If $\cos A = 4/5$ and $\sin B = 5/13$, where $A$ is a acute and $B$ is obtuse, find, without evaluating the angles $A$ and $B$, the values of
a) $\sin (A-B)$
b) $\cos (A+B)$
I'm stuck figuring out the obtuse angle.
I know obtuse is somewhere inbetween 180 and 90 degrees but i'm confused on how to apply on it..
Note that $$\sin(A-B)=\sin(A)\cos(B)-\sin(B)\cos(A)$$ And $$\cos(A+B)=\cos(A)\cos(B)+\sin(A)\sin(B)$$ and further more to gain a sine given a cosine of vice versa remember the pythagorean trigonometric identity, ie $$\sin^2(x)+\cos^2(x)=1$$ Thus $$\sin(x)=\pm\sqrt{1-\cos^2(x)}$$ $$\cos(x)=\pm\sqrt{1-\sin^2(x)}$$ and remember that the cosine of an obtuse angle is always negative.