Two Important Inequalities

Inequality	Statement
Sine bound	$-
Cosine bound	$-

The Inequalities

In this section, we establish the following inequalities for all angles $\theta$ measured in radians:

$$\boxed{-|\theta |\le \sin \theta \le |\theta |\text{and}-|\theta |\le 1-\cos \theta \le |\theta |}$$

For all $\theta$: $-|\theta |\le \sin \theta \le |\theta |$ and $-|\theta |\le 1-\cos \theta \le |\theta |$.

Proof

Proof

Picture $\theta$ as an acute angle in standard position on a unit circle (radius $r=1$).

Let $A=(1,0)$ be the point where the initial side meets the circle, and $P=(\cos \theta ,\sin \theta )$ the point where the terminal side meets the circle. The length of the arc $\stackrel{⌢}{AP}$ equals $r\theta =\theta$ (since $r=1$). The length of the line segment $AP$ is at most the arc length (a chord is never longer than the arc it subtends), so: $$|AP|\le \theta .$$ Drop a perpendicular from $P$ to the $x$-axis, meeting it at the point $H$. In the right triangle $\mathrm{△}OHP$: $$HP=OP\sin \theta =1\cdot \sin \theta =\sin \theta$$ $$OH=OP\cos \theta =1\cdot \cos \theta =\cos \theta .$$ Since $OA=1=OH+HA=\cos \theta +HA$, we get $HA=1-\cos \theta$. In the right triangle $\mathrm{△}AHP$: $$HP=\sin \theta ,HA=1-\cos \theta .$$ By the Pythagorean theorem: $$H{P}^2+H{A}^2=|AP{|}^2$$ $${\sin }^2\theta +(1-\cos \theta {)}^2=|AP{|}^2\le {\theta }^2.$$ On the left side we have two nonnegative quantities whose sum is at most ${\theta }^2$. Each term is therefore at most ${\theta }^2$: $${\sin }^2\theta \le {\theta }^2,(1-\cos \theta {)}^2\le {\theta }^2.$$ Taking square roots (and recalling that $\sqrt{x^2}=|x|$): $$|\sin \theta |\le |\theta |,|1-\cos \theta |\le |\theta |,$$ which is equivalent to: $$-|\theta |\le \sin \theta \le |\theta |,-|\theta |\le 1-\cos \theta \le |\theta |.$$ So far we have proved this for acute $\theta$, i.e., .

• When $\theta =0$: both inequalities are obvious since $\sin 0=0$ and $1-\cos 0=0$.
• When : we have $|\sin \theta |\le 1\le |\theta |$ and $|1-\cos \theta |\le 2\le 2|\theta |/(\pi /2)\approx 1.27|\theta |\le |\theta |\cdot |\theta |/(\pi /2)$... more directly, , so the inequality holds trivially since both $|\sin \theta |\le 1$ and $|1-\cos \theta |\le 2$ while .

Therefore, for all $\theta$: $$-|\theta |\le \sin \theta \le |\theta |,-|\theta |\le 1-\cos \theta \le |\theta |.$$

Why These Inequalities Matter

The inequality $|\sin \theta |\le |\theta |$ is exactly what the squeeze theorem needs to prove:

$$\underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }=1.$$

This limit is the foundation of all the differentiation formulas for trigonometric functions. Without it, we cannot prove that $\frac{d}{dx}\sin x=\cos x$.

The inequality $|1-\cos \theta |\le |\theta |$ similarly implies $\underset{\theta \to 0}{lim}(1-\cos \theta )=0$, i.e., $\underset{\theta \to 0}{lim}\cos \theta =1$, confirming that cosine is continuous at 0.

Frequently Asked Questions

Does $|\sin \theta |\le |\theta |$ mean $\sin \theta$ is always smaller than $\theta$?

Yes, in absolute value. For , we have $0\le \sin \theta \le \theta$. For , we have $\theta \le \sin \theta \le 0$. In both cases, $\sin \theta$ is "closer to zero" than $\theta$. This makes intuitive sense from the graphs: for small positive $\theta$, the sine curve starts below the line $y=\theta$.

What is the squeeze theorem and how does it use these inequalities?

The squeeze theorem says: if $g(x)\le f(x)\le h(x)$ near $x=a$, and $\underset{x\to a}{lim}g(x)=\underset{x\to a}{lim}h(x)=L$, then $\underset{x\to a}{lim}f(x)=L$ as well. To prove $\underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }=1$, one uses the geometric inequality $\cos \theta \le \frac{\sin \theta }{\theta }\le 1$ for small positive $\theta$. As $\theta \to 0$, both $\cos \theta \to 1$ and $1\to 1$, squeezing $\frac{\sin \theta }{\theta }$ to 1.

Why are the inequalities stated in terms of $|\theta |$ rather than just $\theta$?

The inequalities need to hold for all real $\theta$, including negative values. Using $|\theta |$ makes the statement symmetric: the bound $|\theta |$ applies regardless of the sign of $\theta$. The geometric proof naturally handles the case , and by the even-odd properties of sine and cosine, the case follows immediately.