Question

prove that cos x +cos 3x/ cos^2x -sin^2x =2 cos x​

Answer 1

Step-by-step explanation:

We will use well known formula: cos(a)+cos(b)=2cosa−b2cosa−b2.cos⁡(a)+cos⁡(b)=2cos⁡a−b2cos⁡a−b2.

cos5x+cos4x1−2cos3x=−cos2x−cosx

cos⁡5x+cos⁡4x1−2cos⁡3x=−cos⁡2x−cos⁡x

⟺cos5x+cos4x=−cos2x−cosx+2cos2xcos3x+2cosxcos3x

⟺cos⁡5x+cos⁡4x=−cos⁡2x−cos⁡x+2cos⁡2xcos⁡3x+2cos⁡xcos⁡3x

⟺cos5x+cosx+cos4x+cos2x=2cos2xcos3x+2cosxcos3x

⟺cos⁡5x+cos⁡x+cos⁡4x+cos⁡2x=2cos⁡2xcos⁡3x+2cos⁡xcos⁡3x

Answer 2

Answer:

cos5x+cos4x1−2cos3x=−cos2x−cosx

cos⁡5x+cos⁡4x1−2cos⁡3x=−cos⁡2x−cos⁡x

⟺cos5x+cos4x=−cos2x−cosx+2cos2xcos3x+2cosxcos3x

⟺cos⁡5x+cos⁡4x=−cos⁡2x−cos⁡x+2cos⁡2xcos⁡3x+2cos⁡xcos⁡3x

⟺cos5x+cosx+cos4x+cos2x=2cos2xcos3x+2cosxcos3x

⟺cos⁡5x+cos⁡x+cos⁡4x+cos⁡2x=2cos⁡2xcos⁡3x+2cos⁡xcos⁡3x