# Prove that (cosx-sinx)(cosx+sinx)=cos2x?

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L:H:S ≡ (cosx-sinx)(cosx+sinx)

= cos²x - sin²x

= cosx.cosx - sinx.sinx

= cos(x+x)

= cos2x

= R:H:S

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.**