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Property 7

cos 4x sin 3x...

Question

cos 4x sin 3x - cos 2x sin x / sin 4x sin x +cos 6x cos x = tan 2x . prove it

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Solution

$L . H . S . = \frac{\cos 4 x \sin 3 x - \cos 2 x \sin x}{\sin 4 x \sin x + \cos 6 x \cos x} = \frac{2 \cos 4 x \sin 3 x - 2 \cos 2 x \sin x}{2 \sin 4 x \sin x + 2 \cos 6 x \cos x} = \frac{[\sin (4 x + 3 x) - \sin (4 x - 3 x)] - [\sin (2 x + x) - \sin (2 x - x)]}{[\cos (4 x - x) - \cos (4 x + x)] + [\cos (6 x + x) + \cos (6 x - x)]} = \frac{\sin 7 x - \sin x - \sin 3 x + \sin x}{\cos 3 x - \cos 5 x + \cos 7 x + \cos 5 x} = \frac{\sin 7 x - \sin 3 x}{\cos 3 x + \cos 7 x} = \frac{2 \cos (\frac{7 x + 3 x}{2}) \sin (\frac{7 x - 3 x}{2})}{2 \cos (\frac{7 x + 3 x}{2}) \cos (\frac{7 x - 3 x}{2})} = \frac{\sin 2 x}{\cos 2 x} = \tan 2 x = R . H . S H e n c e P r o v e d .$

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\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

y = \frac{cos x + cos 2 x + cos 3 x + cos 4 x + cos 5 x + cos 6 x + cos 7 x}{sin x + sin 2 x + sin 3 x + sin 4 x + sin 5 x + sin 6 x + sin 7 x},

then which of the following hold good?

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

sin x 2 x + 2 sin 4 x + s i n 6 x = 4 {cos}^{2} x . sin 4 x

sin 2 x + 2 sin 4 x + sin 6 x = 4 {cos}^{2} x sin 4 x

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