срочно 50 б Дано: cosα=–0.5cosα=

rustemShamsutd @rustemShamsutd

July 2022 1 13 Report

срочно 50 б

Дано: cosα=–0.5cosα=–0.5 , π<α<3π2,sinβ=0.3,π2<β<π.Найти:sin(β−α),tg(2α).

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Miroslava227

Ответ:

$\cos( \alpha ) = - 0.5 \\ \sin( \beta ) = 0.3$

$\sin( \beta - \alpha ) = \\ = \sin( \beta ) \cos( \alpha ) - \sin( \alpha ) \cos( \beta )$

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

Найдем необходимые тригонометрические единицы:

$\cos( \alpha ) = - 0.5$

угол а принадлежит 3 четверти, значит sina отрицательный.

$\sin( \alpha ) = \sqrt{1 - { \cos}^{2} \alpha } \\ \sin( \alpha ) = 1 \sqrt{1 - 0.25} = - \sqrt{1 - \frac{1}{4} } = \\ = - \sqrt{ \frac{3}{4} } = - \frac{ \sqrt{3} }{2}$

$\sin( \beta ) = 0.3$

угол в принадлежит 2 четверти, значит cosв отрицательный

$\cos( \beta ) = \sqrt{1 - { \sin }^{2} \beta } \\ \cos( \beta ) = - \sqrt{1 - 0.09} = - \sqrt{0.91} = \\ = - \sqrt{ \frac{91}{100} } = - \frac{ \sqrt{91} }{10}$

$\sin( \beta - \alpha ) = 0.3 \times ( - 0.5) - ( - \frac{ \sqrt{3} }{2} ) \times ( - \frac{ \sqrt{91} }{10} ) = \\ = - \frac{15}{100} - \frac{ \sqrt{3 \times 91} }{20} = - \frac{3}{20} - \frac{ \sqrt{ {273} } }{20} = - \frac{3 + \sqrt{273} }{20}$

$\sin(2 \alpha ) = 2 \sin( \alpha ) \cos( \alpha ) = \\ = 2 \times ( - 0.5) \times ( - \frac{ \sqrt{3} }{2} ) = \\ = \sqrt{3} \times \frac{1}{2} = \frac{ \sqrt{3} }{2}$

$\cos( 2\alpha ) = { \cos}^{2} \alpha - { \sin }^{2} \alpha = \\ = 0.25 - \frac{3}{4} = \frac{1}{4} - \frac{3}{4} = - \frac{1}{2}$

$tg (2\alpha ) = \frac{ \frac{ \sqrt{3} }{2} }{ - \frac{1}{2} } = - \frac{ \sqrt{3} }{2} \times 2 = - \sqrt{3} \\$

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