Question

Sa se calculeze tg 2a,sin 2a,cos 2a știind ca tg a=2.​

Answer 1

Explicație pas cu pas:

$\tan(\alpha ) = \frac{ \sin( \alpha ) }{ \cos( \alpha ) }$

$\tan( \alpha ) = 2 = > \frac{ \sin( \alpha ) }{ \cos( \alpha ) } = 2 \\ = > \sin( \alpha ) = 2 \cos( \alpha ) \\ < = > { \sin^{2}( \alpha ) } = 4 \cos^{2}( \alpha )$

$\sin^{2}( \alpha ) + \cos^{2}( \alpha ) = 1$

$= > \sin^{2}( \alpha ) = 1 - \cos^{2}( \alpha )$

$1 - \cos^{2}( \alpha ) = 4\cos^{2}( \alpha ) \\ 5\cos^{2}( \alpha ) = 1 = > \cos^{2}( \alpha ) = \frac{1}{5}$

$\tan( 2\alpha ) = \frac{2 \tan( \alpha ) }{1 - \tan^{2} ( \alpha ) }$

$\tan( 2\alpha ) = \frac{2 \times 2}{1 - {2}^{2} } = \frac{4}{ - 3}$

$= > \tan( 2\alpha ) = - \frac{4}{3}$

$\sin( 2\alpha ) = 2 \sin( \alpha ) \cos( \alpha )$

$\sin( 2\alpha ) = 2(2 \cos( \alpha ) ) \cos( \alpha ) = 4 \cos^{2} ( \alpha ) = 4 \times \frac{1}{5} = \frac{4}{5}$

$= > \sin( 2\alpha ) = \frac{4}{5}$

$\cos(2 \alpha ) = 2 \cos^{2} ( \alpha ) - 1$

$\cos(2 \alpha ) = 2 \times \frac{1}{5} - 1 = \frac{2}{5} - 1 = - \frac{3}{5}$

$= > \cos(2 \alpha ) = - \frac{3}{5}$