4. Найдите:1) cosa, sin2а, соѕ2а, если sina = 0,7 и 0 < a

July 2022 1 6 Report

4. Найдите:
1) cosa, sin2а, соѕ2а, если sina = 0,7 и 0 < a < 90°

2) sina, ctga, tg2a, если cosa = 0,6 и 270° < a < 360°

Answers & Comments

Miroslava227

$\sin( \alpha ) = 0.7 \\ \cos( \alpha ) > 0 \\ \cos( \alpha ) = \sqrt{1 - \sin {}^{2} ( \alpha ) } \\ \cos( \alpha ) = \sqrt{1 -0 .49} = \sqrt{0.51} = \frac{ \sqrt{51} }{10}$

$\sin(2 \alpha ) = 2\sin( \alpha ) \cos( \alpha ) = 2 \times \frac{7}{10} \times \frac{ \sqrt{51} }{10} = \\ = \frac{7 \sqrt{51} }{50}$

$\cos( 2\alpha ) = 1 - 2 \sin {}^{2} ( \alpha ) = \\ = 1 - 2 \times (0.7) {}^{2} = 1 - 0.98 = 0.02$

$\cos( \alpha ) = 0.6 \\ \sin( \alpha ) < 0 \\ \sin( \alpha ) = \sqrt{1 - \cos {}^{2} ( \alpha ) } \\ \sin( \alpha ) = - \sqrt{1 - 0.36} = - \sqrt{0.64} = - 0.8$

$ctg \alpha = \frac{ \cos( \alpha ) }{ \sin( \alpha ) } = \frac{0.6}{ - 0.8} = - \frac{3}{4} = - 0.75 \\$

$tg 2\alpha = \frac{2tg \alpha }{1 - {tg}^{2} \alpha } \\ \\ tg \alpha = \frac{1}{ctg \alpha } = - \frac{4}{3} \\ \\ tg2 \alpha = \frac{2 \times ( - \frac{4}{3} )}{1 - \frac{16}{9} } = - \frac{8}{3} \times ( - \frac{9}{7} ) = \\ = \frac{24}{7}$

$tg \alpha = 5 \\ \sin( \alpha ) < 0 \\ \cos( \alpha ) < 0 \\ \\ 1 + {tg}^{2} \alpha = \frac{1}{ \cos {}^{2} ( \alpha ) } \\ \cos( \alpha ) = \pm \sqrt{ \frac{1}{1 + {tg}^{2} \alpha } } \\ \cos( \alpha ) = - \sqrt{ \frac{1}{1 + 25} } = - \sqrt{ \frac{1}{26} } = - \frac{ \sqrt{26} }{26} \\ \\ \sin( \alpha ) = - \sqrt{1 - \frac{1}{26} } = - \sqrt{ \frac{25}{26} } = - \frac{5 \sqrt{26} }{26}$

$\cos( 2\alpha ) = 2 \cos {}^{2} ( \alpha ) - 1 = \\ = 2 \times \frac{1}{26} - 1 = \frac{1}{13} - 1 = - \frac{12}{13}$

5 votes Thanks 7

Answers & Comments

рекомендуемые вопросы

Helpful Links

Helpful Social