Biết tan α = 2 và 0 ∘ < α < 90 ∘ . a) cot α = 1 2 . b) sin α ⋅ cos α < 0 . c) cos α = 1 √ 5 . d) cos α + sin α = 2 √ 5 5 .
Lời giải
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 \({\rm{sin}}\alpha \,{\rm{ + }}\,{\rm{cos}}\alpha = \frac{{2\sqrt 5 }}{5} + \frac{{\sqrt 5 }}{5} = \frac{{3\sqrt 5 }}{5}\).