Question

Плииз, помогите решить
найти cos 2a, если (sin a + 3 cos a)/(sin a - cos a)=3

Answer 1

$\frac{Sin\alpha+3Cos\alpha}{Sin\alpha-Cos\alpha}=3\\\\Sin\alpha+3Cos\alpha=3(Sin\alpha-Cos\alpha)\\\\Sin\alpha+3Cos\alpha=3Sin\alpha -3Cos\alpha\\\\6Cos\alpha=2Sin\alpha\\\\Sin\alpha=3Cos\alpha|:Cos\alpha\neq 0 \\\\tg\alpha=3\\\\1+tg^{2}\alpha=\frac{1}{Cos^{2}\alpha}\\\\Cos^{2}\alpha=\frac{1}{1+tg^{2} \alpha}=\frac{1}{1+3^{2}}=\frac{1}{10}\\\\Cos2\alpha=2Cos^{2}\alpha-1=2*\frac{1}{10}-1=-\frac{4}{5}=-0,8$