[theta(v)cos(A+B)+sin(A-B)],[=2sin(45^(@)+A]

cos (A +B) + sin (A -B) = 2 sin (45 ^(@) + A) cos ( 45 ^(@) + B).

cos (A + B) + sin (AB) = 2sin (45 ^ (0) + A) cos (45 ^ (@) + B)

Prove the following : cos(A+B)+sin(A-B) 2sin( 45^@ +A)cos( 45^@ +B)

Prove that: i) sin(A+B)cos(A-B)-cos(A+B)sin(A-B)=sin2B ii) cos(45^(@)-A)cos(45^(@)-B)-sin(45^(@)-A)sin(45^(@)-B)=sin(A+B)

Prove that: i) sin(A+B)cos(A-B)-cos(A+B)sin(A-B)=sin2B ii) cos(45^(@)-A)cos(45^(@)-B)-sin(45^(@)-A)sin(45^(@)-B)=sin(A+B)

Prove that: cos(45^(@)-A)cos(45^(@)-B)-sin(45^(@)-A)sin(45^(@)-B)=sin(A+B)

cos (45 ^ (@) - A) cos (45 ^ (@) - B) -sin (45 ^ (@) - A) sin (45 ^ (@) - B) = sin (A + B)

prove that , |{:(sin A ,cos A,sin (A+theta)),(sin B ,cos B,sin (B+theta)),(sin C ,cos C ,sin (C+theta)):}|=0

det [[sin A, cos A, sin (A + theta) sin B, cos B, sin (B + theta) sin C, cos C, sin (C + theta)]] = =

Let A={ theta in R|cos^(2) (sin theta)+sin^(2) (cos theta)=1} and B={ theta in R| cos (sin theta) sin (cos theta)=0} . Then A nn B