Question

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Вычислите sin2a , cos2a, tg2a, если

cosa=4/5 , π

Answer 1

Ответ:

$\cos( \alpha ) = \frac{4}{5}$

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

$\sin( \alpha ) = \sqrt{1 - {( \frac{4}{5} )}^{2} } = \sqrt{1 - \frac{16}{25} } = \sqrt{ \frac{9}{25} } = \frac{3}{5}$

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

$\tan( \alpha ) = \frac{ \frac{3}{5} }{ \frac{4}{5} } = \frac{3}{4}$

1)

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

$\sin(2 \alpha ) = 2 \times \frac{3}{5} \times \frac{4}{5} = \frac{24}{25}$

2)

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

$\cos(2 \alpha ) = {( \frac{4}{5} )}^{2} - {( \frac{3}{5}) }^{2} = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$

3)

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

$\tan(2 \alpha ) = \frac{2 \times \frac{3}{4} }{1 - {( \frac{3}{4}) }^{2} } = \frac{ \frac{3}{2} }{1 - \frac{9}{16} } = \frac{ \frac{3}{2} }{ \frac{7}{16} } = \frac{3}{2} \times \frac{16}{7} = \frac{24}{7}$