Должно получится конкретное число.

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

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

Ответ: 0.