# Thread: Can you prove this identity?

1. ## Can you prove this identity?

cos²x - sin²x = cos2x +sin²x

cos²x - sin²x = cos2x +sin²x

Thanks!
Are they equal..?

$\displaystyle \cos(2x) = \cos(x+x) = \cos(x)\cos(x) - \sin(x)\sin(x) = \cos^2(x) - \sin^2(x)$

So $\displaystyle \cos(2x) + \sin^2(x) = \cos^2(x)$...

3. hm you see I created this identity.. it's a class assignment and we have to see if someone can prove it?

I started off with cos²x = cos²x

and from there you're supposed to keep replacing terms with equivalent equations to make it complex.

Im trying to see if it makes sence and if its possible to prove..

hm you see I created this identity.. it's a class assignment and we have to see if someone can prove it?

I started off with cos²x = cos²x

and from there you're supposed to keep replacing terms with equivalent equations to make it complex.

Im trying to see if it makes sence and if its possible to prove..
$\displaystyle \cos(2x) + \sin^2(x) = \cos^2(x) - \sin^2(x)$

Set $\displaystyle x = \frac{\pi}{2}$.

$\displaystyle \cos(\pi) + \sin^2(\pi/2) = -1 + 1 = 0$

But

$\displaystyle \cos^2(\pi/2) - \sin^2(\pi/2) = 0 - 1^2 = -1$.

Not the same thing am afraid.

Show me how you got it and I'll show you where you went wrong...

5. Okay so I started with..

cos²x = cos²x

1 - sin²x = cos2x + sin²x

cos2x = 1 - 2sin²x + sin²x

cos x cos y - sin x sin y = cos2x + sin²x (after here i didn't know how to break down the right side any further)

cos²x - sin²x = cos2x + sin²x

1 - sin²x = cos2x + sin²x

cos2x = 1 - 2sin²x + sin²x
Are both the left hand sides equal..?

You have put down...

$\displaystyle \cos(2x) = 1 - \sin^2(x)$ and also $\displaystyle \cos(2x) = 1 - 2\sin^2(x)$

7. I'm not sure haa.. I'll just have to try from the start again..