Trigonometric Inequality in Tom Apostol's Book

[Math Amateur]

In Tom Apostol's book "Calculus: Volume 1 (Second Edition) he uses the following inequalities:

\(\displaystyle 0 \lt \cos x \lt \frac{ \sin x }{x} \lt \frac{1}{ \cos x }\) ... ... ... (1)

in order to demonstrate that:

\(\displaystyle \lim_{x \to 0} \frac{ \sin x }{x} = 1\)... ... BUT ... ... how do we prove (1) ...

That is how do we (formally and rigorously) demonstrate that

\(\displaystyle 0 \lt \cos x \lt \frac{ \sin x }{x} \lt \frac{1}{ \cos x }\) ... ...

Can someone please help me with this?

Peter

 

[Euge]

The inequalities are valid in $(0, \pi/2)$, which can be demonstrated by a geometric argument with similar triangles in the first quadrant bounded by the unit circle.

 

[Math Amateur]

The inequalities are valid in $(0, \pi/2)$, which can be demonstrated by a geometric argument with similar triangles in the first quadrant bounded by the unit circle.

Thanks Euge ... ... Do you have a reference or an online source for the argument to which you refer? ...

Just a further question ... In the above argument to which you refer, what is the assumed definition of sine and cos functions ... the power series definition or some other acceptable definition?

(Note - maybe that is a silly question since you refer to a geometrical argument ... Presumably sine and cos are defined in terms of right triangles ... but is that a suitable definition as a basis for real and complex analysis ... ...)Peter

 

[Euge]

Considering that you're trying to prove $\sin x/x \to 1$ as $x\to 0$, and that the author has given those inequalities to prove the result (hinting on the squeeze theorem), I think it's safe to assume that the sine and cosine are defined as you've learned in ordinary trigonometry. You can find an equivalent inequality in several calculus books, but really your skills with analytic geometry will be sufficient.

Draw an arc of the unit circle in the first quadrant. Given $\theta\in (0, \pi/2)$, extend a vertical line from the point $(\cos \theta, \sin \theta)$ on the arc to the $x$-axis. Draw a line from the origin that passes through the point $(\cos \theta, \sin \theta)$ so that it intersects with the vertical line tangent to the arc at $(1,0)$. The length of the vertical line segment can be found to $\tan x$ by similarity of right triangles. The area of the smaller triangle formed is less than the area of the sector subtended by $\theta$, which is less than the area larger triangle with altitude $\tan \theta$. The area of the smaller triangle is $(1/2)\cos \theta \sin \theta$, the area of the sector subtended by $\theta$ is $1/2 * 1^2 * \theta = (1/2)\theta$, and the area of the larger triangle is $1/2 * 1 * \tan \theta = (1/2) \tan \theta$. Therefore $$\cos \theta \sin \theta < \theta < \tan \theta$$ or $$0 < \cos \theta < \frac{\theta}{\sin \theta} < \frac{1}{\cos \theta}$$ that is, $$0 < \cos \theta < \frac{\sin \theta}{\theta} < \frac{1}{\cos \theta}$$

 

[Prove It]

To demonstrate this inequality and find the limit of $\displaystyle \begin{align*} \lim_{x \to 0} \frac{\sin{(x)}}{x} \end{align*}$, you have to use the geometric definition of the trigonometric functions on the unit circle.

https://www.physicsforums.com/attachments/3663

Notice that the area of the sector is a little more than the area of the smaller triangle and a little less than the area of the bigger triangle, so when the angle x is positive, close to 0 and measured in radians we have

$\displaystyle \begin{align*} \frac{\sin{(x)}\cos{(x)}}{2} \leq \frac{x}{2} &\leq \frac{\tan{(x)}}{2} \\ \frac{\sin{(x)}\cos{(x)}}{2} \leq \frac{x}{2} &\leq \frac{\sin{(x)}}{2\cos{(x)}} \\ \sin{(x)} \cos{(x)} \leq x &\leq \frac{\sin{(x)}}{\cos{(x)}}\\ \cos{(x)} \leq \frac{x}{\sin{(x)}} &\leq \frac{1}{\cos{(x)}} \\ \frac{1}{\cos{(x)}} \geq \frac{\sin{(x)}}{x} &\geq \cos{(x)} \\ \cos{(x)} \leq \frac{\sin{(x)}}{x} &\leq \frac{1}{\cos{(x)}} \end{align*}$

and now if $\displaystyle \begin{align*} x \to 0 \end{align*}$ both $\displaystyle \begin{align*} \cos{(x)} \to 1 \end{align*}$ and $\displaystyle \begin{align*} \frac{1}{\cos{(x)}} \to 1 \end{align*}$, and since $\displaystyle \begin{align*} \frac{\sin{(x)}}{x} \end{align*}$ ends up sandwiched between two ones, it must also be that $\displaystyle \begin{align*} \frac{\sin{(x)}}{x} \to 1 \end{align*}$.

Note: Technically speaking we have only proved the right hand limit, but the left hand limit is pretty much identical, just with some negatives chucked in. I'll leave that for the readers :)

 

[Math Amateur]

To demonstrate this inequality and find the limit of $\displaystyle \begin{align*} \lim_{x \to 0} \frac{\sin{(x)}}{x} \end{align*}$, you have to use the geometric definition of the trigonometric functions on the unit circle.

https://www.physicsforums.com/attachments/3663

Notice that the area of the sector is a little more than the area of the smaller triangle and a little less than the area of the bigger triangle, so when the angle x is positive, close to 0 and measured in radians we have

$\displaystyle \begin{align*} \frac{\sin{(x)}\cos{(x)}}{2} \leq \frac{x}{2} &\leq \frac{\tan{(x)}}{2} \\ \frac{\sin{(x)}\cos{(x)}}{2} \leq \frac{x}{2} &\leq \frac{\sin{(x)}}{2\cos{(x)}} \\ \sin{(x)} \cos{(x)} \leq x &\leq \frac{\sin{(x)}}{\cos{(x)}}\\ \cos{(x)} \leq \frac{x}{\sin{(x)}} &\leq \frac{1}{\cos{(x)}} \\ \frac{1}{\cos{(x)}} \geq \frac{\sin{(x)}}{x} &\geq \cos{(x)} \\ \cos{(x)} \leq \frac{\sin{(x)}}{x} &\leq \frac{1}{\cos{(x)}} \end{align*}$

and now if $\displaystyle \begin{align*} x \to 0 \end{align*}$ both $\displaystyle \begin{align*} \cos{(x)} \to 1 \end{align*}$ and $\displaystyle \begin{align*} \frac{1}{\cos{(x)}} \to 1 \end{align*}$, and since $\displaystyle \begin{align*} \frac{\sin{(x)}}{x} \end{align*}$ ends up sandwiched between two ones, it must also be that $\displaystyle \begin{align*} \frac{\sin{(x)}}{x} \to 1 \end{align*}$.

Note: Technically speaking we have only proved the right hand limit, but the left hand limit is pretty much identical, just with some negatives chucked in. I'll leave that for the readers :)

Thanks to both Euge and Prove It for such detailed and helpful posts ... just working through the posts in detail now ...

Thanks again!

Peter