Zoom in close enough on the unit circle and the curved arc, the vertical drop and the straight chord all start to look identical — which is exactly why sin θ, cos θ and tan θ collapse to strikingly simple approximations for small θ.
8 min read3 worked examples
The three approximations
When $\theta$ is small and measured in radians, the three trig functions can be replaced by much simpler polynomial approximations:
These are only valid for $\theta$ close to $0$ — the further $\theta$ strays from $0$, the worse the approximation gets. They are also only valid when $\theta$ is in radians; substituting a degree value (e.g. $\theta=30$ instead of $\theta=\frac{\pi}{6}$) gives numbers with no meaningful relationship to $\sin30°$ at all.
$\sin\theta\approx\theta$ can be seen directly on a sketch: for a very small angle, an arc of the unit circle, its vertical height, and the chord joining its endpoints to the centre are all nearly indistinguishable in length — and $\theta$ itself is the arc length on a unit circle, since $s=r\theta=1\times\theta$. The other two approximations follow from this one algebraically.
How to read the diagram below
It's a circle of radius $r$ centred at $O$. Point $A$ marks where the angle starts ($\theta=0$); point $P$ is the current angle $\theta$, and it moves as the slider changes. Four lengths live in this one picture, one per colour: the curved arc from $A$ to $P$ (orange) has length $r\theta$; dropping straight down from $P$ to the axis (teal) gives $r\sin\theta$; extending the line $OP$ until it hits the vertical line through $A$ (gold) gives $r\tan\theta$; and the horizontal distance from $O$ to the foot of that drop (grey, dashed) gives $r\cos\theta$. Divide every length by $r$ and you get exactly $\theta$, $\sin\theta$, $\tan\theta$ and $\cos\theta$ — watch all four converge as $\theta$ shrinks.
θ (rad)0.611
sin θ0.573
tan θ0.700
cos θ0.819
sin θ ⁄ θ0.9376
θ
Drag the slider — or let it play — to shrink θ toward $0$. Watch the four coloured lengths in the numbers below: as θ shrinks, sin θ, θ itself, and tan θ squeeze together ($\sin\theta<\theta<\tan\theta$ always), while cos θ climbs toward $1$ — which is exactly why $\sin\theta\approx\theta\approx\tan\theta$ and $\cos\theta\approx1$ for small θ.
Derivationcos θ ≈ 1 − θ²/2, from the double angle formula
1
Start from the double angle identity $\cos\theta = 1-2\sin^2\left(\dfrac{\theta}{2}\right)$.
A Year 2 identity that holds for every θ, not just small ones.
2
Since $\theta$ is small, $\dfrac{\theta}{2}$ is small too, so $\sin\left(\dfrac{\theta}{2}\right) \approx \dfrac{\theta}{2}$.
Apply the small angle approximation for sine to the half-angle.
$\tan\theta=\dfrac{\sin\theta}{\cos\theta}\approx\dfrac{\theta}{1-\theta^2/2}\approx\theta$, since the denominator is extremely close to $1$ for small $\theta$.
Combine the two results already found; the θ² term in the denominator is negligible next to 1.
▪ QED
Worked example 1
Checking the approximations numerically
Use small angle approximations to estimate $\sin(0.1)$, $\cos(0.1)$ and $\tan(0.1)$ (θ in radians), and compare with a calculator's exact values.
MethodSubstitute θ = 0.1 directly into each approximation
1
$\sin(0.1)\approx0.1$; calculator: $\sin(0.1)=0.099833\ldots$ — agrees to 3 d.p.
The approximation is θ itself.
2
$\cos(0.1)\approx1-\dfrac{(0.1)^2}{2}=1-0.005=0.995$; calculator: $\cos(0.1)=0.995004\ldots$ — agrees to 5 d.p.
Substitute θ=0.1 into 1−θ²/2.
3
$\tan(0.1)\approx0.1$; calculator: $\tan(0.1)=0.100335\ldots$ — agrees to 3 d.p.
The approximation is θ itself.
▪ solved
Look deeper — the cosine approximation is the best of the three
For the same small $\theta$, $\cos\theta\approx1-\frac{\theta^2}{2}$ is noticeably more accurate than $\sin\theta\approx\theta$ or $\tan\theta\approx\theta$. This is because the "missing" next term in each true expansion is order $\theta^3$ for sine and tangent, but order $\theta^4$ for cosine — a much smaller correction.
Worked example 2
Simplifying a limit-style expression
Given that $\theta$ is small and measured in radians, show that $\dfrac{\sin3\theta}{\theta(1+\cos\theta)} \approx \dfrac{3}{2}$.
MethodApply each approximation term by term, then simplify
1
$3\theta$ is small whenever $\theta$ is, so $\sin3\theta\approx3\theta$.
The sine approximation applies to any small angle — here that angle is 3θ, not θ.
2
$\cos\theta\approx1-\dfrac{\theta^2}{2}$, so $1+\cos\theta\approx2-\dfrac{\theta^2}{2}\approx2$, since $\theta^2$ is negligibly small next to the constant term.
Discard the θ² term as an even smaller correction — this is standard practice once a leading approximation is in hand.
Substitute both approximations and cancel the common factor of θ.
▪ shown
Sanity check
At $\theta=0.05$, the exact value of $\dfrac{\sin3\theta}{\theta(1+\cos\theta)}$ works out to $\approx1.4953$ — already extremely close to $1.5=\dfrac32$, confirming the approximation is doing its job.
Worked example 3
Solving an equation approximately · Challenge
Given that $\theta$ is small and measured in radians, show that $2\cos\theta+3\sin\theta\approx2+3\theta-\theta^2$, and hence find an approximate positive value of $\theta$ satisfying $2\cos\theta+3\sin\theta=2.2$.
MethodApproximate the expression first, then solve the resulting equation
1
$2\cos\theta\approx2\left(1-\dfrac{\theta^2}{2}\right)=2-\theta^2$, and $3\sin\theta\approx3\theta$.
Apply the two approximations separately to each term.
2
Adding: $2\cos\theta+3\sin\theta\approx(2-\theta^2)+3\theta = 2+3\theta-\theta^2$, as required.
Combine like terms.
3
Setting this equal to $2.2$: $2+3\theta-\theta^2=2.2 \Rightarrow \theta^2-3\theta+0.2=0$.
Substitute and rearrange into a standard quadratic.
4
$\theta = \dfrac{3\pm\sqrt{9-0.8}}{2} = \dfrac{3\pm\sqrt{8.2}}{2}$, giving $\theta\approx0.0682$ or $\theta\approx2.93$.
Solve with the quadratic formula.
5
Only $\theta\approx0.0682$ is small enough for the approximation to be trustworthy — the other root, $2.93$, is rejected.
Small angle approximations are only valid near θ=0; a large root of the approximate equation isn't a meaningful solution of the original one.
▪ solved
Look deeper — how good was the approximation?
Solving $2\cos\theta+3\sin\theta=2.2$ exactly (numerically) gives $\theta\approx0.06827$ — within $0.01\%$ of the approximate answer $0.0682$. This is typical: small angle approximations often give startlingly accurate results, provided the resulting root really is small.
Common mistake
Using small angle approximations with $\theta$ in degrees — these formulas come from radian-based arc length ($s=r\theta$) and are simply false for a degree value. Also common: keeping both roots of a quadratic that arises from an approximated equation, without checking which one is actually small enough to trust.
Check yourself
Given that $\theta$ is small and measured in radians, show that $\dfrac{1-\cos2\theta}{\theta\sin\theta} \approx 2$. (Hint: apply the cosine approximation to $2\theta$, not $\theta$, in the numerator.)
Method
1
$\cos2\theta\approx1-\dfrac{(2\theta)^2}{2}=1-2\theta^2$, so $1-\cos2\theta\approx2\theta^2$.
Apply the cosine approximation with 2θ in place of θ.
2
$\sin\theta\approx\theta$, so $\theta\sin\theta\approx\theta\cdot\theta=\theta^2$.
Apply the sine approximation to the denominator.
3
$\dfrac{1-\cos2\theta}{\theta\sin\theta} \approx \dfrac{2\theta^2}{\theta^2} = 2$, as required.