With the usual notation prove that \[\begin{equation*} \frac{a}{\sin A} = 2R \end{equation*}\]

where \(R\) is the radius of the circumcircle of the triangle \(ABC\).

[You need consider only the case when the angle \(A\) is acute.]

Hence or otherwise deduce that

  1. \(a \cos A + b \cos B = c \cos (A-B)\),
  2. \(a \cos A + b \cos B + c \cos C = 4 R \sin A \sin B \sin C\).