How to find sin(α-β)=

Question

I was given this problem and don't know how to solve it.

Answer 1

sinA = 3/5 in Quadrant I, cosA = 4/5

cosB = 12/13 in Quadrant III (that's a typo possibly) cosines are all negative in Quadrant III. Maybe you meant -12/13 or maybe you meant Quadrant IV where cosines >0

sinB = 5/13, or depending on the quadrant -5/13

sin(A-B) = sinAcosB - sinBcosA

= (3/5)(12/13) -(5/13)(4/5)

= (36-20)/65 = 16/65 = about .246

or if cosB =-12/13 then

sin(A-B) = (3/5)(-12/13) - (-5/13)(4/5)

= (-36 +20)/65 = -16/65 = about -.246

check on the solutions

if sinA = 3/5, then A = about 37 degrees

if cosB = 12/13, then B = about 360-23 = 337 degrees in quadrant IV

A -B = 37-337 = -300 degrees

sin-300 = sin60 = .866

or if cosB = -12/13 in quadrant III, then B = 23+180 = 203 degrees

A-B = 37-203 = 166

sin166 = .242 which confirms the 1st solution at the start of this post

Answer 2

See my other answer