`cos^2(beta) - sin^2(beta) = 2cos^2(beta) - 1` Verfiy the identity.

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Chapter 5, 5.2 - Problem 14 - Precalculus (3rd Edition, Ron Larson).
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mathace | (Level 3) Assistant Educator

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Verify the identity: `cos^2(beta)-sin^2(beta)=2cos^2(beta)-1`

Use the pythagorean identity `sin^2(beta)+cos^2(beta)=1.`

If  `sin^2(beta)`  is isolated the pythagorean identity would be `sin^2(beta)=1-cos^2(beta).`

`cos^2(beta)-sin^2(beta)=2cos^2(beta)-1`

`cos^2(beta)-(1-cos^2(beta))=2cos^2(beta)-1`

`cos^2(beta)-1+cos^2(beta)=2cos^2(beta)-1`

` ` `2cos^2(beta)-1=2cos^2(beta)-1`

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