Question

ПОМОГИТЕ. ДАЮ 20б. Математика

Answer 1

1)  $cos\alpha =\frac{5}{13};$     $0<\alpha <\frac{\pi}{2}$  (I четверть)

$sin^2\alpha+ cos^2\alpha =1$

$sin\alpha$ в I четверти положителен

$sin^2\alpha=1- cos^2\alpha$

$sin^2\alpha =1-(\frac{5}{13})^2$

$sin^2\alpha =1-\frac{25}{169}$

$sin^2\alpha =\frac{144}{169}$

$sin\alpha =\sqrt{\frac{144}{169} }=\frac{12}{13}$

$sin\alpha =\frac{12}{13}$

2) $sin\beta =-0,8;$    $\pi <\beta <\frac{3\pi }{2}$  (III четверть)

$sin^2\beta+ cos^2\beta =1$

$cos\beta$  в III четверти отрицателен

$cos^2\beta =1-sin^2\beta$

$cos^2\beta =1-(-0,8)^2$

$cos^2\beta =1-0,64=0,36$

$cos\beta =-\sqrt{0,36}=-0,6$

$cos\beta =-0,6=-\frac{3}{5}$

3) $cos(\alpha-\beta)=cos\alpha* cos\beta+sin\alpha* sin\beta$

 $cos(\alpha-\beta)=\frac{5}{13}*(-\frac{3}{5})+\frac{12}{13}*(-\frac{4}{5})=-\frac{3}{13}+(-\frac{48}{65})=\frac{-15-48}{65}=- \frac{63}{65}$

$cos(\alpha-\beta)=- \frac{63}{65}$