Question: Given: A and B are in quadrant 4, cos A=(15/17) and tan B=-(5/12); How do I find out sin(A+B) and cos(A+B)??
In the fourth quadrant, only cos() is positive. Both sin() and tan() are negative.
cos(A) = 15/17.
Now draw a right angled triangle, with the hypotenuse = 17 and the adjacent side = 15 (that will give you cos() = 15/17 = cos(A) )
The opposite side then = sqrt(17^2 - 15^2) = sqrt(64) = 8
Therefore sin() = opposite/hypotenuse = 8/17.
But A is in the fourth quadrant making sin(A) negative. i.e. sin(A) = -8/17
tan(B) = -5/12.
Now draw a right angled triangle, with the opposite side = 5 and the adjacent side = 12 (that will give you tan() = 5/12 = tan(B).) (ignoring sign for the moment)
The hypotenuse then = sqrt(5^2 + 12^2) = sqrt(169) = 13.
Therefore sin() = opposite/hypotenuse = 5/13.
But B is in the fourth quadrant making sin(B) negative. i.e. sin(B) = -5/13
Also cos() = adjacent/hypotenuse = 12/13.
Again, B is in the fourth quadrant, this time making cos(B) positive. i.e. cos(B) = 12/13
Now we work out sin(A+B) and cos(A+B).
sin(A+B) = sin(A).cos(B) + cos(A).sin(B)
sin(A+B) = (-8/17).(12/13) + (15/17).(-5/13) = -96/221 - 75/221 = -171/221
sin(A+B) =171/221
cos(A+B) = cos(A).cos(B) - sin(A).sin(B)
cos(A+B) = (15/17).(12/13) - (-8/17).(-5/13) = 180/221 - 40/221 = 140/221
cos(A+B) = 140/221