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General Solution of Trigonometric Equation

Evaluate: c...

Question

Evaluate: $\frac{(cos x - cos 3 x) (sin 8 x + sin 2 x)}{(sin 5 x - sin x) (cos 4 x - cos 6 x)}$

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Solution

Given:

$\frac{(cos x - cos 3 x) (sin 8 x + sin 2 x)}{(sin 5 x - sin x) (cos 4 x - cos 6 x)}$

Use the formula:
$sin a + sin b = 2 sin (\frac{a + b}{2}) cos (\frac{a - b}{2})$

$sin a - sin b = 2 cos (\frac{a + b}{2}) sin (\frac{a - b}{2})$

$cos a - cos b = - 2 sin (\frac{a + b}{2}) sin (\frac{a - b}{2})$

$\frac{(cos x - cos 3 x) (sin 8 x + sin 2 x)}{(sin 5 x - sin x) (cos 4 x - cos 6 x)} = \frac{(- 2 sin \frac{4 x}{2} sin \frac{- 2 x}{2}) (2 sin \frac{10 x}{2} cos \frac{6 x}{2})}{(2 cos \frac{6 x}{2} sin \frac{4 x}{2}) (- 2 sin \frac{10 x}{2} sin \frac{- 2 x}{2})}$

$= \frac{(2 sin 2 x sin x) (2 sin 5 x cos 3 x)}{(2 cos 3 x sin 2 x) (2 sin 5 x sin x)}$

$= 1$

Thus, the value is $1$ .

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General Solution of Trigonometric Equation

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