Dear Uncle Colin

How would you prove that $\cos(2x) > 1 - 2x^2$ for $x > 0$?

- Can’t Obviously See It, Need Explanation

Hi, COSINE, and thanks for your message!

I’d start by noting that as soon as $x > 1$, the right hand side is smaller than -1, which is clearly smaller than $\cos(2x)$ - so we only care about $0 < x \le 1$.

I’d then use a trigonometric identity: $\cos(2x) = 1 - 2\sin^2(x)$, which means we need to show $1 - 2\sin^2(x) > 1 - 2x^2$ for $0<x\le1$.

That’s equivalent to showing that $x^2 > \sin^2(x)$ for $0 < x \le 1$.

The area of the sector of a unit circle subtended by an angle of $x$ is $\frac{1}{2}x$, and this is greater than the area of the isosceles triangle with apex angle $x$ and equal sides of length 1, which is $\frac{1}{2}\sin(x)$, certainly for $0<x<1$ ((I imagine it’s true for all positive $x$, but we don’t care here)) - so $x > \sin(x)$ in the domain we care about. If $x<1$, $\sin(x)$ is definitely positive, so we’re done (but for the writing up).

### Writing it up

So, to write this up:

If $0 < x < 1$:

• $x > \sin(x) > 0$
• So $x^2 > \sin^2(x)$
• So $1 - 2\sin^2(x) > 1 - 2x^2$
• So $\cos(2x) > 1 - 2x^2$ for $0 < x < 1$.

If $x > 1$, $1- 2x^2 < -1 < \cos(2x)$ $\blacksquare$

Hope that helps!

- Uncle Colin