Question

prove that cos (2x-3y)-cos(3x-2y)=2sin (sin/2)sin(x+y/2)​

Answer 1

Step-by-step explanation:

(2x-3y) - cos(3x-2y) = 2sin(sin/2)sin(x+y)/2

LHS = cos(2x-3y) - cos(3x-2y)

we know that cosA - cosB = -2sin(A+B)/2 sin(A-B) /2

= cos(2x-3y) - cos(3x -2y)

= - 2sin(2x-3y + 3x-2y) / 2 * sin(2x-3y - 3x+2y) / 2

= -2sin(5x-5y) / 2* sin(-x-y) / 2 (sin(-x) = - sinx)

= -2sin5/2(x-y) * (-sin(x+y))

= 2sin5/2*(x-y) * sin(x+y) /2

I think the question is incomplete in RHS part.

RHS = 2sin(sin/2) sin(x+y) /2

which is not correct since sin/2 should have something like sin(x-y).

please recheck the RHS part, and upload the question again.

hope u understood it