simplify the expression to a single term:

(1-2Sin^2 X)^2 + 4sin^2 X Cos^2 X

I'm not sure how to start this or which identities to use. Any suggestions?

(1-2Sin^2 X)^2 + 4sin^2 X Cos^2 X

(sin^2 x + cos^2 x - 2sin^2 x)^2 + (2sinxcosx)^2
= (cos^2x - sin^2x)^2 + (sin 2x)^2
= (cos 2x)^2 + (sin 2x)^2
= 1

To simplify the expression (1 - 2sin^2X)^2 + 4sin^2Xcos^2X, start by applying the identity for the square of a binomial.

(1 - 2sin^2X)^2 simplifies to (1 - 4sin^4X + 4sin^2X).

Next, expand the expression 4sin^2Xcos^2X using the double angle identity for cosine.

4sin^2Xcos^2X can be rewritten as 4(1-cos^2X)cos^2X.

Distribute the 4:

4 - 4cos^2X + 4cos^4X.

Now, combine the like terms:

(1 - 4sin^4X + 4sin^2X) + 4 - 4cos^2X + 4cos^4X.

Simplify further:

1 - 4sin^4X + 4sin^2X + 4 - 4cos^2X + 4cos^4X.

Combine the constants:

5 - 4sin^4X + 4sin^2X - 4cos^2X + 4cos^4X.

Finally, we can rewrite the expression as a single term:

5 + 4cos^4X - 4sin^4X + 4sin^2X - 4cos^2X.

To simplify the given expression, you'll need to apply trigonometric identities. One useful identity to start with is the Pythagorean identity, which states that sin^2(X) + cos^2(X) = 1.

Let's simplify the given expression step by step:

(1 - 2sin^2(X))^2 + 4sin^2(X)cos^2(X)

First, let's simplify the square of the expression:

(1 - 2sin^2(X))^2 = (1 - 2sin^2(X)) * (1 - 2sin^2(X))

Now, expand the product using the distributive property:

(1 * 1) + (1 * -2sin^2(X)) + (-2sin^2(X) * 1) + (-2sin^2(X) * -2sin^2(X))

Simplify each term:

1 - 2sin^2(X) - 2sin^2(X) + 4sin^4(X)

Combine like terms:

1 - 4sin^2(X) + 4sin^4(X)

Now, let's simplify the second term:

4sin^2(X)cos^2(X)

Since sin^2(X) + cos^2(X) = 1 (Pythagorean identity), we can replace sin^2(X) with (1 - cos^2(X)):

4(1 - cos^2(X))cos^2(X)

Now, distribute the 4:

4cos^2(X) - 4cos^4(X)

Finally, let's combine the simplified terms from both expressions:

(1 - 4sin^2(X) + 4sin^4(X)) + (4cos^2(X) - 4cos^4(X))

Now, you have a single term expression. However, further simplification is not possible since it does not reduce or factorize further.