Prove that cos^4A + sin²A = sin^4A + cos²A

We are tasked with proving the identity:

\[
\cos^4A + \sin^2A = \sin^4A + \cos^2A
\]

### Step 1: Analyze and rearrange
The equation can be rewritten as:

\[
\cos^4A - \sin^4A = \cos^2A - \sin^2A
\]

This is achieved by moving all terms involving sine to one side and cosine terms to the other side. Now we need to prove this new form.

### Step 2: Apply difference of squares
Notice that both sides of the equation can be factored as a difference of squares:

\[
\cos^4A - \sin^4A = (\cos^2A - \sin^2A)(\cos^2A + \sin^2A)
\]

But we know that:

\[
\cos^2A + \sin^2A = 1
\]

So the left-hand side simplifies to:

\[
(\cos^2A - \sin^2A)(1) = \cos^2A - \sin^2A
\]

Thus, we are left with:

\[
\cos^2A - \sin^2A = \cos^2A - \sin^2A
\]

### Step 3: Conclusion
Both sides of the equation are identical, so the identity is proven.

\[
\cos^4A + \sin^2A = \sin^4A + \cos^2A
\]

### Do you have any questions or want further clarification?

Here are five related questions:
1. Can you prove the identity \(\cos^2A + \sin^2A = 1\) using the unit circle?
2. What is the difference of squares formula, and how can it be applied in trigonometry?
3. Can you explore other trigonometric identities using \(\cos^2A - \sin^2A\)?
4. How can the Pythagorean identity \(\sin^2A + \cos^2A = 1\) be used in calculus?
5. How do power-reduction formulas work in trigonometry?

**Tip**: When solving trigonometric identities, factorization techniques, like the difference of squares, often simplify the problem significantly.

Discover a detailed proof for the trigonometric identity cos^4A + sin²A = sin^4A + cos²A using factorization and the Pythagorean identity. This solution covers advanced algebraic manipulation and is suitable for Grades 10-12.

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