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Prove that (cosx-sinx)(cosx+sinx)=cos2x?
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We'll manage the left side because we notice that the special product from the left side returns the difference of two squares:
(cosx-sinx)(cosx+sinx)=cos^2 x - sin^2 x
But this difference of two squares represents the double angle formula:
cos 2x = cos^2 x - sin^2 x
Another way to solve the problem is to remove the brackets from the left side:
(cosx-sinx)(cosx+sinx) = cos x*cos x + cos x*sin x - cos x*sin x - sin x*sin x
We'll eliminate like terms:
(cosx-sinx)(cosx+sinx) = cos x*cos x - sin x*sin x
We'll recognize the identity:
cos x*cos x - sin x*sin x = cos (x + x) = cos 2x
Therefore, the given expression (cosx-sinx)(cosx+sinx) = cos 2x represents an identity.
Posted by giorgiana1976 on September 14, 2011 at 12:09 AM (Answer #1)
Valedictorian, Super Tutor, Tutor, Dean's List
L:H:S ≡ (cosx-sinx)(cosx+sinx)
= cos²x - sin²x
= cosx.cosx - sinx.sinx
Posted by lochana2500 on June 23, 2012 at 3:30 PM (Answer #2)
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