Question

cos²(π\4-A)-sin²(π\4-B)=sin(A+B)cos(A-B) prove

Answer 1

RHS= sin(A+B) cos(A-B)
=(sinAcosB+cosAsinB)(cosAcosB+sinAsinB)
=sinAcosA(cosB)^2+ sinBcosB(sinA)^2+
sinBcosB(cosA)^2+sinAcosA(sinB)^2
=sinAcosA{(cosB)^2+(sinB)^2}+sinBcosB{(sinA)^2+(cosA)^2}
=sinAcosA+sinBcosB
LHS=cos²(π\4-A)-sin²(π\4-B)
=cosπ\4cosA+sinπ\4sinA-{sinπ\4cosB+
cosπ\4sinB
=1\√2cosA+1\√2sinA-1\√2cosB-
1\√2sinB
=(cosA+sinA-cosB-sinB)/√2