The Behaviour of sin x Near the Origin (HSC SSCE Mathematics Advanced): Revision Notes

The Behaviour of sin x Near the Origin

Introduction

This topic establishes a key limit that forms the foundation for finding the derivative of $\sin x$. This limit demonstrates that the curve $y = \sin x$ has gradient 1 when it passes through the origin. In geometric terms, this shows that the line $y = x$ is the tangent to $y = \sin x$ at the origin.

Understanding this behaviour provides the geometric basis for differentiating the trigonometric functions.

A fundamental inequality

To begin, we use a geometric approach to establish an important relationship between $x$, $\sin x$ and $\tan x$ for angles near the origin.

Geometric proof

Part A: Acute angles

Let $x$ be an acute angle.

Draw a circle with centre $O$ and any radius $r$. Create a sector $AOB$ that subtends the angle $x$ at the centre $O$. Let the tangent at point $A$ meet the radius $OB$ at point $M$ (the radius $OB$ may need to be extended). Join the chord $AB$.

In triangle $OAM$:

$\frac{AM}{r} = \tan x$

Therefore:

$AM = r \tan x$

The diagram shows that:

$\text{area } \triangle OAB < \text{area sector } OAB < \text{area } \triangle OAM$

Using the area formulae for triangles and sectors:

$\frac{1}{2}r^2 \sin x < \frac{1}{2}r^2 x < \frac{1}{2}r^2 \tan x$

Dividing through by $\frac{1}{2}r^2$:

$\sin x < x < \tan x$

Part B: Negative angles

Because $x$, $\sin x$ and $\tan x$ are all odd functions, the inequality reverses for negative angles:

$\sin x > x > \tan x, \text{ for } -\frac{\pi}{2} < x < 0$

The main theorem

The inequality established above allows us to prove two fundamental limits that are essential for calculus with trigonometric functions.

Proof of the fundamental limits

When $x$ is acute, we know that:

$\sin x < x < \tan x$

Dividing through by $\sin x$:

$1 < \frac{x}{\sin x} < \frac{1}{\cos x}$

As $x \to 0^+$, $\cos x \to 1$, so:

$\frac{x}{\sin x} \to 1 \text{ as } x \to 0^+$

But $\frac{x}{\sin x}$ is an even function, so:

$\frac{x}{\sin x} \to 1 \text{ as } x \to 0^-$

Combining these two limits:

$\frac{x}{\sin x} \to 1 \text{ as } x \to 0$

Finally:

$\frac{\tan x}{x} = \frac{\sin x}{x} \times \frac{1}{\cos x} \to 1 \times 1 \text{ as } x \to 0$

The behaviour of sin x and tan x near the origin

The line $y = x$ is a tangent to both $y = \sin x$ and $y = \tan x$ at the origin.

When $x = 0$, the derivatives of both $\sin x$ and $\tan x$ are exactly $1$.

Approximations to the trigonometric functions for small angles

For 'small' angles (positive or negative), the limits above provide good approximations for the three trigonometric functions. The angle must be expressed in radians for these approximations to work.