# Verify if cos^4x + 2sin^2x - sin^4x = 1.

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We have to verify if cos^4x + 2sin^2x - sin^4x = 1.

Now cos^4x + 2sin^2x - sin^4x

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

Now use the relation (cos x)^2 = 1 - (sin x)^2

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

Open the brackets

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

cancel the common terms

=> 1

**Therefore we have proved that cos^4x + 2sin^2x - sin^4x = 1**

To verify if cos^4x + 2sin^2x - sin^4x = 1.

We know that LHS : cos^4- - sin^4x = (cos^2x-sin^2x)(cos^2+sin^2x) = cos^2x-sin^2x, as cos^2+sin^2x = 1.

Therefore cos^4x-sin^4 = cos^2x-sin^2x.

cos^4x-sin^4x = (1-sin^2) -sin^2x , as ccos^2 = 1-sin^2x

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

We add 2sin^2sx to both sides and get:

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

**Therefore cos^4x+2sin^2-sin^4x = 1**.

We'll re-write the identity, so that to get both sides differences of squares:

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

We'll write the difference of square from the left side, using the formula:

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

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

We'll recall the fundamental formula of trigonometry:

(cos a)^2 + (sin a)^2 = 1

(cos a)^4 - (sin a)^4 = [(cos a)^2 - (sin a)^2]

We'll substitute (cos a)^2 = 1 - (sin a)^2

(cos a)^4 - (sin a)^4 = 1 - (sin a)^2 - (sin a)^2

We'll combine like terms:

**(cos a)^4 - (sin a)^4 = 1 - 2(sin a)^2 q.e.d**

**We notice that it's no need to transform the right side, since working to the left side, we've get the same expression we have to the right side.**