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Trigonometric Ratios of Compound Angles

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Question

prove that |1 cosA sinA||1 cosB sinB| |1 cosC sinC| = 4 sin(A-B/2) sin(B-C/2) sin(C-A/2)

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Solution

$LHS = |\begin{matrix} 1 & cosA & sinA \\ 1 & cosB & sinB \\ 1 & cosC & sinC \end{matrix}| = |\begin{matrix} 1 & cosA & sinA \\ 0 & cosB - cosA & sinB - sinA \\ 0 & cosC - cosA & sinC - sinA \end{matrix}| (R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1}) = 1 [(cosB - cosA) (sinC - sinA) - (sinB - sinA) (cosC - cosA)] = 2 \sin (\frac{A + B}{2}) \sin (\frac{A - B}{2}) \times 2 \sin (\frac{C - A}{2}) \cos (\frac{C + A}{2}) - 2 \sin (\frac{B - A}{2}) \cos (\frac{B + A}{2}) \times 2 \sin (\frac{A + C}{2}) \sin (\frac{A - C}{2}) = 4 \sin (\frac{A + B}{2}) \sin (\frac{A - B}{2}) \sin (\frac{C - A}{2}) \cos (\frac{C + A}{2}) - 4 \sin (\frac{A - B}{2}) \cos (\frac{A + B}{2}) \sin (\frac{C + A}{2}) \sin (\frac{C - A}{2}) = 4 \sin (\frac{A - B}{2}) \sin (\frac{C - A}{2}) [\sin (\frac{A + B}{2}) \cos (\frac{C + A}{2}) - \cos (\frac{A + B}{2}) \sin (\frac{C + A}{2})] = 4 \sin (\frac{A - B}{2}) \sin (\frac{C - A}{2}) \times \sin [(\frac{A + B}{2}) - (\frac{C + A}{2})] [\sin X \cos Y - \cos X \sin Y = \sin (X - Y)] = 4 \sin (\frac{A - B}{2}) \sin (\frac{C - A}{2}) \sin (\frac{B - C}{2}) = 4 \sin (\frac{A - B}{2}) \sin (\frac{B - C}{2}) \sin (\frac{C - A}{2}) = RHS$

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