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Verifying Double Angle Trig Identities

Prove:

cos(4x) = 1 - 8(sinx)^2(cosx)^2

I have tried and tried and keep ending up with:

(cosx)^4 - 6(cosx)^2(sinx)^2 + (sinx)^4

Here's a hint:

$sin^{2}(x)cos^{2}(x)=\frac{1-cos(4x)}{8}$

Use the identities:

$sin^{2}(x)=\frac{1-cos(2x)}{2}$

$cos^{2}x=\frac{1+cos(2x)}{2}$

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