Posted by Hannah on Monday, March 29, 2010 at 9:38pm.
so the angle u could be in quadrant I inside a 5:12:13 triangle, with 13 as the hypotenuse,
so cos u = 12/13
angle v could be in quadrants II or III and are inside the 3:4:5 triangle , with 5 the hypotenuse, so
sin v = 4/5
draw the triangle in quadrant II with those dimensions
You must by now have learned that
sin(A-B) = sinAcosB - cosAsinB
then sin(u-v) = sinucosv - cosusinv
= (5/13)(-3/5) - (12/13)(4/5) = -63/65
Thank You
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