cos 2x cos x2-cos 3x cos 9x2=2cos 2x cos x2-2cos 3x cos 9x22Use 2cos x cos y=cos x+y+cos x-y=cos 2x+x2+cos 2x-x2-cos 3x+9x2+cos 9x2-3x2=cos 4x+x2+cos 4x-x2-cos 9x+6x2+cos 9x-6x22=cos 5x2+cos 3x2-cos 15x2+cos 3x22=cos 5x2+cos 3x2-cos 15x2-cos 3x22=cos 5x2-cos 15x22Use cos x-cos y=2 sin x+y2 sin y-x2=2sin 15x2+5x22 sin 15x2-5x222=2sin 20x4 sin 10x42=sin 5x sin 5x2Please check your question The answer is isn't matching

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Please solve fast.... — 4stn cos Prove that cos 2xcos cos RHS. 9x — cos 31 cos — = —sin 5x sin 2 2 N We ha ve

\cos 2 x \cos \frac{x}{2} - \cos 3 x \cos \frac{9 x}{2} = \frac{2 \cos 2 x \cos \frac{x}{2} - 2 \cos 3 x \cos \frac{9 x}{2}}{2} U s e 2 \cos x \cos y = \cos (x + y) + \cos (x - y) = \frac{\cos (2 x + \frac{x}{2}) + \cos (2 x - \frac{x}{2}) - \{\cos (3 x + \frac{9 x}{2}) + \cos (\frac{9 x}{2} - 3 x)\}}{2} = \frac{\cos (\frac{4 x + x}{2}) + \cos (\frac{4 x - x}{2}) - \{\cos (\frac{9 x + 6 x}{2}) + \cos (\frac{9 x - 6 x}{2})\}}{2} = \frac{\cos (\frac{5 x}{2}) + \cos (\frac{3 x}{2}) - \{\cos (\frac{15 x}{2}) + \cos (\frac{3 x}{2})\}}{2} = \frac{\cos (\frac{5 x}{2}) + \cos (\frac{3 x}{2}) - \cos (\frac{15 x}{2}) - \cos (\frac{3 x}{2})}{2} = \frac{\cos (\frac{5 x}{2}) - \cos (\frac{15 x}{2})}{2} U s e \cos x - \cos y = 2 \sin \frac{x + y}{2} \sin \frac{y - x}{2} = \frac{2 \sin (\frac{\frac{15 x}{2} + \frac{5 x}{2}}{2}) \sin (\frac{\frac{15 x}{2} - \frac{5 x}{2}}{2})}{2} = \frac{2 \sin (\frac{20 x}{4}) \sin (\frac{10 x}{4})}{2} = \sin 5 x \sin (\frac{5 x}{2}) P l e a s e c h e c k y o u r q u e s t i o n . T h e a n s w e r i s i s n' t m a t c h i n g .

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Please solve fast.... Please solve fast.... — 4stn cos Prove that cos 2xcos cos RHS. 9x — cos 31 cos — = —sin 5x sin 2 2 N We ha ve

Please solve fast....

Please solve fast.... — 4stn cos Prove that cos 2xcos cos RHS. 9x — cos 31 cos — = —sin 5x sin 2 2 N We ha ve