📚 Small Angle Approximations | 小角度近似
In A-Level Mathematics, small angle approximations let you replace sin θ, cos θ and tan θ with simple polynomial expressions whenever θ is measured in radians and is close to zero. These formulas are extremely useful in mechanics, wave theory and numerical estimation, and they follow directly from Maclaurin series.
在 A-Level 数学中，小角度近似允许你在 θ 以弧度为单位且接近零时，用简单的多项式表达式代替 sin θ、cos θ 和 tan θ。这些公式在力学、波动理论和数值估算中非常有用，它们直接来自麦克劳林级数。
1. Radian Measure and the Meaning of "Small" | 弧度制与"小"的含义
The small angle approximations only work when the angle is measured in radians degrees. This is because the limit sin θ / θ → 1 as θ → 0 is only true in radian measure.
小角度近似只在角度以弧度为单位时成立，绝不能用度数。这是因为当 θ → 0 时 sin θ / θ → 1 这一极限只在弧度制下才成立。
If a question gives θ in degrees, convert it first using the standard conversion formula:
如果题目给出的 θ 是度数，必须先用标准换算公式转换：
θ_rad = θ_deg × π / 180
For example, 30° becomes 30 × π / 180 = π/6 ≈ 0.524 radians, which is not yet small enough for a good approximation. That is why exam questions normally keep θ below about 0.1 radians, roughly 5.7°.
例如，30° 换算后是 30 × π / 180 = π/6 ≈ 0.524 弧度，这个值还不够小，近似效果并不好。这就是为什么考题通常把 θ 控制在约 0.1 弧度以内，大约 5.7°。
There is no universal cut-off for "small" rely on the accuracy you need. At 0.1 radians, sin θ ≈ θ is already accurate to about 0.17%, while at 0.5 radians the error grows to roughly 4%.
"小"没有绝对的分界线，它完全取决于你所需要的精度。在 0.1 弧度时，sin θ ≈ θ 的误差已经小到约 0.17%；而在 0.5 弧度时，误差会增大到约 4%。
2. The Standard Approximations | 标准近似公式
There are exactly three approximations you must memorise for the exam. All three are valid for small θ measured in radians:
考试中你必须背下来的近似式正好有三个。当 θ 以弧度为单位且很小时，三者全部成立：
sin θ ≈ θ
cos θ ≈ 1 − θ²/2
tan θ ≈ θ
A very important consequence is that, for small angles, sin θ, tan θ and θ itself are all approximately equal:
一个非常重要的推论是：对小角度而言，sin θ、tan θ 与 θ 本身三者近似相等：
sin θ ≈ tan θ ≈ θ
θ is extremely small, but keeping the θ² term is almost always required for exam accuracy.
还要注意，当 θ 极小时，cos θ ≈ 1 − θ²/2 可以进一步简化为 cos θ ≈ 1，但在考试中为了精度，几乎总是需要保留 θ² 项。
- Use sin θ ≈ θ and tan θ ≈ θ for first-order terms.
- Use cos θ ≈ 1 − θ²/2 whenever a squared term survives elsewhere in the expression.
- Never mix radians and degrees inside the same expression.
- 使用 sin θ ≈ θ 与 tan θ ≈ θ 处理一阶项。
- 当表达式中别处还残留平方项时，使用 cos θ ≈ 1 − θ²/2。
- 绝不要在同一個表达式里混用弧度与度数。
3. Where the Approximations Come From: Maclaurin Series | 近似的来源：麦克劳林级数
Each approximation is simply the first one or two terms of a Maclaurin series. The full expansions, valid for all θ in radians, are:
每一条近似都只是麦克劳林级数的前一到两项。对所有弧度值 θ 都成立的完整展开式为：
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 + …
, the θ² term is the first non-constant term, so it must be kept, giving cos θ ≈ 1 − θ²/2.
如果 θ 很小，那么 θ³、θ⁵ 以及所有更高次幂相比 θ 都可以忽略。把它们丢掉就得到 sin θ ≈ θ 与 tan θ ≈ θ。对余弦而言，θ² 项是第一个非常数项，必须保留，于是得到 cos θ ≈ 1 − θ²/2。
The factorial notation matters: 3! = 3 × 2 × 1 = 6, and 5! es after the first surviving term, the error is of the order of the next term you removed.
阶乘符号很重要：3! = 3 × 2 × 1 = 6，5! = 120。当你在第一个存留项之后截断级数时，误差的量级就是你删掉的下一个项。
4. Accuracy and Error Analysis | 精度与误差分析
The table below compares the exact value of sin θ with the approximation θ over a range of angles. Study how quickly the percentage error grows.
下表在若干角度上比较了 sin θ 的精确值与近似值 θ，请留意百分比误差增长得有多快。
| θ (rad) | θ (deg) | sin θ | Error in sin θ ≈ θ |
|---|---|---|---|
| 0.05 | 2.86° | 0.049979 | 0.042% |
| 0.10 | 5.73° | 0.099833 | 0.167% |
| 0.20 | 11.46° | 0.198669 | 0.665% |
| 0.50 | 28.65° | 0.479426 | 4.115% |
.1, cos θ = 0.99500417 while 1 − θ²/2 = 0.995, an absolute error of only about 4.2 × 10⁻⁶.
对余弦而言，近似效果要好得多。当 θ = 0.1 时，cos θ = 0.99500417，而 1 − θ²/2 = 0.995，绝对误差仅约 4.2 × 10⁻⁶。
he key message for exam answers is that the leading error in sin θ ≈ θ comes from the θ³/3! term, so the relative error is roughly θ²/6.
考试作答时的关键是：sin θ ≈ θ 的主要误差来自 θ³/3! 项，因此相对误差约为 θ²/6。
5. Using Approximations in Limits | 在极限中使用近似
Small angle approximations are the fastest way to evaluate trigonometric limits without L'Hopital's rule. The three standard results you should recognise instantly are:
小角度近似是不用洛必达法则就能求三角极限的最快方法。以下三个标准结果你应该一眼认出：
lim (θ → 0) sin θ / θ = 1
lim (θ → 0) tan θ / θ = 1
lim (θ → 0) (1 − cos θ) / θ² = 1/2
from substituting cos θ ≈ 1 − θ²/2, which gives (1 − (1 − θ²/2)) / θ² = (θ²/2) / θ² = 1/2.
第三个结果直接来自代入 cos θ ≈ 1 − θ²/2，得到 (1 − (1 − θ²/2)) / θ² = (θ²/2) / θ² = 1/2。
A very common exam instruction is "given that θ is small, show that ..." followed by a messy fraction. The technique is always the same: replace every sine and tangent by θ, replace every cosine by 1 − θ²/2, then simplify algebraically.
考试中非常常见的一句指令是"已知 θ 很小，证明……"，后面跟着一个复杂的分数式。方法永远一样：把每个正弦和正切换成 θ，把每个余弦换成 1 − θ²/2，然后做代数化简。
6. Geometric Interpretation | 几何意义
small angles the chord, the arc and the tangent line all have essentially the same length.
想象一个半径为 r 的圆，圆心处有一个小角 θ。弧长为 s = rθ，弦长为 2r sin(θ/2) ≈ rθ。因此对小角度而言，弦、弧与切线段的长度本质上相同。
s(θ/2) ≈ r, which explains why a small sector is almost indistinguishable from a triangle.
从圆心到弦的垂线长度为 r cos(θ/2) ≈ r，这就解释了为什么小扇形几乎与三角形无法区分。
ight-angled triangle is shared, while the hypotenuse and the adjacent side both shrink towards r.
这一几何图像正是 sin θ ≈ tan θ 的原因：直角三角形的对边是公共的，而斜边与邻边都收缩到接近 r。
- Arc length: s = rθ
- Chord length ≈ rθ
- Sector area: A = ½ r²θ
- Small sector ≈ isosceles triangle of area ½ r²θ
- 弧长：s = rθ
- 弦长 ≈ rθ
- 扇形面积：A = ½ r²θ
- 小扇形 ≈ 面积为 ½ r²θ 的等腰三角形
7. Applications in Mechanics: Pendulums and Oscillations | 力学应用：单摆与振动
endulum is the classic mechanics application. The restoring force is −mg sin θ, and Newton's second law gives ma = −mg sin θ, that is a = −g sin θ.
单摆是力学中经典的应用。回复力为 −mg sin θ，由牛顿第二定律得 ma = −mg sin θ，即 a = −g sin θ。
pplying sin θ ≈ θ does it become a = −gθ = −(g/L)x, which is the defining equation of SHM.
这并不是简谐运动，因为加速度与 sin θ 成正比而非与 θ 成正比。只有应用 sin θ ≈ θ 之后，方程才变成 a = −gθ = −(g/L)x，这才是简谐运动的定义式。
T = 2π √L / g
y valid for small oscillations. For a pendulum of length 0.8 m with g = 9.81 m s⁻², T = 2π √0.8 / 9.81 ≈ 1.79 s.
你应当始终说明该周期公式只适用于小振幅摆动。对摆长 0.8 m、g = 9.81 m s⁻² 的单摆，T = 2π √0.8 / 9.81 ≈ 1.79 s。
8. Applications in Waves and Optics | 波动与光学中的应用
D, where y is the distance from the central maximum and D is the slit-to-screen distance.
在双缝实验中，条纹间距公式的推导使用了 sin θ ≈ tan θ ≈ y / D，其中 y 是到中央明纹的距离，D 是缝到屏的距离。
y = n λ D / d
urs at sin θ = λ / a, so the angular width of the central maximum is approximately 2λ / a.
对单缝衍射，第一极小出现在 sin θ = λ / a，因此中央明纹的角宽度约为 2λ / a。
Snell' comes n₁ θ₁ ≈ n₂ θ₂, which is the basis of the paraxial approximation used in lens theory.
斯涅尔定律也会漂亮地简化。对小入射角，n₁ sin θ₁ = n₂ sin θ₂ 变为 n₁ θ₁ ≈ n₂ θ₂，这正是透镜理论中傍轴近似的基础。
9. Related Approximations You Should Know | 其他相关近似
Small angle approximations belong to a wider family of first-order approximations that appear throughout A-Level. Recognising the pattern helps you spot when an approximation is legitimate.
小角度近似属于 A-Level 中随处可见的一阶近似大家族。识别这一模式有助于你判断何时使用近似是合理的。
- Binomial: (1 + x)ⁿ ≈ 1 + nx for small x
- Logarithmic: ln(1 + x) ≈ x for small x
- Exponential: eˣ ≈ 1 + x for small x
- Rational: (1 + x) / (1 − x) ≈ 1 + 2x for small x
- 二项式：(1 + x)ⁿ ≈ 1 + nx（x 很小）
- 对数：ln(1 + x) ≈ x（x 很小）
- 指数：eˣ ≈ 1 + x（x 很小）
- 分式：(1 + x) / (1 − x) ≈ 1 + 2x（x 很小）
These are all first-order Taylor or Maclaurin truncations. The unifying idea is that when a variable is small, its square and higher powers contribute far less than the variable itself.
它们都是一阶泰勒或麦克劳林截断。统一的思想是：当变量很小时，它的平方及更高次幂的贡献远小于变量本身。
10. Worked Exam-Style Examples | 考试风格例题精讲
Example A. Given that θ is small, show that (1 − cos 2θ) / (θ sin θ) ≈ 2.
例 A。已知 θ 很小，证明 (1 − cos 2θ) / (θ sin θ) ≈ 2。
ore 1 − cos 2θ ≈ 2θ². Also θ sin θ ≈ θ × θ = θ². Dividing gives 2θ² / θ² = 2, as required.
由于 2θ 也很小，cos 2θ ≈ 1 − (2θ)²/2 = 1 − 2θ²，所以 1 − cos 2θ ≈ 2θ²。又 θ sin θ ≈ θ × θ = θ²。相除得 2θ² / θ² = 2，证毕。
Example B. Given that θ is small, find an approximate solution of sin 3θ = 0.24.
例 B。已知 θ 很小，求 sin 3θ = 0.24 的近似解。
et 3θ ≈ 0.24, so θ ≈ 0.08 radians. This is consistent with the assumption that θ is small.
使用 sin 3θ ≈ 3θ，得 3θ ≈ 0.24，故 θ ≈ 0.08 弧度。这与 θ 很小的假设自洽。
Example C. Evaluate lim (θ → 0) tan θ / (sin θ + θ).
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