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cos⁡2xsin⁡3x−sin⁡x−1\frac{\cos 2x}{\sin 3x - \sin x} - 1

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Full solution

Q. cos⁡2xsin⁡3x−sin⁡x−1\frac{\cos 2x}{\sin 3x - \sin x} - 1
  1. Apply identities: Apply the sum-to-product identities to simplify the denominator sin⁡3x−sin⁡x\sin 3x - \sin x.\newlinesin⁡3x−sin⁡x=2cos⁡(2x)sin⁡(x)\sin 3x - \sin x = 2\cos(2x)\sin(x)
  2. Rewrite expression: Rewrite the original expression using the simplified denominator.\newlinecos⁡2x2cos⁡(2x)sin⁡(x)−1\frac{\cos 2x}{2\cos(2x)\sin(x)}-1
  3. Cancel common factors: Simplify the expression by canceling out common factors. \newline12sin⁡(x)−1\frac{1}{2\sin(x)} - 1

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