# Verify the identity sin^4x-cos^4x=2sin^2x-1

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We have to prove : sin^4x - cos^4x = 2sin^2x - 1

Let's start from the left

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

=> (sin x)^4 - [ 1 - (sin x)^2]^2

=> (sin x)^4 - 1 - (sin x)^4 + 2*(sin x)^2

=> - 1 + 2*(sin x)^2

=> 2*(sin x)^2 - 1

which is the right hand side.

**This proves that sin^4x - cos^4x = 2sin^2x - 1**

Given sin^4 x - cos^4 x = 2sin^2 x - 1

==> We will rearrange terms.

==> sin^4 x - 2sin^2 x + 1 = cos^4 x

Now we will factor the left side.

==> (sin^2 x -1)^2 = cos^ 4 x

But we know that sin62 x = 1- cos^2 x

==> (1-cos^2 x -1 )^2 = cos^4 x

==> (-cos^2 x)^2 = cos^4 x

==> cos^2 4 x = cos^4 x

**Then, we have proved that sin^4 x - cos^4 x = 2sin^2 x - 1.....**

We'll shift (sin x)^4 to the right:

- (cos x)^4 = -(sin x)^4 + 2(sin x)^2 - 1

We'll multiply by -1:

(cos x)^4 = (sin x)^4 - 2(sin x)^2 + 1

We recognize in the expression from the right side a perfect square:

(cos x)^4 = [(sin x)^2 - 1]^2 = {-[1 - (sin x)^2]}^2 = [1 - (sin x)^2]^2

But, from Pythagorean identity, we'll get:

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

We'll raise to square both sides:

(cos x)^4 = [1 - (sin x)^2]^2

**We notice that we,ve get LHS = RHS, therefore the given identity (sin x)^4- (cos x)^4 = 2(sin x)^2 - 1 is true.**