Question

求sin∧2a·sin∧2β+cos∧2a·cos∧2β-1/2cos2a·cos2β..

Answer 1

(sina)^2*(sinb)^2+(cosa)^2*(cosb)^2-(1/2)(cos2acos2b)=(sinasinb)^2+(cosacosb)^2-(1/2)[cos2acos2b)=(1/4)(cos(a-b)-cos(a+b))^2+(1/4)cos(a-b)+cos(a+b))^2-(1/4)cos(2a+2b)-(1/4)cos(2a-2b)=(1/2)cos(a+b)^2+(...