Question

Найти cos2a, если tga=1/3, а sinа=1/√10

Answer 1

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

$\cos( \alpha ) = \frac{ \frac{1}{ \sqrt{10} } }{ \frac{1}{3} } = \frac{1}{ \sqrt{10} } \times 3 = \frac{3}{ \sqrt{10} }$

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

$\cos(2 \alpha ) = \frac{9}{10} - \frac{1}{10} = \frac{8}{10} = 0.8$