Вычислить :1) 3arccos(cos2) 2)cos(arccos(4/5)-arccos(3/5))

Решите задачу:

$3arccos(cos2)=3*2=6\\\\cos(arccos(4/5)-arccos(3/5))=\\=cos(arccos(4/5))*cos(arccos(3/5))+\\+sin(arccos(4/5))*sin(arccos(3/5))=\\=(4/5)*(3/5)+ \sqrt{1-(4/5)^2}* \sqrt{1-(3/5)^2}=\\=(12/25)+ \sqrt{1-(16/25)}* \sqrt{1-(9/25)}=\\=(12/25)+ \sqrt{(9/25)}* \sqrt{(16/25)}=\\=(12/25)+(3/5)*(4/5)=(12/25)+(12/25)=24/25$