I have a question on an online exercise that doesn't provide an answer or an explanation. I believe I know how to do it, but the answer I get every time is marked wrong when I submit it. I need a 2nd set of eyeballs.
Find sin(A+B) if cos(A) = 3/5 and sin(B) = -4/5 and A is in QI and B is in QIII.
using identities:
sin(A+B) = sin(A)cos(B) + cos(A)sin(B)
plugging in what we know we start with:
sin(A)cos(B) + (3/5)*(-4/5) = sin(A)cos(B) - 12/5
Both of these are 3,4,5 right triangles. 3 and 4 are positive for angle A (QI) and 3 and 4 are negative for angle B (QIII).
sin(A) = 4/5
cos(B) = -3/5
again plugging in we have:
(4/5) * (-3/5) - 12/5
-12/5 - 12/5 = -24/5
That's not the answer, so where did I go wrong?
Thanks
$$\sin(A+B)=\sin A\cos B+\cos A\sin B$$
Putting the values in
$$\sin(A+B)=(4/5)(-3/5)+(3/5)(-4/5)$$
$$\sin(A+B)= -24/25$$