With usual notations,in a triangle ABC,a cos( | vedclass

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(A) We know that $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.Substituting these in the expression $S = a \cos(B - C) + b \cos(C - A) + c \cos(

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$\frac{abc}{R^2}$

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(A) We know that $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.Substituting these in the expression $S = a \cos(B - C) + b \cos(C - A) + c \cos(A - B)$:$S = 2R \sin A \cos(B - C) + 2R \sin B \cos(C - A) + 2R \sin C \cos(A - B)$.Since $A = 180^{\circ} - (B + C)$,$\sin A = \sin(B + C)$.$S = 2R [\sin(B + C) \cos(B - C) + \sin(C + A) \cos(C - A) + \sin(A + B) \cos(A - B)]$.Using $2 \sin X \cos Y = \sin(X + Y) + \sin(X - Y)$:$S = R [(\sin 2B + \sin 2C) + (\sin 2C + \sin 2A) + (\sin 2A + \sin 2B)]$.$S = 2R (\sin 2A + \sin 2B + \sin 2C)$.Using the identity $\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C$:$S = 2R (4 \sin A \sin B \sin C) = 8R \sin A \sin B \sin C$.Since $\sin A = \frac{a}{2R}$,$\sin B = \frac{b}{2R}$,and $\sin C = \frac{c}{2R}$:$S = 8R \left(\frac{a}{2R}\right) \left(\frac{b}{2R}\right) \left(\frac{c}{2R}\right) = \frac{8abc}{8R^2} = \frac{abc}{R^2}$.

With usual notations,in a triangle $ABC$,$a \cos(B - C) + b \cos(C - A) + c \cos(A - B)$ is equal to

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