Question

`cos^4x+sin^2x=cos^2x+sin^4x`?

Answer 1

Please see below.

Explanation:

.

`cos^4x+sin^2x=cos^2x+sin^4x`

`cos^4x=(cos^2x)^2`

Let's plug this in:

`cos^4x+sin^2x=(cos^2x)^2+sin^2x`

We know that we have the identity:

`sin^2x+cos^2x=1`

Let's solve for `cos^2x` and plug it in:

`cos^2x=1-sin^2x`

`(1-sin^2x)^2=1-2sin^2x+sin^4x`, using `(a-b)^2=a^2-2ab+b^2`

Let's plug it in:

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

Now, we substitute `sin^2x+cos^2x` for `1`:

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

Answer 2

Kindly refer to a Proof in the Explanation.

Explanation:

`cos^4x+sin^2x=(cos^2x)^2+sin^2x`,

`=(1-sin^2x)^2+sin^2x`,

`=(1-ul(2sin^2x)+sin^4x)+ul(sin^2x)`,

`=ul(1-sin^2x)+sin^4x`,

`=cos^2x+sin^4x`, as desired!

Answer 3

Another Proof.

Explanation:

A very simple way to solve is as under :

`cos^4x-sin^4x`,

`=(cos^2x+sin^2x)(cos^2x-sin^2x)`,

`i.e., cos^4x-sin^4x=cos^2x-sin^2x`.

`:. cos^4x+sin^2x=cos^2x+sin^4x`.