We are asked to prove the trigonometric identity $\sin^4 A - \cos^4 A = 2\sin^2 A - 1$.

1. Problem Description

We are asked to prove the trigonometric identity $\sin^4 A - \cos^4 A = 2\sin^2 A - 1$.

2. Solution Steps

We will start with the left-hand side (LHS) of the equation and try to simplify it to match the right-hand side (RHS).

LHS = $\sin^4 A - \cos^4 A$

We can factor this expression as a difference of squares:

$a^2 - b^2 = (a - b)(a + b)$

In our case, $a = \sin^2 A$ and $b = \cos^2 A$.

So,

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

We know the Pythagorean identity:

$\sin^2 A + \cos^2 A = 1$

So, our expression simplifies to:

$(\sin^2 A - \cos^2 A)(1) = \sin^2 A - \cos^2 A$

We can rewrite $\cos^2 A$ using the Pythagorean identity as $\cos^2 A = 1 - \sin^2 A$. Substituting this into our expression:

$\sin^2 A - (1 - \sin^2 A) = \sin^2 A - 1 + \sin^2 A = 2\sin^2 A - 1$

Thus, we have shown that $\sin^4 A - \cos^4 A = 2\sin^2 A - 1$, which is the right-hand side (RHS) of the original equation.

3. Final Answer

$\sin^4 A - \cos^4 A = 2\sin^2 A - 1$