# Prove the following identity 1-2sin^2a=cos^4a-sin^4a.

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1 - 2sin62 a = cos^4 a - sin^4 a

First we will start from the right side and prove the left sides.

==> cos^4 a - sin^4 a

Let us factor.

We know that:

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

But sin^2 a + cos^2 a = 1

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

But we know that:

cos^2 a + sin^2 a = 1

==> sin^2 a = 1- cos^2 a

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

= cos^2 a -1 + cos^2 a

= 2cos^2 a -1

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

We have to prove that 1 - 2(sin a)^2 = (cos a)^4 - (sin a)^4.

We know that (sin a)^2 + (cos a)^2 = 1

(cos a)^4 - (sin a)^4

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

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

=> 1 - (sin a)^2 - (sin a)^2

=1- 2 (sin a )^2

**Therefore 1 - 2(sin a)^2 = (cos a)^4 - (sin a)^4.**

L:H:S ≡ 1 - 2sin²A

= 1 x ( 1-2sin²A)

**⇒ use 1 = sin²A + cos²A**

= (sin²A + cos²A ) (sin²A + cos²A - 2sin²A)

= (cos²A + sin²A ) (cos²A - sin²A)

**⇒ use (x+y)(x-y) = x² - y²**

= (cos²A)² - (sin²A)²

= cos⁴A - sin⁴A

Hence L:H:S ≡ R:H:S

To prove the identity 1-2sin^2a=cos^4a-sin^4a

We start from the cos^4a-sin^4a.

cos^4x-sin^4x = (cos^2x)^2- (sin^2x)^2 = (cos^2x+sin^2x)(cos^2x-sin^2x), as a^2-b^2 = (a+b)(a-b).

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

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

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

Therefore 1-2sin^2a=cos^4a-sin^4a.

We'll write the difference of square from the right 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.**