Show that sin 5 A + 2 sin 8 A + sin 11 A / sin. 8 A + 2 sin 11 A + sin 14 A = sin. 8 A / sin 11 A?

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Solution

$L . H . S . = \frac{\sin 5 A + 2 \sin 8 A + \sin 11 A}{\sin 8 A + 2 \sin 11 A + \sin 14 A} = \frac{(\sin 5 A + \sin 11 A) + 2 \sin 8 A}{(\sin 8 A + \sin 14 A) + 2 \sin 11 A} = \frac{2 \sin (\frac{5 A + 11 A}{2}) \cos (\frac{5 A - 11 A}{2}) + 2 \sin 8 A}{2 \sin (\frac{8 A + 14 A}{2}) \cos (\frac{8 A - 14 A}{2}) + 2 \sin 11 A} = \frac{2 \sin (8 A) \cos (3 A) + 2 \sin 8 A}{2 \sin (11 A) \cos (3 A) + 2 \sin 11 A} = \frac{2 \sin (8 A) (\cos (3 A) + 1)}{2 \sin (11 A) (\cos (3 A) + 1)} = \frac{\sin 8 A}{\sin 11 A} = R . H . S . Hence Proved .$

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