Question

Tg x = 1/3Найти cos4x

Answer 1

$tgx=\frac{1}{3}\\\\1+tg^{2}x=\frac{1}{Cos^{2}x } \\\\Cos^{2}x=\frac{1}{1+tg^{2}x } =\frac{1}{1+(\frac{1}{3})^{2}}=\frac{1}{1+\frac{1}{9}}=\frac{1}{\frac{10}{9}}=\frac{9}{10}\\\\Cos4x=(2Cos^{2}x -1)^{2}-4Cos^{2} x(1-Cos^{2}x)=(2*\frac{9}{10}-1)^{2}-4*\frac{9}{10}(1-\frac{9}{10})=\\\\=\frac{16}{25}-\frac{18}{5}*\frac{1}{10}=\frac{16}{25}-\frac{9}{25}=\frac{7}{25}=0,28\\\\\boxed{Cos4x=0,28}$