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1616. If sin⁡2θ+sin⁡θ=1\sin ^2 \theta+\sin \theta=1, then show that cos⁡2θ+cos⁡4θ=1\cos ^2 \theta+\cos ^4 \theta=1

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Q. 1616. If sin⁡2θ+sin⁡θ=1\sin ^2 \theta+\sin \theta=1, then show that cos⁡2θ+cos⁡4θ=1\cos ^2 \theta+\cos ^4 \theta=1
  1. Rewrite Equation: Rewrite the given equation using the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1.sin⁡2θ+sin⁡θ=1\sin^2 \theta + \sin \theta = 1sin⁡2θ+sin⁡θ+cos⁡2θ=1+cos⁡2θ\sin^2 \theta + \sin \theta + \cos^2 \theta = 1 + \cos^2 \theta
  2. Subtract to Isolate: Subtract sin⁡θ\sin \theta from both sides to isolate cos⁡2θ\cos^2 \theta.sin⁡2θ+cos⁡2θ=1−sin⁡θ\sin^2 \theta + \cos^2 \theta = 1 - \sin \theta
  3. Replace with Identity: Use the Pythagorean identity again to replace sin⁡2θ+cos⁡2θ\sin^2 \theta + \cos^2 \theta with 11.1=1−sin⁡θ1 = 1 - \sin \theta
  4. Add to Find Value: Add sin⁡θ\sin \theta to both sides to find the value of sin⁡θ\sin \theta.sin⁡θ=0\sin \theta = 0

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