acos A + b cos B + c cos C = 2b sin A sin C We can observe that we all the terms present in the equation to be proved are not showing any resemblance with known formula but the term is RHS side has sine terms, so there is a possibility that sine formula can solve our problem Theangle sum trigonometric identity in sine function is written in several forms but it is popularly expressed in the following three forms. ( 1). sin ( A + B) = sin A cos B + cos A sin B. ( 2). sin ( x + y) = sin x cos y + cos x sin y. ( 3). sin ( α + β) = sin α cos β + cos α sin β. Thenit's just a matter of using algebra. so sin (alpha) = x/B and sin (beta) = x/A. So in less math, splitting a triangle into two right triangles makes it so that perpendicular equals both A * sin (beta) and B * sin (alpha). Then you can further rearange this to get the law of sines as we know it. Thekey Angle Sum and Difference Formulas are: cos(a + b) = cos(a)*cos(b) - sin(a)*sin(b) cos(a - b) = cos(a)*cos(b) + sin(a)*sin(b) -sin(A-B)sin(B) results in cos(C)cos(B)-sin(C)sin(B) which is equivalent to cos(C+B) by the trig identity for the cosine of a sum of two angles. Now substitute A-B for C into the last expression producing cos Iknow that cos is even while sin is odd, and I know $\cos(\pi)=\sin((\pi/2)-x)$, but I still can't figure the derivation of $\sin (a+b)$ from $\cos(a+b)=\cos(a)\cos OiKK.

sin a sin b sin c formula