Question

найти 25cos ( 2arccos 0 + 2arccos 0,8 )

Answer 1

$25*(cos(2arccos0)*cos(2arccos0.8) - sin(2arccos0)*sin(2arccos0.8))$$=25*(cos(2* frac{ pi }{2})*cos(2arccos0.8)-sin(2* frac{ pi }{2})*sin(2arccos0.8))$$=25*(cos(pi)*cos(2arccos0.8)-sin( pi)*sin(2arccos0.8))=$$-25*cos(2arccos0.8)=-25*(2cos^{2}(arccos0.8)-1)=$$-50cos^{2}(arccos0.8)+25=-50*0.8^{2}+25=-32+25=-7$