75cos(2 arccos 4/5) найдите значение выражения

[latex]75\cdot cos(2\cdot arccos\frac{4}{5})=75\cdot (cos^2(arccos\frac{4}{5})-sin^2(arccos\frac{4}{5}))=\\\\=75\cdot ((\frac{4}{5})^2-(\frac{3}{5})^2)=75\cdot \frac{16-9}{25}=3\cdot 7=21\\\\\\sin(arccos\frac{4}{5})=sin \alpha ,\; \alpha =arccos\frac{4}{5}\; \to \; cos \alpha =\frac{4}{5}\\\\sin^2 \alpha =1-cos^2 \alpha =1-(\frac{4}{5})^2=\frac{3}{5}[/latex]