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Тригонометријске функције двоструког угла 1

Тригонометријске функције двоструког угла. Примена на једноставне примере.

Задаци

Текст задатака објашњених у видео лекцији.

Доказати следећу једнакост:

Пр.1

\[{\sin ^2}\alpha  = \frac{{1 - \cos 2\alpha }}{2}\]

Пр.2

\[\sin 2\alpha  = \frac{{2tg\alpha }}{{1 + t{g^2}\alpha }}\]

Пр.3

\[\sin 3\alpha  = 3\sin \alpha  - 4{\sin ^3}\alpha \]

Упростити селедећи израз:

Пр.4

\[\cos 4\alpha  + 2{\sin ^2}2\alpha  = \]

Пр.1

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

Пр.2

\[\begin{gathered}
\sin 2\alpha = \frac{{2tg\alpha }}{{1 + t{g^2}\alpha }} \hfill \\
2\sin \alpha cos\alpha = \frac{{2\frac{{\sin \alpha }}{{\cos \alpha }}}}{{1 + \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }}}} \hfill \\
2\sin \alpha cos\alpha = \frac{{2\frac{{\sin \alpha }}{{\cos \alpha }}}}{{\frac{{{{\cos }^2}\alpha + {{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }}}} \hfill \\
2\sin \alpha cos\alpha = 2\sin \alpha cos\alpha \hfill \\
\end{gathered} \]

Пр.3

\[\begin{gathered}
\sin 3\alpha = 3\sin \alpha - 4{\sin ^3}\alpha \hfill \\
\sin 3\alpha = \sin \left( {\alpha + 2\alpha } \right) = \sin \alpha \cdot \cos 2\alpha + \cos \alpha \cdot \sin 2\alpha = \hfill \\
= \sin \alpha \left( {{{\cos }^2}\alpha - {{\sin }^2}\alpha } \right) + \cos \alpha \cdot 2\sin \alpha \cos \alpha = \hfill \\
= \sin \alpha {\cos ^2}\alpha - {\sin ^3}\alpha + 2\sin \alpha {\cos ^2}\alpha = \hfill \\
= 3\sin \alpha {\cos ^2}\alpha - {\sin ^3}\alpha = 3\sin \alpha \left( {1 - {{\sin }^2}\alpha } \right) - {\sin ^3}\alpha = \hfill \\
= 3\sin \alpha - 3{\sin ^3}\alpha - {\sin ^3}\alpha = 3\sin \alpha - 4{\sin ^3}\alpha \hfill \\
\end{gathered} \]

Пр.4

\[\begin{gathered}
\cos 4\alpha + 2{\sin ^2}2\alpha = \cos \left( {2 \cdot 2\alpha } \right) + 2{\left( {2\sin \alpha \cos \alpha } \right)^2} = \hfill \\
= {\cos ^2}2\alpha - {\sin ^2}2\alpha + 2 \cdot 4{\sin ^2}\alpha {\cos ^2}\alpha = \hfill \\
= {\left( {{{\cos }^2}\alpha - {{\sin }^2}\alpha } \right)^2} - {\left( {2\sin \alpha \cos \alpha } \right)^2} + 8{\sin ^2}\alpha {\cos ^2}\alpha = \hfill \\
= {\cos ^4}\alpha - 2{\sin ^2}\alpha {\cos ^2}\alpha + {\sin ^4}\alpha - 4{\sin ^2}\alpha {\cos ^2}\alpha + 8{\sin ^2}\alpha {\cos ^2}\alpha = \hfill \\
= {\cos ^4}\alpha + 2{\sin ^2}\alpha {\cos ^2}\alpha + {\sin ^4}\alpha = \hfill \\
= {\left( {{{\cos }^2}\alpha + {{\sin }^2}\alpha } \right)^2} = 1 \hfill \\
\end{gathered} \]