Тригонометријски идентитети – примери

Текст задатака објашњених у видео лекцији.

Пр.7   Упростити израз $\left( {1 + \sin \alpha } \right)\left( {tg\alpha  + ctg\alpha } \right)\left( {1 - \sin \alpha } \right)$.

Пр.8   Доказати идентитете: $\frac{{1 -2 {{\cos }^2}\alpha }}{{\sin \alpha  \cos \alpha }} = tg\alpha  - ctg\alpha $

                                             $\frac{{{{\sin }^2}x}}{{\sin x - \cos x}} + \frac{{\sin x + \cos x}}{{1 - t{g^2}x}} = \sin x + \cos x$

Пр.7   

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

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

Пр.8  

\[\begin{gathered}
\frac{{1 - 2{{\cos }^2}\alpha }}{{\sin \alpha \cos \alpha }} = tg\alpha - ctg\alpha \hfill \\
\frac{{1 - 2{{\cos }^2}\alpha }}{{\sin \alpha \cos \alpha }} = \frac{{\sin \alpha }}{{\cos \alpha }} - \frac{{\cos \alpha }}{{\sin \alpha }} \hfill \\
\frac{{1 - 2{{\cos }^2}\alpha }}{{\sin \alpha \cos \alpha }} = \frac{{{{\sin }^2}\alpha - {{\cos }^2}\alpha }}{{\sin \alpha \cos \alpha }} \hfill \\
\frac{{1 - 2{{\cos }^2}\alpha }}{{\sin \alpha \cos \alpha }} = \frac{{1 - {{\cos }^2}\alpha - {{\cos }^2}\alpha }}{{\sin \alpha \cos \alpha }} \hfill \\
\frac{{1 - 2{{\cos }^2}\alpha }}{{\sin \alpha \cos \alpha }} = \frac{{1 - 2{{\cos }^2}\alpha }}{{\sin \alpha \cos \alpha }} \hfill \\
\end{gathered} \]

\[\begin{gathered}
\frac{{{{\sin }^2}x}}{{\sin x - \cos x}} + \frac{{\sin x + \cos x}}{{1 - t{g^2}x}} = \sin x + \cos x \hfill \\
\frac{{{{\sin }^2}x}}{{\sin x - \cos x}} + \frac{{\sin x + \cos x}}{{1 - {{\left( {\frac{{\operatorname{sinx} }}{{\operatorname{cosx} }}} \right)}^2}}} = \hfill \\
= \frac{{{{\sin }^2}x}}{{\sin x - \cos x}} + \frac{{\sin x + \cos x}}{{\frac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\cos }^2}x}}}} = \hfill \\
= \frac{{{{\sin }^2}x}}{{\sin x - \cos x}} + \frac{{\left( {\sin x + \cos x} \right) \cdot {{\cos }^2}x}}{{{{\cos }^2}x - {{\sin }^2}x}} = \hfill \\
= \frac{{{{\sin }^2}x}}{{\sin x - \cos x}} + \frac{{\left( {\sin x + \cos x} \right) \cdot {{\cos }^2}x}}{{\left( {\sin x + \cos x} \right) \cdot \left( {\cos x - \sin x} \right)}} = \hfill \\
= \frac{{{{\sin }^2}x}}{{\sin x - \cos x}} - \frac{{co{s^2}x}}{{\sin x - \cos x}} = \hfill \\
= \frac{{{{\sin }^2}x - co{s^2}x}}{{\sin x - \cos x}} = \frac{{\left( {\sin x + \cos x} \right) \cdot \left( {\sin x - \cos x} \right)}}{{\sin x - \cos x}} = \sin x + \cos x \hfill \\
\end{gathered} \]