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Свођење на први квадрант 2

Свођење вредности тригонометријских функција на први квадрант. Једноставни примери.

Задаци

Текст задатака објашњених у видео лекцији.

Решити следећи задатак.

пр.1)   $\frac{{\sin \left( {\frac{\pi }{2} + \alpha } \right)\cos \left( {\frac{{3\pi }}{2} - \alpha } \right)}}{{\cos \left( {\pi  + \alpha } \right)}} - \frac{{\sin \left( {2\pi  - \alpha } \right)\cos \left( {\frac{\pi }{2} + \alpha } \right)}}{{\sin \left( {2\pi  + \alpha } \right)}}=$

пр.2)   $\frac{{\sin \left( {\frac{\pi }{2} - \alpha } \right) \cdot \cos \left( {\alpha  - \frac{{3\pi }}{2}} \right)}}{{t{g^2}\left( {\alpha  - \pi } \right) \cdot ctg\left( {\alpha  - 2\pi } \right)}}=$

пр.3)   $\frac{{{{\cos }^3}\left( {{{270}^ \circ } + \alpha } \right) - {{\sin }^3}\left( {\alpha  - \frac{{3\pi }}{2}} \right)}}{{1 + \cos \left( {\alpha  - \frac{{3\pi }}{2}} \right)\sin \left( {\alpha  - {{90}^ \circ }} \right)}} - \sin \left( {\alpha  - \frac{{5\pi }}{2}} \right) = $

Пр.1

\[\begin{gathered}
\frac{{\sin \left( {\frac{\pi }{2} + \alpha } \right)\cos \left( {\frac{{3\pi }}{2} - \alpha } \right)}}{{\cos \left( {\pi + \alpha } \right)}} - \frac{{\sin \left( {2\pi - \alpha } \right)\cos \left( {\frac{\pi }{2} + \alpha } \right)}}{{\sin \left( {2\pi + \alpha } \right)}} = \hfill \\
= \frac{{\cos \alpha \left( { - \sin \alpha } \right)}}{{ - \cos \alpha }} - \frac{{ - \sin \alpha \left( { - \sin \alpha } \right)}}{{\sin \alpha }} = 0 \hfill \\
\end{gathered} \]

Пр.2

\[\begin{gathered}
\frac{{\sin \left( {\frac{\pi }{2} - \alpha } \right) \cdot \cos \left( {\alpha - \frac{{3\pi }}{2}} \right)}}{{t{g^2}\left( {\alpha - \pi } \right) \cdot ctg\left( {\alpha - 2\pi } \right)}} = \frac{{\cos \alpha \left( { - \sin \alpha } \right)}}{{t{g^2}\alpha \cdot ctg\alpha }} = \hfill \\
= \frac{{ - \cos \alpha \cdot \sin \alpha }}{{\frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} \cdot \frac{{\cos \alpha }}{{\sin \alpha }}}} = \frac{{ - \cos \alpha \cdot \sin \alpha }}{{\frac{{\sin \alpha }}{{\cos \alpha }}}} = \frac{{ - {{\cos }^2}\alpha \cdot \sin \alpha }}{{\sin \alpha }} = \hfill \\
= - {\cos ^2}\alpha \hfill \\
\end{gathered} \]

пр.3 

$\frac{{{{\cos }^3}\left( {{{270}^ \circ } + \alpha } \right) - {{\sin }^3}\left( {\alpha  - \frac{{3\pi }}{2}} \right)}}{{1 + \cos \left( {\alpha  - \frac{{3\pi }}{2}} \right)\sin \left( {\alpha  - {{90}^ \circ }} \right)}} - \sin \left( {\alpha  - \frac{{5\pi }}{2}} \right) = $

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