Show that for any triangle ABC, we have sin A cos C + A cos B > 0.
Solution
The result is obviously true unless B or C is obtuse. If C is obtuse, then |cos C| < cos B (because |cos C| = cos(A + B) < cos B). But sin A < A, so the result follows. So assume that B is obtuse.
We have A acute and hence A < tan A. So A cos A < sin A. But C is acute, so cos C is positive and A cos A cos C < sin A cos C. Now A sin A sin C is positive, so A cos(A + C) = A cos A cos C - A sin A sin C < A cos A cos C < sin A cos C. But cos(A + C) = -cos B, so - A cos B < sin A cos C.
© John Scholes
jscholes@kalva.demon.co.uk
4 Dec 1999