Prove that frac{cos 9x - cos 5x}{sin 17x - si | vedclass

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We use the trigonometric identities:$\cos A - \cos B = -2 \sin \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right)$$\sin A - \sin B = 2 \co

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We use the trigonometric identities:$\cos A - \cos B = -2 \sin \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right)$$\sin A - \sin B = 2 \cos \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right)$$L.H.S. = \frac{\cos 9x - \cos 5x}{\sin 17x - \sin 3x}$Applying the identities:$= \frac{-2 \sin \left( \frac{9x+5x}{2} \right) \sin \left( \frac{9x-5x}{2} \right)}{2 \cos \left( \frac{17x+3x}{2} \right) \sin \left( \frac{17x-3x}{2} \right)}$$= \frac{-2 \sin(7x) \sin(2x)}{2 \cos(10x) \sin(7x)}$Canceling the common term $\sin(7x)$ and the constant $2$:$= -\frac{\sin 2x}{\cos 10x}$$= R.H.S.$

Prove that $\frac{\cos 9x - \cos 5x}{\sin 17x - \sin 3x} = -\frac{\sin 2x}{\cos 10x}$

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