Question

Cos4(alpha + beta) = ?еслиcos(alpha + 3beta)+cos(3alpha + beta)= 1cos(alpha + 3beta)-cos(3alpha + beta)= 1/3

Answer 1

$zamena\\ \left \{ {{cos(a+3b)=x} \atop {cos(3a+b)=y}} \right. \\ \left \{ {{x+y=1} \atop {x-y=\frac{1}{3}}} \right. \\ \left \{ {{1-2y=\frac{1}{3}} \atop {x=\frac{1}{3}+y}} \right.\\ \\ y=\frac{1}{3}\\ x=\frac{2}{3}\\ \\ cos(a+3b)=\frac{2}{3}\\ cos(3a+b) = \frac{1}{3}$
$cos(arccos(\frac{2}{3})-3arccos(\frac{1}{3}))\\ tak\ kak\\ cos(a-b)=sina*sinb+cosa*cosb\\ poluchaem\\ \\ sin(arccos(\frac{2}{3}))*sin(3arccos(\frac{1}{3}))+cos(arccos(\frac{2}{3}))*cos(3arccos(\frac{1}{3}))=\\ \frac{\sqrt{5}}{3}*\frac{-5}{9}*\frac{\sqrt{8}}{3}+\frac{2}{3}*\frac{1}{27}=\frac{2}{81}-\frac{10\sqrt{10}}{81}$
$a+3b=arccos(\frac{2}{3})\\ 3a+b=arccos(\frac{1}{3})\\ \\ a=arccos(\frac{2}{3})-3b\\ 3arccos(\frac{2}{3})-8b=arccos\frac{1}{3} \\ b=\frac{arccos\frac{1}{3}-3arccos\frac{2}{3}}{-8}\\ a=arccos\frac{2}{3}-3(\frac{arccos\frac{1}{3}-3arccos\frac{2}{3}}{-8})\\ \\ cos(4(a+b))=cos(4a+4b)\\ cos(4a+4b)=cos(4arccos(\frac{2}{3})+12\frac{arccos(\frac{1}{3})-3arccos(\frac{2}{3}))}{8})+4*\frac{((arccos(\frac{2}{3})-3arccos( \frac{1}{3}))}{-8}))$