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Prove that: sin^2 A+cos^2 A*cos2B = sin^2 B - cos^2 B*cos2ADo it by using multiple...
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You should remember that cos `2 A = 2cos^2 A - 1` and `cos 2B = 2cos^2 B- 1` .
You need to substitute `2cos^2 B - 1` for `cos 2B` and `2cos^2 A - 1` for `cos 2 A` such that:
`sin^2 A + cos^2A*(2cos^2 B - 1) = sin^2B+ cos^2B*(2cos^2A - 1)`
You should open the brackets such that:
`sin^2 A + 2cos^2A*cos^2B - cos^2A = sin^2 B + 2cos^2A*cos^2B - cos^2B`
Reducing `2cos^2A*cos^2B` both sides yields:
`sin^2 A - cos^2A = sin^2 B - cos^2B`
`-cos 2A = - cos 2B`
Hence, the identity may be verified if the angles A and B have equal measures.
Posted by sciencesolve on June 27, 2012 at 3:12 PM (Answer #1)
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