Тригонометријске функције оштрог угла – примери 3

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Пр.5   Израчунати вредност израза $\frac{{{{\sin }^2}\alpha  - 3{{\cos }^2}\alpha }}{{2{{\sin           }^2}\alpha  + {{\cos }^2}\alpha }}$ , ако је $tg\alpha  = 3$.

Пр.6  Ако је $\frac{{3\sin \alpha  - \cos \alpha }}{{\sin \alpha  + 2\cos \alpha }} = 1$, одреди $tg\alpha $.

Пр.5

\[\begin{gathered}
tg\alpha = 3 \hfill \\
tg\alpha = \frac{{\sin \alpha }}{{\cos \alpha }} \hfill \\
\frac{{\sin \alpha }}{{\cos \alpha }} = \left. 3 \right| \cdot \cos \alpha \hfill \\
\sin \alpha = 3 \cdot \cos \alpha \hfill \\
{\sin ^2}\alpha + {\cos ^2}\alpha = 1 \hfill \\
{\left( {3 \cdot \cos \alpha } \right)^2} + {\cos ^2}\alpha = 1 \hfill \\
9 \cdot {\cos ^2}\alpha + {\cos ^2}\alpha = 1 \hfill \\
10{\cos ^2}\alpha = 1 \hfill \\
{\cos ^2}\alpha = \frac{1}{{10}} \hfill \\
\cos \alpha = \frac{{\sqrt {10} }}{{10}} \hfill \\
\sin \alpha = \frac{{3\sqrt {10} }}{{10}} \hfill \\
\end{gathered} \]

$\frac{{{{\sin }^2}\alpha - 3{{\cos }^2}\alpha }}{{2{{\sin }^2}\alpha + {{\cos }^2}\alpha }} = \frac{{{{\left( {\frac{{3\sqrt {10} }}{{10}}} \right)}^2} - 3 \cdot \frac{1}{{10}}}}{{2{{\left( {\frac{{3\sqrt {10} }}{{10}}} \right)}^2} + \frac{1}{{10}}}} $ $= \frac{{\frac{{9 \cdot 10}}{{100}} - \frac{3}{{10}}}}{{2\frac{{9 \cdot 10}}{{100}} + \frac{1}{{10}}}} = \frac{{\frac{9}{{10}} - \frac{3}{{10}}}}{{\frac{{18}}{{10}} + \frac{1}{{10}}}} = \frac{{\frac{6}{{10}}}}{{\frac{{19}}{{10}}}} = \frac{6}{{19}}$

Пр.6

\[\begin{gathered}
\left. {\frac{{3\sin \alpha - \cos \alpha }}{{\sin \alpha + 2\cos \alpha }} = 1} \right| \cdot \left( {\sin \alpha + 2\cos \alpha } \right) \hfill \\
3\sin \alpha - \cos \alpha = \sin \alpha + 2\cos \alpha \hfill \\
3\sin \alpha - \sin \alpha = 2\cos \alpha + \cos \alpha \hfill \\
2\sin \alpha = 3\cos \alpha \hfill \\
\sin \alpha = \frac{3}{2}\cos \alpha \hfill \\
\end{gathered} \]

\[\begin{gathered}
{\sin ^2}\alpha + {\cos ^2}\alpha = 1 \hfill \\
{\left( {\frac{3}{2}\cos \alpha } \right)^2} + {\cos ^2}\alpha = 1 \hfill \\
\frac{9}{4}{\cos ^2}\alpha + {\cos ^2}\alpha = 1 \hfill \\
{\cos ^2}\alpha \left( {\frac{9}{4} + 1} \right) = 1 \hfill \\
{\cos ^2}\alpha = \frac{4}{{19}} \hfill \\
\cos \alpha = \frac{{2\sqrt {13} }}{{13}} \hfill \\
\end{gathered} \]

\[\begin{gathered}
\sin \alpha = \frac{3}{2} \cdot \frac{{2\sqrt {13} }}{{13}} = \frac{{3\sqrt {13} }}{{13}} \hfill \\
tg\alpha = \frac{{\sin \alpha }}{{\cos \alpha }} = \frac{{\frac{{3\sqrt {13} }}{{13}}}}{{\frac{{2\sqrt {13} }}{{13}}}} = \frac{3}{2} \hfill \\
\end{gathered} \]