Question 1131550
{{{cos(A+B)cos(A-B)=cos^2(A)+cos^2(B)-1 }}}

start with left side:


{{{cos(A+B)cos(A-B)}}}


={{{(cos(A) cos(B) - sin(A) sin(B))(sin(A) sin(B) + cos(A) cos(B))}}}


={{{cos(A)*cos(B)*sin(A)*sin(B)+cos^2(A)* cos^2(B) - sin^2(A) *sin^2(B)- sin(A)* sin(B)*cos(A)*cos(B)}}}


={{{cross(cos(A) cos(B)sin(A) sin(B))+cos^2(A) cos^2(B) - sin^2(A) sin^2(B) - cross(sin(A) sin(B)cos(A) cos(B))}}}


={{{cos^2(A) cos^2(B) - sin^2(A) sin^2(B))}}}..............since {{{sin^2(A)=1-cos^2(A)}}} and {{{sin^2(B)=1-cos^2(B)}}}, we have


={{{cos^2(A) cos^2(B) - (1-cos^2(A)) (1-cos^2(B))}}}


={{{cos^2(A) cos^2(B) -  (1-cos^2(B)-cos^2(A)+cos^2(A)cos^2(B))}}}


={{{cos^2(A) cos^2(B) -  1+cos^2(B)+cos^2(A)-cos^2(A)cos^2(B)}}}


={{{cross(cos^2(A) cos^2(B)) -  1+cos^2(B)+cos^2(A)-cross(cos^2(A)cos^2(B))}}}


={{{ -  1+cos^2(B)+cos^2(A)}}} rearrange the terms


={{{cos^2(A)+cos^2(B)-  1 }}}-> identity is proven