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cos 3x-cos 5x=sin 4x

cos⁡3x−cos⁡5x=sin⁡4x \cos 3 x-\cos 5 x=\sin 4 x

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Full solution

Q. cos⁡3x−cos⁡5x=sin⁡4x \cos 3 x-\cos 5 x=\sin 4 x
  1. Apply sum-to-product identities: Use the sum-to-product identities to simplify the left side of the equation.\newlineThe sum-to-product identities state that cos⁡A−cos⁡B=−2sin⁡(A+B2)sin⁡(A−B2)\cos A - \cos B = -2 \sin\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right).\newlineSo, cos⁡3x−cos⁡5x\cos 3x - \cos 5x can be written as −2sin⁡(3x+5x2)sin⁡(3x−5x2)-2 \sin\left(\frac{3x + 5x}{2}\right) \sin\left(\frac{3x - 5x}{2}\right).
  2. Use identity to simplify: Apply the sum-to-product identity to the left side of the equation.\newlinecos⁡3x−cos⁡5x=−2sin⁡(3x+5x2)sin⁡(3x−5x2)\cos 3x - \cos 5x = -2 \sin\left(\frac{3x + 5x}{2}\right) \sin\left(\frac{3x - 5x}{2}\right)\newline=−2sin⁡(4x)sin⁡(−x)= -2 \sin(4x) \sin(-x)\newlineSince sin⁡(−x)=−sin⁡(x)\sin(-x) = -\sin(x), this simplifies to:\newline=2sin⁡(4x)sin⁡(x)= 2 \sin(4x) \sin(x)
  3. Substitute simplified expression: Substitute the simplified left side back into the original equation.\newline2sin⁡(4x)sin⁡(x)=sin⁡4x2 \sin(4x) \sin(x) = \sin 4x
  4. Divide by sin⁡(4x)\sin(4x): Divide both sides of the equation by sin⁡(4x)\sin(4x), assuming sin⁡(4x)\sin(4x) is not zero.\newlinesin⁡(4x)sin⁡(4x)=2sin⁡(4x)sin⁡(x)sin⁡(4x)\frac{\sin(4x)}{\sin(4x)} = \frac{2 \sin(4x) \sin(x)}{\sin(4x)}\newline1=2sin⁡(x)1 = 2 \sin(x)
  5. Isolate sin⁡(x)\sin(x): Divide both sides by 22 to isolate sin⁡(x)\sin(x).12=sin⁡(x)\frac{1}{2} = \sin(x)
  6. Find general solution: Find the general solution for xx.\newlineSince sin⁡(x)=12\sin(x) = \frac{1}{2}, xx can be any angle whose sine is 12\frac{1}{2}.\newlineThis occurs at x=π6+2πnx = \frac{\pi}{6} + 2\pi n or x=5π6+2πnx = \frac{5\pi}{6} + 2\pi n, where nn is any integer.

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