Question

cos ( π/4-A)-cos (π/4+4)=√2sinA

Answer 1

Since cosC-cosD = -sin[(C+D)/2]sin[(C-D)/2]

So

cos ( π/4-A)-cos (π/4+A) = -sin[(π/4-A+π/4+A)/2]sin[(π/4-A-π/4-A)/2]

                                       = -sin[(π/2)/2]sin[(-2A)/2]

                                      = sin(π/4)sin(A)                 { since sin(-θ) = -sinθ ]

                                      = (1/√2)sinA                     [ sin(π/4) = 1/√2]

cos ( π/4-A)-cos (π/4+A) = sinA/√2