Question

let sinA =3/5 in QII and sinB = -5/13 in QIII find sin(A-B), cos(A-B),and tan (A-B). What quadrant does (A-B) terminate in?

Answer 1

\[\sin(A-B)=\sin A\cos B-\cos B\sin A=-3/5\cdot\sqrt{1-(5/13)^2}+\sqrt{1-(3/5)^2}\cdot5/13\]

\[\cos(A-B)=\cos A\cos B+\sin B\sin A=\sqrt{1-(3/5)^2}\cdot\sqrt{1-(5/13)^2}-5/13 \cdot 3/5\]

\[\tan(A-B)=\sin(A-B)/\cos(A-B)\]

Answer 2

The last fraction of sin(A-B) is 5/13.

Answer 3

The final angle is in the Q IV, because the sin(A-B)<0 and cos(A-B)>0.