We use

sin(α − β) = sin α cos β − cos α sin β

We firstly need to find `cos\ α` and `sin\ β`.

If `sin\ α = 4/5`, then we can draw a triangle and find the value of the unknown side using Pythagoras' Theorem (in this case, 3):

We do the same thing for `cos\ β = 12/13`, and we obtain the following triangle.

Note 1: We are using the positive value `12/13` to calculate the required reference angle relating to `beta`.

Note 2: The sine ratio is positive in both Quadrant I and Quadrant II.

Note 3: We have used Pythagoras' Theorem to find the unknown side, 5.

Now for the unknown ratios in the question:

`cos\ α = 3/5 `

(positive because in quadrant I)

(positive because in quadrant II)

We are now ready to find the required value, sin(α − β):

`{:(sin(alpha-beta),=sin\ alpha\ cos\ beta-cos\ alpha\ sin\ beta),(,=4/5(-12/13)-3/5(5/13)),(,(=-48-15)/65),(,=(-63)/65):}`

This is the exact value for sin(α − β).