Đáp án:
\(\begin{array}{l}
4,\\
P = \dfrac{{13}}{4}\\
5,\\
a,\\
\cos \beta = \dfrac{{\sqrt {15} }}{4};\,\,\,\,\tan \beta = \dfrac{{\sqrt {15} }}{{15}}\\
b,\\
\left[ \begin{array}{l}
\sin \alpha = \dfrac{{2\sqrt 2 }}{3};\,\,\tan x = - 2\sqrt 2 ;\,\,\,\cot x = - \dfrac{{\sqrt 2 }}{4}\\
\sin \alpha = - \dfrac{{2\sqrt 2 }}{3};\,\,\tan x = 2\sqrt 2 ;\,\,\,\cot x = \dfrac{{\sqrt 2 }}{4}
\end{array} \right.\\
c,\\
\left[ \begin{array}{l}
\sin x = \dfrac{{2\sqrt 2 }}{3};\,\,\cos x = \dfrac{1}{3}\\
\sin x = - \dfrac{{2\sqrt 2 }}{3};\,\,\,\cos x = - \dfrac{1}{3}
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
4,\\
\cos x = \dfrac{1}{2}\\
{\sin ^2}x + {\cos ^2}x = 1\\
P = 3{\sin ^2}x + 4{\cos ^2}x\\
= \left( {3{{\sin }^2}x + 3{{\cos }^2}x} \right) + {\cos ^2}x\\
= 3.\left( {{{\sin }^2}x + {{\cos }^2}x} \right) + {\cos ^2}x\\
= 3.1 + {\left( {\dfrac{1}{2}} \right)^2}\\
= 3 + \dfrac{1}{4} = \dfrac{{13}}{4}\\
5,\\
a,\\
0 < \beta < 90^\circ \Rightarrow 0 < \cos \beta < 1\\
{\sin ^2}\beta + {\cos ^2}\beta = 1\\
\Leftrightarrow {\left( {\dfrac{1}{4}} \right)^2} + {\cos ^2}\beta = 1\\
\Leftrightarrow \dfrac{1}{{16}} + {\cos ^2}\beta = 1\\
\Leftrightarrow {\cos ^2}\beta = \dfrac{{15}}{{16}}\\
0 < \cos \beta < 1 \Rightarrow \cos \beta = \dfrac{{\sqrt {15} }}{4}\\
\tan \beta = \dfrac{{\sin \beta }}{{\cos \beta }} = \dfrac{1}{{\sqrt {15} }} = \dfrac{{\sqrt {15} }}{{15}}\\
b,\\
{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\
\Leftrightarrow {\sin ^2}\alpha + {\left( { - \dfrac{1}{3}} \right)^2} = 1\\
\Leftrightarrow {\sin ^2}\alpha + \dfrac{1}{9} = 1\\
\Leftrightarrow {\sin ^2}\alpha = \dfrac{8}{9}\\
\Rightarrow \left[ \begin{array}{l}
\sin \alpha = \dfrac{{2\sqrt 2 }}{3}\\
\sin \alpha = - \dfrac{{2\sqrt 2 }}{3}
\end{array} \right.\\
*)\\
\sin \alpha = \dfrac{{2\sqrt 2 }}{3} \Rightarrow \left\{ \begin{array}{l}
\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }} = \dfrac{{2\sqrt 2 }}{{ - 1}} = - 2\sqrt 2 \\
\cot \alpha = \dfrac{{\cos \alpha }}{{\sin \alpha }} = \dfrac{{ - 1}}{{2\sqrt 2 }} = \dfrac{{ - \sqrt 2 }}{4}
\end{array} \right.\\
*)\\
\sin \alpha = - \dfrac{{2\sqrt 2 }}{3} \Rightarrow \left\{ \begin{array}{l}
\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }} = \dfrac{{ - 2\sqrt 2 }}{{ - 1}} = 2\sqrt 2 \\
\cot \alpha = \dfrac{{\cos \alpha }}{{\sin \alpha }} = \dfrac{{ - 1}}{{ - 2\sqrt 2 }} = \dfrac{{\sqrt 2 }}{4}
\end{array} \right.\\
c,\\
\tan x = 2\sqrt 2 \Leftrightarrow \dfrac{{\sin x}}{{\cos x}} = 2\sqrt 2 \Leftrightarrow \sin x = 2\sqrt 2 \cos x\\
{\sin ^2}x + {\cos ^2}x = 1\\
\Leftrightarrow {\left( {2\sqrt 2 \cos x} \right)^2} + {\cos ^2}x = 1\\
\Leftrightarrow 8{\cos ^2}x + {\cos ^2}x = 1\\
\Leftrightarrow 9{\cos ^2}x = 1\\
\Leftrightarrow {\cos ^2}x = \dfrac{1}{9}\\
\Rightarrow \left[ \begin{array}{l}
\cos x = \dfrac{1}{3};\,\,\,\sin x = 2\sqrt 2 \cos x = \dfrac{{2\sqrt 2 }}{3}\\
\cos x = - \dfrac{1}{3};\,\,\sin x = 2\sqrt 2 \cos x = - \dfrac{{2\sqrt 2 }}{3}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\sin x = \dfrac{{2\sqrt 2 }}{3};\,\,\cos x = \dfrac{1}{3}\\
\sin x = - \dfrac{{2\sqrt 2 }}{3};\,\,\,\cos x = - \dfrac{1}{3}
\end{array} \right.
\end{array}\)
