Question

решите 2,5, 8,9,10! очень надо

Answer 1

$2); ; cos6 alpha +1=2cos^23 alpha$

$Formyla:\\cos^2x=frac{1+cos2x}{2}; ; to ; ; 1+cos2x=2cos^2x; ; (x=3 alpha )\\5); ; cos alpha =frac{2}{3}; ,; ; frac{3pi}{2} textless alpha textless 2pi \\sin beta =frac{1}{3}; ,; ; frac{pi}{2} textless beta textless pi \\\sin alpha =-sqrt{1-cos^2 alpha }=-sqrt{1-frac{4}{9}}=-frac{sqrt5}{3}\\cos beta =-sqrt{1-sin^2 beta }=-sqrt{1-frac{1}{9}}=-frac{2sqrt2}{3}\\cos2 beta =cos^2 beta -sin^2 beta =frac{8}{9}-frac{5}{9}=frac{1}{3}$

$sin2 beta =2sin beta cdot cos beta =2cdot frac{1}{3} cdot frac{-2sqrt2}{3} =-frac{4sqrt2}{9}$


$cos( alpha -2 beta )=cos alpha cdot cos2 beta +sin alpha cdot sin2 beta =$

$= frac{2}{3} cdot frac{1}{3} +frac{sqrt5}{3} cdot frac{4sqrt2}{9}=frac{2}{9}+ frac{4sqrt{10}}{27}= frac{6+4sqrt{10}}{27}= frac{2(3+2sqrt{10})}{27}$