# How does `(cos^2x - sin^2x) / (cos^2x + sin^2x)` get to `cos (2x)` ?

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### 2 Answers

Replacing `sin^2 x + cos^2 x` with 1, the given expression becomes:

cos^2 x - sin^2 x

You may write `cos^2 x` as `cos x*cos x` and `sin^2 x` as `sin x*sin x` such that:

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

You need to use the following trigonometric identity, such that:

`cos alpha*cos beta - sin alpha*sin beta = cos (alpha + beta)`

Reasoning by analogy yields:

`cos x*cos x - sin x*sin x = cos(x + x) = cos (2x)`

**Thus, evaluating the equivalent expression of the fraction `(cos^2 x - sin^2 x)/(sin^2 x + cos^2 x)` yields that it may be converted into `cos 2x` .**

`(cos^2x - sin^2x) /(cos^2x + sin^2x)`

To show that the given expression is equal to cos (2x), apply the Pythagorean identity which states that `cos^2x + sin^2x = 1` .

`= (cos^2x - sin^2x) /1`

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

Next, apply the double angle identity of cosine which is `cos (2x) = cos^2x - sin^2x` .

`= cos(2x)`

**Hence, `(cos^2x-sin^2x)/(cos^2x+sin^2x) = cos(2x)` .**