# Proof of a cosine inequality

#### mech-eng

Problem Statement
Proof of the inequality $cosx \ge (1-x^2/2$
Relevant Equations
$1\ge cosx \ge 0$
How can the inequality $cosx \ge(1-x^2/2)$ be proved? Would you please explain how to prove this inequality?

This is the only equation that I could think of. $1\ge cosx \ge 0$ but I cannot use it here.

Source: Thomas's Calculus, this is from an integration question there.

Thank you.

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#### Ray Vickson

Homework Helper
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How can the inequality $cosx \ge(1-x^2/2)$ be proved? Would you please explain how to prove this inequality?

This is the only equation that I could think of. $1\ge cosx \ge 0$ but I cannot use it here.

Source: Thomas's Calculus, this is from an integration question there.

Thank you.
PF rules require an effort on your part.

#### mech-eng

I know it intuitively but I cannot establish a good way. I cannot express it in a good form. For less than -2, $1-x^2/2$ is less than -1 so this is valid. For more than 2, $1-x^2/2$ is less than -1 this is also valid. Between -1 and 1, $1-x^2/2$ is between 1/2 and 1 and the equation is also valid.

Thank you.

#### Mark44

Mentor
Since the question is in a section about integration, one approach would be to look at the graph of $y = \cos(x) - 1 + \frac {x^2} 2$, and see if the integral of this function is positive or negative.

#### Orodruin

Staff Emeritus
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Since the question is in a section about integration, one approach would be to look at the graph of $y = \cos(x) - 1 + \frac {x^2} 2$, and see if the integral of this function is positive or negative.
What does the integral have to do with whether or not the function itself is positive, which is what is relevant to this question?

#### Mark44

Mentor
What does the integral have to do with whether or not the function itself is positive, which is what is relevant to this question?
If the integral is positive on some interval where cos(x) and 1 - x^2/2 don't intersect, it means that cos(x) >= 1 - x^2/2. That was my thinking. The OP mentioned that the problem was in a section on integration, which I presume comes in a section before Taylor's series are covered.

#### StoneTemplePython

Gold Member
How can the inequality $cosx \ge(1-x^2/2)$ be proved? Would you please explain how to prove this inequality?

This is the only equation that I could think of. $1\ge cosx \ge 0$ but I cannot use it here.

Source: Thomas's Calculus, this is from an integration question there.

Thank you.
I see the ingredients of a solution sitting here though we are guessing a bit since we don't know the background... I assume 1st and 2nd derivatives are ok at this point.

I found "$1\ge \cos x \ge 0$" interesting because when I set $x := \pi$

I get something that violates this. You are right that the magnitude of cosine of x is bounded in 0 and 1.

or less than -2, $1-x^2/2$ is less than -1 so this is valid. For more than 2, $1-x^2/2$ is less than -1 this is also valid.
Yes -- so why not use this fact to truncate your domain to $x \in [-2,2]$.

Note this interval is closed and bounded (compact in reals) and thus any continuous function has a minimum in this interval. So check for values of the function $g(x)$ the border at $x=-2$ and $x=2$ and also use 1st (and 2nd) derivatives to find the minimum value of the function $g$... I'd suggest using the one that Mark gave, i.e.
$g(x) = \cos(x) - 1 + \frac {x^2} 2$ again for $x \in [-2,2]$

#### mech-eng

Yes -- so why not use this fact to truncate your domain to x∈[−2,2]x∈[−2,2]x \in [-2,2]
Yes, for the domain [-2,2] this is valid but in the original question it is said that this is valid for all real number which means the domain is [$-\infty, \infty$]. So this is confusing.

#### StoneTemplePython

Gold Member
Yes, for the domain [-2,2] this is valid but in the original question it is said that this is valid for all real number which means the domain is [$-\infty, \infty$]. So this is confusing.

View attachment 241883
ok, let's be clear -- the original domain is $(-\infty, \infty)$ that's an open interval that doesn't actually include +/- $\infty$

Second, let's partition that into 3 pieces -- i.e.

$(-\infty, \infty) = (-\infty, -2) \cup [-2, 2] \cup (2, \infty)$

so we have 3 different pieces to consider:
$\{(-\infty, -2) , [-2, 2], (2, \infty)\}$

for $(-\infty, -2)$ and for $(2, \infty)$

apply this argument

For less than -2, $1-x^2/2$ is less than -1 so this is valid. For more than 2, $1-x^2/2$ is less than -1 this is also valid.
The argument needs cleaned up a little bit but it is a key insight.

what remains is to consider the case of $x \in [-2,2]$ which was what I was pushing you to focus on in post 7.

#### mech-eng

Okay. Thank you. I got it. For all x beyond the interval [-2,2] $1-x^2/2$ is less than -1 which is also less than the minumum of cos x for all x.

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