If $R$ is the circumradius of triangle $ABC$, can we show $a\:/\sin A = 2R$?

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