Question

найдите 9cos2α если cosα=$\frac{1}{3}$

Answer 1

$cos2 \alpha =cos^2 \alpha -sin^2 \alpha$
$sin^2 \alpha=1-cos^2 \alpha =1- (\frac{1}{3})^2=1- \frac{1}{9}= \frac{8}{9}$
$9cos2 \alpha =9(cos^2 \alpha -sin^2 \alpha )=9( \frac{1}{9}- \frac{8}{9})=9*(- \frac{7}{9})=-7$