \(\cos {44^0} < \sin {50^0} < \sin {70^0} < \cos {55^0}\) B \(\cos{44^0} < \cos {55^0} < \sin {50^0} < \sin {70^0}\) C \(\cos {55^0} < \cos {44^0} < \sin {50^0} < \sin {70^0}\) D \(\cos {55^0} < \cos {44^0} < \sin {70^0} < \sin {50^0}\)

\(\cos {\rm{ }}{44^o},{\rm{ sin }}{50^o},{\rm{ sin }}{70^o},{\rm{ cos }}{55^o}\)

A \(\cos {44^0} < \sin {50^0} < \sin {70^0} < \cos {55^0}\)

B \(\cos{44^0} < \cos {55^0} < \sin {50^0} < \sin {70^0}\)

C \(\cos {55^0} < \cos {44^0} < \sin {50^0} < \sin {70^0}\)

D \(\cos {55^0} < \cos {44^0} < \sin {70^0} < \sin {50^0}\)

Hướng dẫn Chọn đáp án là: C

Phương pháp giải:

Áp dụng \(0 < \alpha < \beta < {90^0} \Rightarrow \left\{ \begin{array}{l}\sin \alpha < \sin \beta \\cos\alpha > cos\beta \end{array} \right..\)

Ta có: \(\alpha + \beta = {90^0} \Rightarrow \left\{ \begin{array}{l}\sin \alpha = \cos \beta \\\cos \alpha = \sin \beta \end{array} \right.\)

Lời giải chi tiết:

\(\cos {\rm{ }}{44^o},{\rm{ sin }}{50^o},{\rm{ sin }}{70^o},{\rm{ cos }}{55^o}\)

Ta có: \(\left\{ \begin{array}{l}\cos {\rm{ }}{44^0} = \cos {\rm{ }}\left( {{{90}^0} – {{46}^0}} \right) = {\rm{sin 4}}{{\rm{6}}^0}\\\cos {\rm{ 5}}{{\rm{5}}^0} = \cos {\rm{ }}\left( {{{90}^0} – {{35}^0}} \right) = {\rm{sin 3}}{{\rm{5}}^0}\end{array} \right.\)

Vì \({35^0} < {46^0} < {50^0} < {70^0}\)\( \Rightarrow {\rm{sin 3}}{5^o} < \sin {\rm{ }}{46^o} < {\rm{sin }}{50^o} < {\rm{sin }}{70^o}\)

\( \Rightarrow {\rm{cos }}{55^o} < \cos {\rm{ }}{44^o} < {\rm{sin }}{50^o} < {\rm{sin }}{70^o}.\)

Chọn C.