Question

Должно получится конкретное число.

Answer 1

$(cos62+cos28)^2=cos^262+2cos62cos28+cos^228 \ (cos62+cos28)^2-2cos62cos28=cos^262+cos^228 \ \ cos^262+cos^228=(cos62+cos28)^2-2cos62cos28= \ =(2cos frac{62+28}{2}cos frac{62-28}{2} )^2-2* frac{cos(62-28)+cos(62+28)}{2}= \ \ =(2cos45cos17)^2-(cos34+cos90)= \ =(2* frac{ sqrt{2} }{2}cos17 )^2-cos34=( sqrt{2}cos17 )^2-cos(2*17)=$
=$=2cos^217-(cos^217-sin^217)=2cos^217-cos^217+sin^217= \ =cos^217+sin^217=1$

$frac{cos^262+cos^228}{4}- frac{1}{2}cos60= frac{1}{4}- frac{1}{2}* frac{1}{2}= frac{1}{4}- frac{1}{4}=0$

Ответ: 0.