Ctg х = 1/3Найти: sin 4x

Решите задачу:

$ctgx=\frac{1}{3}\\ \frac{cosx}{sinx}=\frac{1}{3}\\ 3cosx=sinx\\ 3\sqrt{1-sin^2x}=sinx\\ 9(1-sin^2x)=sin^2x\\ 10sin^2x=9\\ sinx=\frac{3}{\sqrt{10}}\\ x=arcsin\frac{3}{\sqrt{10}}\\ x=arcos(\frac{1}{\sqrt{10}})\\ sin4x=2sin2xcos2x=4sinxcosx(2cos^2x-1)=\\ 4sin(arcsin\frac{3}{\sqrt{10}})*cos(arcos(\frac{1}{\sqrt{10}}))(2cos(arcos(\frac{1}{\sqrt{10}}))^2-1)=\frac{12}{\sqrt{10}}*\frac{1}{\sqrt{10}}(2*\frac{1}{10}-1)=\frac{12}{10}(\frac{2}{10}-1)=-\frac{24}{25}$