Câu 1. C
Ta có
\({\tan ^2}\alpha + {\cot ^2}\alpha \)
\(= {\left( {\tan \alpha + \cot \alpha } \right)^2} - 2\tan \alpha \cot \alpha \)
\(= 4 - 2 = 2\).
Câu 2. D
\(P = 3{\sin ^2}\alpha + 4{\cos ^2}\alpha \)
\(\;\;\;\;= 3\left( {1 - {{\cos }^2}\alpha } \right) + 4{\cos ^2}\alpha \)
\( \;\;\;\;= 3 + {\cos ^2}\alpha = 3 + \dfrac{1}{4} = \dfrac{{13}}{4}.\)
Câu 3. B
\(S = {\cos ^2}1^\circ + {\cos ^2}12^\circ + {\cos ^2}78^\circ + {\cos ^2}89^\circ \)
\(\;\;\; = {\cos ^2}1^\circ + {\cos ^2}12^\circ + {\sin ^2}12^\circ + {\sin ^2}1^\circ\)
\(\;\;\; = 2\)
Câu 4. A
Ta có:
\(1 = {\sin ^2}\alpha + {\cos ^2}\alpha \)
\(\;\;\;= {\sin ^2}\alpha + {\left( {\dfrac{1}{5} - \sin \alpha } \right)^2} \)
\(\;\;\;= 2{\sin ^2}\alpha - \dfrac{2}{5}\sin \alpha + \dfrac{1}{{25}}\)
\( \Leftrightarrow 2{\sin ^2}\alpha - \dfrac{2}{5}\sin \alpha - \dfrac{{24}}{{25}} = 0 \)
\(\Leftrightarrow \left[ \begin{array}{l}\sin \alpha = \dfrac{4}{5}\\\sin \alpha = - \dfrac{3}{5}\end{array} \right.\)
Do \(0 \le x \le \pi \) nên \(\sin \alpha \ge 0\). Chọn \(\sin \alpha = \dfrac{4}{5}\).
Suy ra \(\cos \alpha = \dfrac{1}{5} - \dfrac{4}{5} = - \dfrac{3}{5}\).
Vậy \(\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }} = - \dfrac{4}{3}\).
Câu 5. D
Ta có: \(\dfrac{1}{{{{\cos }^2}\alpha }} = 1 + {\tan ^2}\alpha = 1 + 7 = 8\).
Suy ra \({\cos ^2}\alpha = \dfrac{1}{8}\). Do đó \(\cos \alpha = \pm \dfrac{1}{{2\sqrt 2 }}\).
Vậy \(\sin \alpha = \tan \alpha \cos \alpha = \pm \dfrac{{\sqrt {14} }}{4}\).
Câu 6. C
Ta có \(\dfrac{1}{{\sin 18^\circ }} - \dfrac{1}{{\sin 54^\circ }} = \dfrac{{\sin 54^\circ - \sin 18^\circ }}{{\sin 18^\circ \sin 54^\circ }} \)\(\,= \dfrac{{2\cos 36^\circ \sin 18^\circ }}{{\sin 18^\circ \sin 54^\circ }} = 2\)
Câu 7. C
Ta có: \(\sin 6x + \cos 4x = 0\)
\(\Leftrightarrow \sin 6x + \sin \left( {90^\circ - 4x} \right) = 0\)
\( \Leftrightarrow 2\sin \left( {x + 45^\circ } \right)\cos \left( {5x - 45^\circ } \right) = 0.\)
Với \(x = 27^\circ \) thì \(5x - 45^\circ = 90^\circ \) nên \(\cos \left( {5x - 45^\circ } \right) = 0.\)
Câu 8. D
Ta có: \(\tan \left( {x + y} \right) = \dfrac{{{\mathop{\rm tanx}\nolimits} + \tan y}}{{1 - {\mathop{\rm tanx}\nolimits} {\mathop{\rm tany}\nolimits} }}\)\(\, = \dfrac{{\dfrac{1}{2} + \dfrac{1}{3}}}{{1 - \dfrac{1}{2}.\dfrac{1}{3}}} = \dfrac{{\dfrac{5}{6}}}{{\dfrac{5}{6}}} = 1\).
Do \(x,y \in \left( {0;\dfrac{\pi }{2}} \right)\) nên \(0 < x + y < \pi \). Suy ra \(x + y = \dfrac{\pi }{4}\).
Câu 9. B
Ta có \(1 = {\sin ^2}x + {\cos ^2}x = 9{\cos ^2}x + {\cos ^2}x \)\(\,= 10{\cos ^2}x\)
\( \Rightarrow {\cos ^2}x = \dfrac{1}{{10}} \Rightarrow \cos x = \pm \dfrac{1}{{\sqrt {10} }}\)
Suy ra \({\mathop{\rm sinx}\nolimits} = \pm \dfrac{3}{{\sqrt {10} }}\).
Vậy \(\sin 2x = 2\sin x\cos x = \dfrac{6}{{10}} = \dfrac{3}{5}.\)
Câu 10. A
Ta có:
\(F = 6\left( {1 - {{\sin }^2}x} \right) + 6\sin x - 2 \)
\(\;\;\;= 4 + 6\sin x - 6{\sin ^2}x\)
\(\;\;\; = 4 - 6\left( {{{\sin }^2}x - {\mathop{\rm sinx}\nolimits} } \right) \)
\(\;\;\;= 4 - 6\left[ {\left( {{\mathop{\rm sinx}\nolimits} - \dfrac{1}{2}} \right) - \dfrac{1}{4}} \right]\)
\(\;\;\;= \dfrac{{11}}{2} - 6{\left( {{\mathop{\rm sinx}\nolimits} - \dfrac{1}{2}} \right)^2}\).
Suy ra giá trị lớn nhất của F là \(\dfrac{{11}}{2}\) đạt được khi \({\mathop{\rm sinx}\nolimits} = \dfrac{1}{2}\).
