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Адиционе формуле 2

Адиционе формуле. Примена на сложеније примере.

Задаци

Текст задатака објашњених у видео лекцији.

Пр.5 Одредити \[\cos \left( {\frac{{17\pi }}{3} - \alpha } \right)=\] ако jе \[tg\alpha  =  - \frac{4}{3},\alpha  \in \left( {\frac{{3\pi }}{2},2\pi } \right)\]

Пр.6 Ако  jе \[tg\alpha  = \frac{{\sqrt 2  + 1}}{{\sqrt 2  - 1}},tg\beta  = \frac{1}{{\sqrt 2 }},\] и $\alpha ,\beta  \in \left( {0,\frac{\pi }{2}} \right),$ доказати да jе \[\alpha  - \beta  = \frac{\pi }{4}\]


Пр.5

\[\begin{gathered}
\cos \left( {\frac{{17\pi }}{3} - \alpha } \right) = \cos \frac{{17\pi }}{3} \cdot \cos \alpha - \sin \frac{{17\pi }}{3} \cdot \sin \alpha \hfill \\
\frac{{17\pi }}{3} = 5\pi + \frac{2}{3}\pi \hfill \\
\sin \frac{{17\pi }}{3} = - \frac{{\sqrt 3 }}{2},\cos \frac{{17\pi }}{3} = \frac{1}{2} \hfill \\
tg\alpha = - \frac{4}{3} \hfill \\
\frac{{\sin \alpha }}{{\cos \alpha }} = - \frac{4}{3} \hfill \\
\sin \alpha = - \frac{4}{3} \cdot \cos \alpha \hfill \\
{\sin ^2}\alpha + {\cos ^2}\alpha = 1 \hfill \\
{\left( { - \frac{4}{3} \cdot \cos \alpha } \right)^2} + {\cos ^2}\alpha = 1 \hfill \\
\frac{{16}}{9}{\cos ^2}\alpha + {\cos ^2}\alpha = 1 \hfill \\
{\cos ^2}\alpha \left( {\frac{{16}}{9} + 1} \right) = 1 \hfill \\
{\cos ^2}\alpha \cdot \frac{{25}}{9} = 1\left| { \cdot \frac{9}{{25}}} \right. \hfill \\
{\cos ^2}\alpha = \frac{9}{{25}} \hfill \\
\cos \alpha = \pm \frac{3}{5} \hfill \\
\alpha \in \left( {\frac{{3\pi }}{2},2\pi } \right) \Rightarrow \cos \alpha = \frac{3}{5} \hfill \\
\sin \alpha = - \frac{4}{3} \cdot \frac{3}{5} = - \frac{4}{5} \hfill \\
\cos \left( {\frac{{17\pi }}{3} - \alpha } \right) = \cos \frac{{17\pi }}{3} \cdot \cos \alpha - \sin \frac{{17\pi }}{3} \cdot \sin \alpha = \hfill \\
= \frac{1}{2} \cdot \frac{3}{5} + \left( { - \frac{{\sqrt 3 }}{2}} \right) \cdot \left( { - \frac{4}{5}} \right) = \frac{3}{{10}} + \frac{{4\sqrt 3 }}{{10}} = \frac{{3 + 4\sqrt 3 }}{{10}} \hfill \\
\end{gathered} \]

Пр.6

Нека jе $\alpha  - \beta  = \frac{\pi }{4}$ онда  

\[\begin{gathered}
tg\left( {\alpha - \beta } \right) = tg\frac{\pi }{4} \hfill \\
tg\left( {\alpha - \beta } \right) = 1 \hfill \\
\frac{{tg\alpha - tg\beta }}{{1 + tg\alpha tg\beta }} = 1 \hfill \\
\frac{{\frac{{\sqrt 2 + 1}}{{\sqrt 2 - 1}} - \frac{1}{{\sqrt 2 }}}}{{1 + \frac{{\sqrt 2 + 1}}{{\sqrt 2 - 1}} \cdot \frac{1}{{\sqrt 2 }}}} = 1 \hfill \\
\frac{{\frac{{\sqrt 2 \left( {\sqrt 2 + 1} \right) - \left( {\sqrt 2 - 1} \right)}}{{\sqrt 2 \left( {\sqrt 2 - 1} \right)}}}}{{1 + \frac{{\sqrt 2 + 1}}{{\sqrt 2 \left( {\sqrt 2 - 1} \right)}}}} = 1 \hfill \\
\frac{{\frac{{2 + \sqrt 2 - \sqrt 2 + 1}}{{\sqrt 2 \left( {\sqrt 2 - 1} \right)}}}}{{\frac{{\sqrt 2 \left( {\sqrt 2 - 1} \right) + \sqrt 2 + 1}}{{\sqrt 2 \left( {\sqrt 2 - 1} \right)}}}} = 1 \hfill \\
\frac{3}{{2 - \sqrt 2 + \sqrt 2 + 1}} = 1 \hfill \\
\frac{3}{3} = 1 \hfill \\
\end{gathered} \]

 дошли смо до таутологиjе. Онда  jе $\alpha  - \beta  = \frac{\pi }{4}$