Question

Find sin (A-B) if sin A = 4/5 with A between 90 and 180 and if cos B = 3/5 with B between 0 and 90

Answer 1

sin(A-B) = 24/25

Explanation:

The trig identity for the differnce of angles tells you ...

sin(A -B) = sin(A)cos(B) -sin(B)cos(A)

We are given that sin(A) = 4/5 in quadrant II, so cos(A) = -√(1-(4/5)^2) = -3/5.

And we are given that cos(B) = 3/5 in quadrant I, so sin(B) = 4/5.

Then ...

sin(A-B) = (4/5)(3/5) -(4/5)(-3/5) = 12/25 + 12/25 = 24/25

The desired sine is 24/25.