Текст задатака објашњених у видео лекцији.

Израчунати:

Пр.1   \[\sin 105^\circ  = \]

Пр.2   \[tg15^\circ  = \]

Одредити вредност следећег израза:

Пр.3  \[\cos \frac{\pi }{7} \cdot \sin \frac{{8\pi }}{7} - \sin \frac{\pi }{7} \cdot \cos \frac{{8\pi }}{7} = \]

Пр.4  \[\frac{{\sin 35^\circ  \cdot \cos 20^\circ  + \cos 35^\circ  \cdot \sin 20^\circ }}{{\cos 46^\circ  \cdot \cos 29^\circ  - \sin 46^\circ  \cdot \sin 29^\circ }} = \]

Пр.1 

\[\begin{gathered}
\sin 105^\circ = \sin \left( {60^\circ + 45^\circ } \right) = \hfill \\
= \sin 60^\circ \cdot \cos 45^\circ + \cos 60^\circ \cdot \sin 45^\circ = \hfill \\
= \frac{{\sqrt 3 }}{2} \cdot \frac{{\sqrt 2 }}{2} + \frac{1}{2} \cdot \frac{{\sqrt 2 }}{2} = \frac{{\sqrt 6 }}{4} + \frac{{\sqrt 2 }}{4} = \frac{{\sqrt 6 + \sqrt 2 }}{4} \hfill \\
\end{gathered} \]

Пр.2

\[\begin{gathered}
tg15^\circ = tg\left( {60^\circ - 45^\circ } \right) = \frac{{tg60^\circ - tg45^\circ }}{{1 + tg60^\circ tg45^\circ }} = \hfill \\
= \frac{{\sqrt 3 - 1}}{{1 + \sqrt 3 \cdot 1}} \cdot \frac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}} = \frac{{\sqrt 3 - 3 - 1 + \sqrt 3 }}{{{1^2} - {{\left( {\sqrt 3 } \right)}^2}}} = \frac{{2\sqrt 3 - 4}}{{ - 2}} = \hfill \\
= \frac{{ - 2\left( {2 - \sqrt 3 } \right)}}{{ - 2}} = 2 - \sqrt 3 \hfill \\
\end{gathered} \]

Пр.3

\[\begin{gathered}
\cos \frac{\pi }{7} \cdot \sin \frac{{8\pi }}{7} - \sin \frac{\pi }{7} \cdot \cos \frac{{8\pi }}{7} = \hfill \\
= - \left( {\sin \frac{\pi }{7} \cdot \cos \frac{{8\pi }}{7} - \cos \frac{\pi }{7} \cdot \sin \frac{{8\pi }}{7}} \right) = - \sin \left( {\frac{\pi }{7} - \frac{{8\pi }}{7}} \right) = \hfill \\
= - \sin \left( { - \pi } \right) = \sin \pi = 0 \hfill \\
\end{gathered} \]

Пр.4

\[\begin{gathered}
\frac{{\sin 35^\circ \cdot \cos 20^\circ + \cos 35^\circ \cdot \sin 20^\circ }}{{\cos 46^\circ \cdot \cos 29^\circ - \sin 46^\circ \cdot \sin 29^\circ }} = \frac{{\sin \left( {35^\circ - 20^\circ } \right)}}{{\cos \left( {46^\circ + 29^\circ } \right)}} = \hfill \\
= \frac{{\sin 15^\circ }}{{\cos 75^\circ }} = \frac{{\sin \left( {60^\circ - 45^\circ } \right)}}{{\cos \left( {30^\circ + 45^\circ } \right)}} = \hfill \\
= \frac{{\sin 60^\circ \cdot \cos 45^\circ - \cos 60^\circ \cdot \sin 45^\circ }}{{\cos 30^\circ \cdot \cos 45^\circ - \sin 30^\circ \cdot \sin 45^\circ }} = \frac{{\frac{{\sqrt 3 }}{2} \cdot \frac{{\sqrt 2 }}{2} - \frac{1}{2} \cdot \frac{{\sqrt 2 }}{2}}}{{\frac{{\sqrt 3 }}{2} \cdot \frac{{\sqrt 2 }}{2} - \frac{1}{2} \cdot \frac{{\sqrt 2 }}{2}}} = 1 \hfill \\
\end{gathered} \]