1.05e - Small angle approximations | OCR A-Level Mathematics A

Replace trigonometric functions by simple expressions near zero, and understand why the angle must be in radians. You will keep the right powers, handle cancellation and decide whether an approximation is accurate enough for a geometric model.

Why small angles behave simply

How close is a short chord to the arc joining its ends? On a circle of radius $2\,\mathrm m$ , an angle of $0.12$ radians gives arc length $0.24\,\mathrm m$ . The straight chord is a little shorter. By the end you will calculate how much shorter, without treating the two lengths as equal.

You already know

For an angle $\theta$ in radians, arc length is $r\theta$ and sector area is $r^2\theta/2$ . The unit-circle point is $(\cos\theta,\sin\theta)$ , so its squared coordinates add to $1$ .

At $0.1$ radians, a calculator gives $\sin(0.1)=0.0998334\ldots$ , $\cos(0.1)=0.995004\ldots$ and $\tan(0.1)=0.100335\ldots$ . Sine and tangent almost equal the angle; cosine almost equals $1$ . Radian measure makes these simple numerical comparisons possible.

Small angle approximation

A small angle approximation replaces a trigonometric function by a simpler expression that becomes relatively accurate as its radian argument approaches zero. It is local: a formula that works near zero need not work far from zero.

There is geometry behind the calculator values. For $0<\theta<\pi/2$ on a unit circle, an inscribed triangle lies inside the sector, which lies inside the enclosing tangent triangle. Their areas give

\frac12\sin\theta<\frac12\theta<\frac12\tan\theta

Dividing by positive quantities gives

\cos\theta<\frac{\sin\theta}{\theta}<1

As $\theta$ tends to zero, $\cos\theta$ tends to $1$ , so the middle ratio tends to $1$ too. Also $\tan\theta/\theta=(\sin\theta/\theta)/\cos\theta$ tends to $1$ . Odd symmetry extends these results to negative angles approaching zero. For cosine,

1-\cos\theta =\frac{1-\cos^2\theta}{1+\cos\theta} =\frac{\sin^2\theta}{1+\cos\theta} \approx\frac{\theta^2}{2}

The first two steps are exact identities near zero; only the final step is approximate.

Small angle approximations

For $|\theta|$ small, with $\theta$ measured in radians,

\sin\theta\approx\theta \qquad \cos\theta\approx1-\frac{\theta^2}{2} \qquad \tan\theta\approx\theta

In the formula sheet.

Here is a useful surprise: halving a small angle makes $1-\cos\theta$ roughly four times smaller, because its leading expression contains $\theta^2$ . At $0.1$ and $0.05$ radians the exact differences are $0.00499583\ldots$ and $0.00124974\ldots$ , almost exactly in the ratio $4:1$ .

Key idea

“Small” describes the radian argument of the function. The formulas give exact endpoint values at zero, but they do not define a quotient whose denominator is zero.

The distinction between units and accuracy is worth explaining before doing any algebra.

Substitution and retained powers

Apply each formula to the whole argument, then keep only powers supported by the approximations you used.

For example, $\cos(3x)\approx1-9x^2/2$ , because the whole argument $3x$ is squared. The condition is that $|3x|$ is small. A tiny $x$ does not help if a huge multiplier makes the actual argument large.

Cancellation changes what must be retained. For small nonzero $x$ ,

\frac{1-\cos x}{\sin x} \approx\frac{x^2/2}{x} =\frac x2

Replacing cosine by $1$ would erase the leading numerator. The simplified approximation does not make the starting quotient defined at $x=0$ .

If both sides of a quotient vanish quadratically, their leading coefficients decide its leading value. For small nonzero $x$ ,

\frac{1-\cos(2x)}{x\sin(3x)}\approx\frac{2x^2}{3x^2}=\frac23

This requires both $2x$ and $3x$ to be small in radians; it still excludes $x=0$ .

Worked example 1: keeping terms below cubic order

Question. Find an approximate expression for $\dfrac{\sin(3x)}{1+\cos x}$ , neglecting terms in $x^3$ and above, for small $x$ in radians.

Plan. Substitute the small angle formulas, then examine the order of the denominator's correction.

\begin{aligned} \frac{\sin(3x)}{1+\cos x} &\approx\frac{3x}{2-x^2/2} && \text{whole arguments}\\ &=\frac{3x}{2}\left(1-\frac{x^2}{4}\right)^{-1} && \text{factor the denominator}\\ &\approx\frac{3x}{2}\left(1+\frac{x^2}{4}\right) && \text{binomial approximation}\\ &\approx\frac{3x}{2} && \text{discard cubic terms} \end{aligned}

Why this step? The denominator correction first contributes a cubic term after multiplication by $x$ . Retaining it while ignoring the cubic correction to $\sin(3x)$ would be inconsistent.

Answer. The required approximation is $3x/2$ ; there is no quadratic term.

Check. At $x=0.01$ , the quotient is $0.0149981\ldots$ , close to $0.015$ .

Change just the cosine's argument from $x$ to $2x$ . This changes the discarded correction, but not the leading answer.

Worked example 2: changing a denominator argument

Question. Approximate $\dfrac{\sin(3x)}{1+\cos(2x)}$ , neglecting terms in $x^3$ and above, when $|3x|$ is small in radians.

Plan. Square the complete cosine argument, then identify the first retained power.

\begin{aligned} 1+\cos(2x)&\approx2-2x^2 && \text{square }2x\\ \frac{\sin(3x)}{1+\cos(2x)} &\approx\frac{3x}{2}(1-x^2)^{-1} && \text{factor }2\\ &\approx\frac{3x}{2}(1+x^2) && \text{expand near zero}\\ &\approx\frac{3x}{2} && \text{discard cubic terms} \end{aligned}

Why this step? The requested order decides which terms survive, rather than the number of terms visible in a calculator display.

Answer. Again the approximation is $3x/2$ .

Check. At $x=0.01$ , the exact quotient is $0.0149992\ldots$ , again close to $0.015$ .

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1.05e - Small angle approximations

Why small angles behave simply

Substitution and retained powers

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