If Sin A=3/5 and cos B= -12/13, find the value of sin(A-B) if Angle A is in Quadrant I and angle B is in Quadrant III.

in QI, sinA=3/5 means cosA = 4/5

in QIII, cosB = -12/13 means sinB = -5/13

sin(A-B) = sinAcosB - cosAsinB = (3/5)(-12/13) - (4/5)(-5/13) = (-36+20)/65 = -16/65

To find the value of sin(A-B), we can use the trigonometric identity sin(A-B) = sin A cos B - cos A sin B.

Given that sin A = 3/5 and cos B = -12/13, we need to find the values of cos A and sin B in order to proceed.

Since angle A is in Quadrant I, both sine and cosine will be positive. Therefore, we can find cos A using the Pythagorean identity sin^2 A + cos^2 A = 1:
(3/5)^2 + cos^2 A = 1
9/25 + cos^2 A = 1
cos^2 A = 16/25
cos A = ±4/5

Since angle B is in Quadrant III, cosine will be negative and sine will be positive. Therefore, we can find sin B using the Pythagorean identity sin^2 B + cos^2 B = 1:
sin^2 B + (-12/13)^2 = 1
sin^2 B + 144/169 = 1
sin^2 B = 25/169
sin B = ±5/13

Now that we have the values of sin A, cos A, sin B, and cos B, we can substitute them into the formula sin(A-B) = sin A cos B - cos A sin B:

sin(A-B) = (3/5)(-12/13) - (4/5)(5/13)
= -36/65 - 20/65
= -56/65

Therefore, sin(A-B) = -56/65.