Small-angle approximations for sine, cosine and tangent

When an angle $x$ is close to zero and measured in radians, its trigonometric functions have the approximations

$$\boxed{\sin x\approx x}, \qquad \boxed{\tan x\approx x}, \qquad \boxed{\cos x\approx1-\frac{x^2}{2}}.$$

These replace curved functions locally by simple polynomials. They are useful for estimating values, simplifying models and evaluating limits which initially have the indeterminate form $0/0$.

Prerequisites

You should be able to:

• convert between degrees and radians;
• use $\sin^2x+\cos^2x=1$ and $\tan x=\sin x/\cos x$;
• expand brackets and simplify algebraic fractions;
• understand that a limit describes behaviour near a value, not necessarily at that value.

Review radians, arcs and sectors and trigonometric identities if needed.

Why radians are essential

For a unit circle, an angle of $x$ radians subtends an arc of length $x$. When $x$ is tiny, the arc and the perpendicular height $\sin x$ are almost equal. This is the geometric reason that $\sin x\approx x$.

The statement would be false in degrees. For example, with $x=1^\circ$,

$$\sin1^\circ\approx0.01745,$$

not $1$. Converting first gives $1^\circ=\pi/180\approx0.01745$ radians, which agrees with the approximation.

Check the angle unit

Small-angle approximations use radians. If an angle is given in degrees, multiply it by $\pi/180$ before applying them.

Worked example 1: estimate a sine

Estimate $\sin0.04$ without a calculator.

The angle is already in radians and is close to zero, so

$$\boxed{\sin0.04\approx0.04}.$$

The calculator value is approximately $0.0399893$, so the absolute error is about $0.0000107$.

Worked example 2: start from degrees

Estimate $\tan2^\circ$.

First convert to radians:

$$2^\circ=\frac{2\pi}{180}=\frac{\pi}{90}.$$

Since $\tan x\approx x$,

$$\boxed{\tan2^\circ\approx\frac{\pi}{90}\approx0.0349}.$$

Writing $\tan2^\circ\approx2$ would confuse degrees with radians.

Self-check 1

Use a small-angle approximation to estimate:

1. $\sin0.015$;
2. $\tan3^\circ$.

Answer

$$\sin0.015\approx\boxed{0.015}.$$

Also, $3^\circ=\pi/60$ radians, so

$$\tan3^\circ\approx\boxed{\frac{\pi}{60}}\approx0.0524.$$

Understanding the three approximations

The basic result is

$$\lim_{x\to0}\frac{\sin x}{x}=1.$$

Therefore $\sin x$ and $x$ become equal in relative terms as $x\to0$, which is written $\sin x\approx x$ for small $x$.

Since

$$\tan x=\frac{\sin x}{\cos x}$$

and $\cos x\to1$ as $x\to0$,

$$\tan x\approx\frac{x}{1}=x.$$

To obtain the cosine approximation, use the exact identity

$$1-\cos x=2\sin^2\frac{x}{2}.$$

For small $x$, the angle $x/2$ is also small, so

$$\sin\frac{x}{2}\approx\frac{x}{2}.$$

Hence

$$1-\cos x =2\sin^2\frac{x}{2} \approx2\left(\frac{x}{2}\right)^2 =\frac{x^2}{2},$$

which rearranges to

$$\boxed{\cos x\approx1-\frac{x^2}{2}}.$$

This quadratic term matters. The rough statement $\cos x\approx1$ describes the limiting value, but it loses the leading change in cosine and is often too crude for limit questions.

Why sine and cosine have different leading changes

Near $x=0$, the graph of $y=\sin x$ has gradient $1$, so its leading change is linear. The graph of $y=\cos x$ has a horizontal tangent and maximum value $1$, so there is no linear term. Its leading change is quadratic and negative:

$$\sin x=x+\text{smaller terms}, \qquad \cos x=1-\frac{x^2}{2}+\text{smaller terms}.$$

This is also consistent with the derivatives of sine and cosine at zero.

Substituting a small expression

The approximations apply to the whole angle. If $x\to0$, then $kx\to0$ for any fixed constant $k$, so

$$\sin(kx)\approx kx, \qquad \tan(kx)\approx kx, \qquad \cos(kx)\approx1-\frac{k^2x^2}{2}.$$

Notice that the square applies to $kx$, giving $k^2x^2$.

Worked example 3: approximate a compound expression

Find a small-angle approximation for

$$3\sin(2x)-2\tan(3x)$$

as $x\to0$.

Replace each trigonometric function by its leading approximation:

$$\begin{aligned} 3\sin(2x)-2\tan(3x) &\approx3(2x)-2(3x)\\ &=6x-6x\\ &=0. \end{aligned}$$

The leading terms cancel. This answer says only that the expression is smaller than a typical linear term. A more accurate nonzero approximation would require higher order series terms, beyond the three basic approximations.

Worked example 4: cosine with a multiple angle

Find a polynomial approximation for $5-5\cos(4x)$ when $x$ is small.

Apply the cosine approximation to the complete angle $4x$:

$$\cos(4x)\approx1-\frac{(4x)^2}{2}=1-8x^2.$$

Therefore

$$\begin{aligned} 5-5\cos(4x) &\approx5-5(1-8x^2)\\ &=\boxed{40x^2}. \end{aligned}$$

Square the whole angle

$\cos(4x)\approx1-(4x)^2/2=1-8x^2$, not $1-4x^2/2$.

Self-check 2

For small $x$, simplify:

1. $2\sin(5x)+3\tan x$;
2. $1-\cos(6x)$.

Answer

$$2\sin(5x)+3\tan x \approx2(5x)+3x =\boxed{13x}.$$ $$1-\cos(6x) \approx1-\left(1-\frac{(6x)^2}{2}\right) =\boxed{18x^2}.$$

Evaluating limits

Direct substitution into a trigonometric limit often gives $0/0$. This does not mean that the limit is zero or undefined. It means more analysis is required. Replace each small-angle factor by its leading approximation, then simplify.

Worked example 5: a scaled sine limit

Evaluate

$$\lim_{x\to0}\frac{\sin(7x)}{3x}.$$

As $x\to0$, $\sin(7x)\approx7x$. Thus

$$\frac{\sin(7x)}{3x} \approx\frac{7x}{3x} =\frac73.$$

Therefore

$$\boxed{\lim_{x\to0}\frac{\sin(7x)}{3x}=\frac73}.$$

Worked example 6: a ratio of two functions

Evaluate

$$\lim_{x\to0}\frac{\tan(2x)}{\sin(5x)}.$$

Both complete angles approach zero:

$$\tan(2x)\approx2x, \qquad \sin(5x)\approx5x.$$

Hence

$$\boxed{\lim_{x\to0}\frac{\tan(2x)}{\sin(5x)} =\lim_{x\to0}\frac{2x}{5x} =\frac25}.$$

Worked example 7: a cosine limit

Evaluate

$$\lim_{x\to0}\frac{1-\cos(3x)}{x^2}.$$

Use the quadratic cosine approximation:

$$\cos(3x)\approx1-\frac{(3x)^2}{2} =1-\frac{9x^2}{2}.$$

Therefore

$$1-\cos(3x)\approx\frac{9x^2}{2},$$

so

$$\boxed{\lim_{x\to0}\frac{1-\cos(3x)}{x^2} =\frac92}.$$

Using only $\cos(3x)\approx1$ would turn the numerator into zero and discard the term that determines the limit.

Worked example 8: combine approximations and algebra

Evaluate

$$\lim_{x\to0}\frac{\sin(4x)\tan(3x)}{1-\cos(2x)}.$$

Use

$$\sin(4x)\approx4x, \qquad \tan(3x)\approx3x,$$

and

$$1-\cos(2x) \approx\frac{(2x)^2}{2}=2x^2.$$

Then

$$\frac{\sin(4x)\tan(3x)}{1-\cos(2x)} \approx\frac{(4x)(3x)}{2x^2} =6.$$

Thus the limit is

$$\boxed{6}.$$

Self-check 3

Evaluate:

1. $\displaystyle\lim_{x\to0}\frac{\sin(9x)}{\tan(4x)}$;
2. $\displaystyle\lim_{x\to0}\frac{1-\cos(5x)}{x^2}$;
3. $\displaystyle\lim_{x\to0}\frac{x\sin(2x)}{1-\cos(4x)}$.

Answer

1. Since $\sin(9x)\approx9x$ and $\tan(4x)\approx4x$,

   $$\boxed{\frac94}.$$

2. Since $1-\cos(5x)\approx(5x)^2/2$,

   $$\boxed{\frac{25}{2}}.$$

3. The numerator is approximately $x(2x)=2x^2$, while

   $$1-\cos(4x)\approx\frac{(4x)^2}{2}=8x^2.$$

   Therefore the limit is

   $$\boxed{\frac14}.$$

Accuracy and sensible use

An approximation is not an identity. The symbol $\approx$ means the values are close under the stated condition. It must not be replaced by $=$ in a numerical estimate.

For small positive $x$,

$$\sin x<x<\tan x.$$

Thus $x$ slightly overestimates $\sin x$ and slightly underestimates $\tan x$. The errors grow as $|x|$ grows.

$x$ in radians	$\sin x$	approximation $x$	absolute error
$0.01$	$0.00999983$	$0.01$	about $1.67\times10^{-7}$
$0.1$	$0.0998334$	$0.1$	about $1.67\times10^{-4}$
$0.5$	$0.479426$	$0.5$	about $0.0206$

There is no universal numerical boundary for the word small. Required accuracy determines whether an approximation is acceptable. In a limit as $x\to0$, the angle becomes arbitrarily small, so the method is exact for the limiting value even though the intermediate relation uses $\approx$.

Common misconceptions

• Using degrees: $\sin x\approx x$ requires $x$ in radians.
• Treating an approximation as an identity: it is valid near zero, not for every $x$.
• Approximating the wrong angle: $\sin(3x)\approx3x$, not $x$.
• Forgetting the square: $\cos(kx)\approx1-k^2x^2/2$.
• Using $\cos x\approx1$ in a cancellation: for $1-\cos x$, retain $-x^2/2$.
• Substituting $x=0$ too soon: $0/0$ is an indeterminate form, not an answer.
• Ignoring cancellation: if leading terms cancel, the basic approximations may not reveal the next nonzero term.

Final self-check

For each statement, decide whether it is correct and explain briefly.

1. For small $x$, $\cos(2x)\approx1-2x^2$.
2. For small $x$, $\sin(x^2)\approx x^2$.
3. $\sin5^\circ\approx5$.
4. $\displaystyle\lim_{x\to0}\frac{\sin x-x}{x}=0$.

Answer

1. Correct, because $1-(2x)^2/2=1-2x^2$.

2. Correct. If $x\to0$, then the complete angle $x^2\to0$.

3. Incorrect. In radians, $5^\circ=\pi/36$, so $\sin5^\circ\approx\pi/36$.

4. Correct. Since $\sin x\approx x$, the numerator is smaller than a linear term. More formally,

   $$\frac{\sin x-x}{x}=\frac{\sin x}{x}-1\to1-1=0.$$

Next steps

Next, apply radian based trigonometry in differentiating standard functions and combine approximations with the methods in trigonometric proof and modelling.