Cho \(\sin \alpha = \frac{4}{5}\) \(\left( {0^\circ < \alpha < 90^\circ } \right)\). Tính \(\cos \alpha \).

Ta có \[P = \sin \left( {90^\circ  - \alpha } \right) - \cos \left( {180^\circ  - \alpha } \right) = \cos \alpha  - \left( { - \cos \alpha } \right) = 2\cos \alpha \].

Mặt khác \({\sin ^2}\alpha  + {\cos ^2}\alpha  = 1 \Rightarrow {\cos ^2}\alpha  = 1 - {\sin ^2}\alpha  = 1 - {\left( {\frac{1}{3}} \right)^2} = \frac{8}{9} \Leftrightarrow \left[ \begin{array}{l}\cos \alpha  = \frac{{2\sqrt 2 }}{3}\\\cos \alpha  =  - \frac{{2\sqrt 2 }}{3}\end{array} \right.\).

Lại có \(0^\circ  < \alpha  < 90^\circ \) nên \(\cos \alpha  > 0\), từ đó ta được \(\cos \alpha  = \frac{{2\sqrt 2 }}{3}\).

Vậy \[P = 2\cos \alpha  = \frac{{4\sqrt 2 }}{3} \approx 1,89\].

Đáp án: \(1,89\).

a) Đúng. Ta có \(\cot \alpha  = \frac{1}{{\tan \alpha }} = \frac{1}{2}\).

b) Sai. \(\tan \alpha  = \frac{{\sin \alpha }}{{{\rm{cos}}\alpha }} = 2 > 0 \Rightarrow \sin \alpha  \cdot {\rm{cos}}\alpha  > 0\).

c) Đúng. Vì \(0^\circ  < \alpha  < 90^\circ \) nên \({\rm{cos}}\alpha  > 0\).

Ta có \(1 + {\tan ^2}\alpha  = \frac{1}{{{\rm{co}}{{\rm{s}}^2}\alpha }} \Rightarrow {\rm{co}}{{\rm{s}}^2}\alpha  = \frac{1}{{1 + {2^2}}} = \frac{1}{5} \Rightarrow {\rm{cos}}\alpha  = \frac{{\sqrt 5 }}{5} = \frac{1}{{\sqrt 5 }}\).

d) Sai. Ta có \(\tan \alpha  = \frac{{\sin \alpha }}{{{\rm{cos}}\alpha }} \Rightarrow \sin \alpha  = \tan \alpha  \cdot {\rm{cos}}\alpha  = \frac{{2\sqrt 5 }}{5}\).

Suy ra \({\rm{sin}}\alpha \,{\rm{ + }}\,{\rm{cos}}\alpha  = \frac{{2\sqrt 5 }}{5} + \frac{{\sqrt 5 }}{5} = \frac{{3\sqrt 5 }}{5}\).