Question

How do you verify the identity: `cos^2x-sin^2x=1-2sin^2x`?

Answer 1

From the basic definitions of `sin` and `cos` and the Pythagorean Theorem
`cos^2(x)+sin^2(x) = 1`

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

So
`cos^2(x) - sin^2(x)`

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

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

Answer 2

There are many acceptable (correct) ways to do this.

Here's one:

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

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

`color(white)"ssssssssssss"` `=cos^2x+sin^2x-sin^2x-sin^2x`

`color(white)"ssssssssssss"` `=cos^2x-sin^2x`

Answer 3

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

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

Answer 4

See below:

Explanation:

Recall the Pythagorean Identity

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

Which can be manipulated into this form:

`color(blue)(cos^2x=1-sin^2x)`

In our equation, we can replace `cos^2x` with this to get

`color(blue)(1-sin^2x)-sin^2x`, which simplifies to

`1-2sin^2x`. We have just verified the identity

`bar( ul(|color(white)(2/2)cos^2x-sin^2x=1-2sin^2x color(white)(2/2)|))`

With the use of the Pythagorean Identity.

Hope this helps!