Question

Va rog, ajutor la acest exercitiu. Coroana:)

Answer 1

$sin^{2} x +cos^{2}x=1 =\ \textgreater \ cosx= \sqrt{1-sin^{2}x} = \sqrt{ \frac{25}{25}- \frac{9}{25} } =\ \textgreater \ cosx= +/-\frac{4}{5}   x$∈[tex]( 0, \frac{ \pi }{2} ) =\ \textgreater \ cox= - \frac{4}{5} [/tex]
$tg \frac{x}{2} = \frac{sinx}{1-cosx} =\ \textgreater \ tg \frac{x}{2} = \frac{3}{5} * \frac{5}{9} =\ \textgreater \ tg \frac{x}{2} = \frac{1}{3}$