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Sum of Trigonometric Ratios in Terms of Their Product

Solve: sin 8...

Question

Solve:
$\frac{sin 8 x cos x - sin 6 x cos 3 x}{cos 2 x cos x - sin 3 x sin 4 x} = tan 2 x$

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Solution

$\begin{matrix} L H S = \frac{sin 8 x cos x - sin 6 x cos 3 x}{cos 2 x cos x - sin 3 x sin 4 x} = \frac{\frac{1}{2} (2 sin 8 x cos x) - \frac{1}{2} (2 sin 6 x cos 3 x)}{\frac{1}{2} (2 cos 2 x cos x) - \frac{1}{2} (2 sin 4 x sin 3 x)} = \frac{\frac{1}{2} {sin (8 x + x) + sin (8 x - x)} - \frac{1}{2} {sin (6 x + 3 x) + sin (6 x - 3 x)}}{\frac{1}{2} {cos (2 x + x) + cos (2 x - x)} - \frac{1}{2} {cos (4 x + 3 x) + cos (4 x - 3 x)}} = \frac{(sin 9 x + sin 7 x) - (sin 9 x + sin 3 x)}{(cos 3 x + cos x) - (cos x - cos 7 x)} = \frac{sin 9 x + sin 7 x - sin 9 x - sin 3 x}{cos 3 x + cos x - cos x + cos 7 x} = \frac{sin 7 x - sin 3 x}{cos 7 x + cos 3 x} = \frac{2 cos (\frac{7 x + 3 x}{2}) sin (\frac{7 x - 3 x}{2})}{2 cos (\frac{7 x - 3 x}{2}) cos (\frac{7 x - 3 x}{2})} = \frac{2 cos 5 x sin 2 x}{2 cos 5 x cos 2 x} = \frac{sin 2 x}{cos 2 x} = tan 2 x = R H S \end{matrix}$

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Sum of Trigonometric Ratios in Terms of Their Product

Standard XII Mathematics

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