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(cos^4)x-(sin^4)x=cos2x

1 Answer

2 votes
Identities needed:
1. A^2-B^2=(A+B)(A-B)
2. sin^2(x)+cos^2(x)=1
3. cos^2(x)-sin^2(x)=cos(2x)

First, set A=cos^2(X), B=sin^2(x) and apply identity (1)
A^2-B^2=(A+B)(A-B) =>
(cos^2(x))^2-(sin^2(x))^2=(cos^2(x)+sin^2(x))*(cos^2(x)-sin^2(x))

Then apply identity (2) to the left term
(cos^2(x)+sin^2(x))*(cos^2(x)-sin^2(x))
=1*(cos^2(x)-sin^2(x))
=(cos^2(x)-sin^2(x))

Finally, apply identity (3)
(cos^2(x)-sin^2(x))=cos(2x)

All three identities are standard algebraic and trigonometric identities. Proofs are available readily on request, or search the web.
User Sadije
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