# Math whizzes: help me prove this, please

**URL:** <https://boards.straightdope.com/t/math-whizzes-help-me-prove-this-please/32354>\
**Category:** Factual Questions\
**Created:** [September 7, 2000, 1:04pm UTC](https://boards.straightdope.com/t/math-whizzes-help-me-prove-this-please/32354 "2000-09-07T13:04:09Z")\
**Posts on this page:** 8\
**Page:** 1

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**Author:** ![AWB](https://avatars.discourse-cdn.com/v4/letter/a/bb73d2/32.png) [@AWB](https://boards.straightdope.com/u/AWB)\
**Post date:** [September 7, 2000, 1:04pm UTC](https://boards.straightdope.com/t/math-whizzes-help-me-prove-this-please/32354/1 "2000-09-07T13:04:09Z")

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I’ve been trying to prove this inequality:

```auto

     ___________
θ > √ 2(1-cos θ)

for 0 < θ <= π; θ being expressed in radians

```

for a couple of days. I’ve done it empirically, but that doesn’t count in my mind.

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**Author:** ![Cabbage](https://avatars.discourse-cdn.com/v4/letter/c/f07891/32.png) [@Cabbage](https://boards.straightdope.com/u/Cabbage)\
**Post date:** [September 7, 2000, 1:21pm UTC](https://boards.straightdope.com/t/math-whizzes-help-me-prove-this-please/32354/2 "2000-09-07T13:21:11Z")

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I’m having a hard time reading that; it may just be that my browser isn’t displaying the characters properly. I’m seeing an “a”, an “s”, both with some kind of carat thing over them. And I’m seeing 0 \< I (with carats) \<= I(and some character I don’t recognize at all). What are these things?

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**Author:** ![AWB](https://avatars.discourse-cdn.com/v4/letter/a/bb73d2/32.png) [@AWB](https://boards.straightdope.com/u/AWB)\
**Post date:** [September 7, 2000, 1:23pm UTC](https://boards.straightdope.com/t/math-whizzes-help-me-prove-this-please/32354/3 "2000-09-07T13:23:45Z")

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> [@](#):
>
> \*Originally posted by Cabbage \*  
> **I’m having a hard time reading that; it may just be that my browser isn’t displaying the characters properly.**

☹ probably. I’ll restate my statement by using standard characters.

```auto

         _______________
theta > √ 2(1-cos theta)

for 0 < theta <= pi; theta being expressed in radians

```

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**Author:** ![Cabbage](https://avatars.discourse-cdn.com/v4/letter/c/f07891/32.png) [@Cabbage](https://boards.straightdope.com/u/Cabbage)\
**Post date:** [September 7, 2000, 1:27pm UTC](https://boards.straightdope.com/t/math-whizzes-help-me-prove-this-please/32354/4 "2000-09-07T13:27:33Z")

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That definitely helps, I’m still not sure what “a” and “s” are, though; I’m seeing

theta \> a^s 2(1 - cos theta).

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**Author:** ![Nen](https://avatars.discourse-cdn.com/v4/letter/n/c5a1d2/32.png) [@Nen](https://boards.straightdope.com/u/Nen)\
**Post date:** [September 7, 2000, 1:50pm UTC](https://boards.straightdope.com/t/math-whizzes-help-me-prove-this-please/32354/5 "2000-09-07T13:50:07Z")

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AWB, you must define parameters for a & s. If, for example, a = 2 and s = 3, then, for theta = pi/2, the inequality does not maintain validity.

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**Author:** ![AWB](https://avatars.discourse-cdn.com/v4/letter/a/bb73d2/32.png) [@AWB](https://boards.straightdope.com/u/AWB)\
**Post date:** [September 7, 2000, 1:51pm UTC](https://boards.straightdope.com/t/math-whizzes-help-me-prove-this-please/32354/6 "2000-09-07T13:51:37Z")

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sigh… remind me not to use those special characters anymore:

Using computer formula format:

_theta \> sqrt(2_(1-cos(theta)))

0 \< theta \<= pi  
\*

Better? 🙂

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<div class="post-metadata">

**Author:** ![AWB](https://avatars.discourse-cdn.com/v4/letter/a/bb73d2/32.png) [@AWB](https://boards.straightdope.com/u/AWB)\
**Post date:** [September 7, 2000, 1:53pm UTC](https://boards.straightdope.com/t/math-whizzes-help-me-prove-this-please/32354/7 "2000-09-07T13:53:11Z")

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BTW, “sqrt” is the square root function, if you didn’t know.

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**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\
**Post date:** [September 7, 2000, 2:07pm UTC](https://boards.straightdope.com/t/math-whizzes-help-me-prove-this-please/32354/8 "2000-09-07T14:07:03Z")

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This a direct result of the Maclaurin series, which states that:

Cos(A) = 1 - A[sup]2[/sup]/2! + A[sup]4[/sup]/4! - A[sup]6[/sup]/6! ···

A is in radians. It’s not terribly difficult to show that for the range you specified, the sum of all the terms after A[sup]2[/sup]/2! is positive, so you get:

Cos(A) \< 1 - A[sup]2[/sup]/2!

And after that, all you need is a little Algebra. Maybe someone else has a better solution, that doesn’t require Calculus.