📚 A-Level Mathematics: Small Angle Approximations | A-Level 数学：小角度近似
When you meet trigonometric expressions in mechanics, pure mathematics, or even physics, a tiny angle measured in radians can unlock a powerful set of simplifying tools. These are the small angle approximations, and they replace awkward sine, cosine and tangent functions with simple powers of the angle itself.
在力学、纯数学甚至物理中，当遇到一个以弧度为单位的小角度时，小角度近似可以大大简化问题。它把讨厌的 sin、cos、tan 函数替换成关于这个角度的简单幂函数，这也是 A-Level 数学中一个高频且易拿分的考点。
1. What Are Small Angle Approximations? | 什么是小角度近似
For an angle θ measured in radians, when θ is very close to zero, the graphs of y = sin θ, y = tan θ and y = θ almost coincide. Likewise, the graph of y = cos θ is almost the parabola y = 1 − θ²/2. These observations are made rigorous by the Maclaurin series of the trigonometric functions.
对于以弧度为单位的角度 θ，当 θ 非常接近零时，曲线 y = sin θ、y = tan θ 与直线 y = θ 几乎重合。类似地，y = cos θ 的图象几乎与抛物线 y = 1 − θ²/2 吻合。这些几何直觉可以由三角函数的麦克劳林级数严格验证。
The three standard approximations are:
标准的三个近似公式是：
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sin θ ≈ θ
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tan θ ≈ θ
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cos θ ≈ 1 − θ²/2
They are valid only when θ is small and always measured in radians, not degrees.
这些公式仅在 θ 很小并且一定以弧度为单位时才成立，绝不能直接代入角度制的数值。
2. The Geometric Idea Behind the Approximations | 近似公式背后的几何直观
Consider a unit circle. The length of the arc subtended by angle θ is exactly θ radians. The vertical coordinate of the point on the circle is sin θ, and the tangent segment drawn from the x-axis has length tan θ.
考虑单位圆。半径 1 的圆中，角度 θ 对应的弧长正好是 θ。点的纵坐标是 sin θ，过该点作切线并延长到 x 轴，切线段长为 tan θ。
When θ is very small, the arc is almost a straight vertical line, so the arc length, the sine and the tangent are all nearly equal. Also, the x-coordinate of the point, cos θ, is almost 1, but the second-order correction −θ²/2 becomes important in many calculations.
当 θ 很小时，弧段几乎是一条竖直线，因此弧长、sin θ 与 tan θ 三者近似相等。同时点的横坐标 cos θ 几乎等于 1，但在很多高阶计算中，二阶修正项 −θ²/2 不可忽略。
This geometric picture also explains why the approximations fail for large angles: once the arc curls around the circle, it is no longer close to a straight line.
这幅几何图也解释了大角度时近似失效的原因：一旦弧线明显弯曲，它就不再接近一条直线。
3. Derivation from Maclaurin Series | 从麦克劳林级数推导
The Maclaurin series for sine, cosine and tangent are:
sin、cos、tan 的麦克劳林级数分别为：
sin θ = θ − θ³/3! + θ⁵/5! − θ⁷/7! + …
cos θ = 1 − θ²/2! + θ⁴/4! − θ⁶/6! + …
tan θ = θ + θ³/3 + 2θ⁵/15 + …
If we keep only the leading term in each series, we obtain sin θ ≈ θ, tan θ ≈ θ and cos θ ≈ 1. However, the approximation cos θ ≈ 1 is rarely accurate enough because it ignores the first-order change in angle; the second-order term −θ²/2 is essential whenever products or limits are involved.
若每个级数只保留首项，就得到 sin θ ≈ θ、tan θ ≈ θ 和 cos θ ≈ 1。但 cos θ ≈ 1 往往不够精确，因为它忽略了角度变化的首要效应；在求极限、乘积或二阶量时，必须保留 −θ²/2 这一项。
Sometimes the next-order terms are needed:
有时也需要更高阶的项：
sin θ ≈ θ − θ³/6, tan θ ≈ θ + θ³/3, cos θ ≈ 1 − θ²/2 + θ⁴/24
These extended forms are not always required by the exam, but they help with error analysis and with problems that involve cancellation of the leading term.
这些扩展形式不一定是考试硬性要求，但有助于误差分析以及处理首项抵消的问题。
4. The Standard Small Angle Approximations | 标准小角度近似公式
For an angle θ in radians that is close to zero:
当弧度 θ 接近于零时，有下列标准近似：
| Function | Approximation | Hidden terms |
| sin θ | θ | − θ³/6 + θ⁵/120 − … |
| tan θ | θ | + θ³/3 + 2θ⁵/15 + … |
| cos θ | 1 − θ²/2 | + θ⁴/24 − θ⁶/720 + … |
Always remember that θ must be in radians. If the problem gives degrees, convert using θrad = θdeg × π/180 before applying any approximation.
始终牢记 θ 的单位必须是弧度。如果题目给的是角度制，先用 θrad = θdeg × π/180 转换，再使用近似公式。
5. Worked Example: Direct Evaluation | 例题：直接求近似值
Example 1. Estimate sin(0.12) and compare with the exact value.
例 1：估计 sin(0.12)，并与精确值比较。
Using sin θ ≈ θ, we get sin(0.12) ≈ 0.12.
由 sin θ ≈ θ，得 sin(0.12) ≈ 0.12。
The exact value is sin(0.12) = 0.119712…, so the error is about 0.0024, or roughly 0.24%. This is an excellent approximation for most practical purposes.
精确值 sin(0.12) = 0.119712…，误差约为 0.0024，即约 0.24%。对大多数实际用途来说已经非常精确。
Example 2. Estimate cos(0.2) using the second-order approximation.
例 2：用二阶近似估计 cos(0.2)。
cos(0.2) ≈ 1 − (0.2)²/2 = 1 − 0.04/2 = 1 − 0.02 = 0.98
The exact value is cos(0.2) = 0.980067…, so the approximation is accurate to about 0.007%.
精确值为 cos(0.2) = 0.980067…，近似值精确到约 0.007%。
Example 3. For small θ, find a simple expression for (1 − cos 2θ)/θ².
例 3：当 θ 很小时，化简 (1 − cos 2θ)/θ²。
Since cos 2θ ≈ 1 − (2θ)²/2 = 1 − 2θ², we have:
因为 cos 2θ ≈ 1 − (2θ)²/2 = 1 − 2θ²，所以：
(1 − cos 2θ)/θ² ≈ (2θ²)/θ² = 2
This type of question often appears as a limit calculation: limθ→0 (1 − cos 2θ)/θ² = 2.
这类问题常以极限形式出现：limθ→0 (1 − cos 2θ)/θ² = 2。
6. Combining Approximations in Expressions | 多个近似同时使用的化简问题
When an expression contains several trigonometric functions, apply each approximation before simplifying. Be careful with the order of smallness: products, quotients and powers all affect which terms survive.
当一个式子包含多个三角函数时，先分别代入近似，再化简。注意“小量的阶”：乘积、商和幂都会影响最终哪些项被保留。
Example 4. Given that θ is small, expand sin θ + cos θ in powers of θ up to θ².
例 4：已知 θ 很小，将 sin θ + cos θ 按 θ 的幂展开到 θ²。
sin θ + cos θ ≈ θ + (1 − θ²/2) = 1 + θ − θ²/2
So the coefficients are: constant term 1, linear coefficient 1, quadratic coefficient −1/2.
因此常数项为 1，一次项系数为 1，二次项系数为 −1/2。
Example 5. Simplify tan θ / sin θ for small θ.
例 5：当 θ 很小时化简 tan θ / sin θ。
Using tan θ ≈ θ and sin θ ≈ θ, we get:
用 tan θ ≈ θ 与 sin θ ≈ θ，得：
tan θ / sin θ ≈ θ / θ = 1
This matches the exact limit limθ→0 tan θ / sin θ = 1, which can also be shown by writing tan θ = sin θ / cos θ.
这与精确极限 limθ→0 tan θ / sin θ = 1 一致，也可以利用 tan θ = sin θ / cos θ 验证。
Example 6. For small θ, simplify (sin θ + tan θ)/(1 − cos θ).
例 6：当 θ 很小时，化简 (sin θ + tan θ)/(1 − cos θ)。
sin θ ≈ θ, tan θ ≈ θ, 1 − cos θ ≈ θ²/2
Therefore:
因此：
(sin θ + tan θ)/(1 − cos θ) ≈ (θ + θ)/(θ²/2) = 2θ / (θ²/2) = 4/θ
This result tells us that the expression behaves like 4/θ for very small θ, which tends to infinity as θ tends to zero. Always check whether your denominator is a second-order small quantity — it can turn the whole expression into a first-order divergent term.
这个结果说明原式在 θ 很小时约等于 4/θ，当 θ 趋于 0 时趋于无穷大。一定要检查分母是否是二阶小量，它会把整个表达式变成一阶发散量。
7. Solving Equations Using Small Angle Approximations | 利用小角度近似求解方程
Sometimes a trigonometric equation is difficult to solve exactly, but if you know the root is small, you can replace the trigonometric functions with their approximations and solve a polynomial equation instead.
有些三角方程很难精确求解，但如果已知根很小，就可以把三角函数替换成近似多项式，从而转化为解多项式方程。
Example 7. Show that the small positive solution of tan θ = 0.6θ + 0.1 is approximately θ = 0.25.
例 7：证明方程 tan θ = 0.6θ + 0.1 的小正根约为 θ = 0.25。
For small θ, tan θ ≈ θ. The equation becomes:
对于小 θ，tan θ ≈ θ。原方程变为：
θ ≈ 0.6θ + 0.1 ⇒ 0.4θ ≈ 0.1 ⇒ θ ≈ 0.25
Check: tan(0.25) = 0.2553, while the right side 0.6(0.25) + 0.1 = 0.25. The two sides differ by only 0.0053, about 2%, so the approximation is reasonable.
验证：tan(0.25) = 0.2553，而右端 0.6(0.25) + 0.1 = 0.25。两者仅相差 0.0053，约 2%，因此近似合理。
Example 8. Find the small positive root of x² + 1 − cos x = 0.3x.
例 8：求方程 x² + 1 − cos x = 0.3x 的小正根。
Use cos x ≈ 1 − x²/2:
代入 cos x ≈ 1 − x²/2：
x² + 1 − (1 − x²/2) ≈ 0.3x ⇒ 1.5x² ≈ 0.3x
Since x ≠ 0, divide by x:
因为 x ≠ 0，两边同除以 x：
1.5x ≈ 0.3 ⇒ x ≈ 0.2
The exact root of the original equation is approximately 0.187, so the approximation x ≈ 0.2 is quite acceptable.
原方程的精确根约为 0.187，因此近似值 x ≈ 0.2 已经相当不错。
8. Application: Period of a Simple Pendulum | 应用：单摆的周期
In mechanics, the equation of motion for a simple pendulum of length L is:
在力学中，长度为 L 的单摆运动方程为：
d²θ/dt² = −(g/L) sin θ
This equation is nonlinear and difficult to solve exactly. However, for small oscillations, we may replace sin θ by θ:
该方程为非线性方程，难以精确求解。然而，对于小幅摆动，可以用 θ 代替 sin θ：
d²θ/dt² ≈ −(g/L) θ
This is the standard equation of simple harmonic motion. The angular frequency is ω = √(g/L), so the period is:
这正是标准简谐运动方程。角频率为 ω = √(g/L)，因此周期为：
T = 2π√(L/g)
This famous formula is exactly what you get when the small angle approximation is applied. It shows that, to first order, the period is independent of the amplitude of the swing.
这个著名公式正是小角度近似的直接产物。它表明，在一阶近似下，单摆周期与摆动的振幅无关。
If the initial amplitude is larger, the true period becomes slightly longer; a more accurate formula is T ≈ 2π√(L/g)(1 + θ₀²/16), where θ₀ is the initial angle in radians.
如果初始振幅较大，真实周期会比公式稍长；更高精度的近似为 T ≈ 2π√(L/g)(1 + θ₀²/16)，其中 θ₀ 是初始角（弧度）。
9. Truncation Error and When the Approximation Is Valid | 截断误差与近似的适用范围
The small angle approximations are nothing but truncated Taylor series. The omitted terms give a measure of the error.
小角度近似本质上是截断的泰勒级数。被省略的项给出了误差的量级。
| Approximation | Leading error term | Error magnitude |
| sin θ ≈ θ | − θ³/6 | θ³/6 for positive θ |
| tan θ ≈ θ | + θ³/3 | θ³/3 for positive θ |
| cos θ ≈ 1 − θ²/2 | + θ⁴/24 | θ⁴/24 |
For example, when θ = 0.1 rad, the error in sin θ ≈ θ is at most 0.001/6 ≈ 0.000167. When θ = 0.5 rad, the error bound rises to 0.125/6 ≈ 0.0208, which is about 4% of the value 0.5. So θ = 0.5 rad is near the edge of what is acceptable.
例如，当 θ = 0.1 rad 时，sin θ ≈ θ 的误差至多为 0.001/6 ≈ 0.000167。当 θ = 0.5 rad 时，误差界上升到 0.125/6 ≈ 0.0208，约为数值 0.5 的 4%。因此 θ = 0.5 rad 已接近近似可接受的边缘。
A useful rule of thumb: for exam questions, θ should generally be below 0.2 rad (about 11.5°), and often the question will explicitly state “θ is small” or “for small values of x.”
一个实用经验：在考试题中，θ 通常应小于 0.2 rad（约 11.5°）。题目一般会明确说明 “θ is small” 或 “for small values of x”。
10. Exam Traps and Final Checklist | 考试陷阱与检查清单
Here are the most common mistakes students make, and how to avoid them.
以下是学生最常见的错误以及如何避免它们。
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Forgetting radians: Always convert degrees to radians before applying the approximation. If the problem says 3°, write 3 × π/180 ≈ 0.05236 rad first. | 忘记弧度制：使用近似前必须把角度制转换为弧度。遇到 3° 时应先写成 3 × π/180 ≈ 0
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