📚 Differentiation of Trigonometric Functions | 三角函数的求导方法
The derivatives of trigonometric functions form a cornerstone of calculus and appear frequently in physics, engineering, and advanced mathematics. Mastering these derivative rules is essential for solving optimisation problems, analysing periodic motion, and understanding rates of change in oscillatory systems. This article provides a systematic and exam-focused guide to differentiating sine, cosine, tangent, and their reciprocal functions, with worked examples following the A-Level syllabus.
三角函数的导数是微积分的重要基石，在物理、工程以及高等数学中频繁出现。掌握这些求导法则，对于解决最优化问题、分析周期运动、理解振动系统中的变化率都至关重要。本文将按照 A-Level 考纲要求，系统讲解正弦、余弦、正切及其倒数函数的求导方法，并配以详细的例题。
1. The Six Basic Trigonometric Derivatives | 六个基本三角函数的导数
Before tackling complex problems, it is essential to memorise the derivative of each of the six trigonometric functions. The derivatives of sine and cosine are the most fundamental, and all others can be derived from them using the quotient rule and reciprocal identities.
在解决复杂问题之前，必须牢记六个基本三角函数的导数公式。正弦和余弦的导数是最基础的，其余四个函数都可以通过商法则和倒数恒等式由它们推导出来。
| Function 函数 | Derivative 导数 |
| sin x | cos x |
| cos x | −sin x |
| tan x | sec² x |
| cosec x | −cosec x · cot x |
| sec x | sec x · tan x |
| cot x | −cosec² x |
A common mistake among students is forgetting the negative signs attached to the derivatives of cos x, cosec x, and cot x. A useful mnemonic is that the “co-functions” (cosine, cosecant, cotangent) all produce negative derivatives.
学生常犯的一个错误是忘记 cos x、cosec x 和 cot x 导数前的负号。一个有用的记忆技巧是：”co 函数”（余弦、余割、余切）的导数都带有负号。
2. Deriving the Derivative of sin x from First Principles | 用第一性原理推导 sin x 的导数
Understanding where the derivative formula comes from strengthens conceptual clarity and is a frequent examination question. We begin with the formal definition of the derivative:
理解导数公式的来源能够加深概念理解，这也是考试中常见的问题。我们从导数的形式定义开始：
f ‘(x) = lim (h→0) [f(x+h) − f(x)] / h
For f(x) = sin x, we write:
对于 f(x) = sin x，我们有：
d/dx (sin x) = lim (h→0) [sin(x+h) − sin x] / h
Using the compound angle identity sin(x+h) = sin x cos h + cos x sin h, the numerator becomes sin x(cos h − 1) + cos x sin h. Dividing by h and taking the limit, we use two special limits: lim(h→0) sin h / h = 1 and lim(h→0) (cos h − 1)/h = 0. This yields the result d/dx (sin x) = cos x.
利用和角公式 sin(x+h) = sin x cos h + cos x sin h，分子变为 sin x(cos h − 1) + cos x sin h。除以 h 后取极限，我们需要用到两个特殊极限：lim(h→0) sin h / h = 1 和 lim(h→0) (cos h − 1)/h = 0。由此得到 d/dx (sin x) = cos x。
Similarly, applying the same process to f(x) = cos x yields the derivative −sin x. These two special limits are worth memorising, as examiners sometimes ask students to evaluate them directly.
类似地，对 f(x) = cos x 应用同样的过程，可以得到导数 −sin x。这两个特殊极限值得牢记，因为考官有时会直接要求同学们求它们的值。
3. Derivative of tan x via the Quotient Rule | 利用商法则求 tan x 的导数
Since tan x = sin x / cos x, we can apply the quotient rule to derive its derivative. The quotient rule states that if y = u/v, then dy/dx = (v·du/dx − u·dv/dx) / v².
因为 tan x = sin x / cos x，我们可以应用商法则来推导其导数。商法则指出，若 y = u/v，则 dy/dx = (v·du/dx − u·dv/dx) / v²。
d/dx (tan x) = [cos x · cos x − sin x · (−sin x)] / cos² x = (cos² x + sin² x) / cos² x = 1 / cos² x = sec² x
Notice how the Pythagorean identity sin² x + cos² x = 1 simplifies the numerator beautifully. This elegant derivation illustrates the interconnectedness of trigonometric identities and calculus.
注意毕达哥拉斯恒等式 sin² x + cos² x = 1 如何巧妙地将分子化简。这个优雅的推导过程展示了三角恒等式与微积分之间的内在联系。
The derivatives of sec x, cosec x, and cot x can be obtained similarly by writing them as reciprocal functions: sec x = 1/cos x, cosec x = 1/sin x, and cot x = cos x/sin x, then applying the quotient rule or the chain rule.
sec x、cosec x 和 cot x 的导数可以通过类似的倒数关系得到：sec x = 1/cos x，cosec x = 1/sin x，cot x = cos x/sin x，然后应用商法则或链式法则。
4. The Chain Rule with Trigonometric Functions | 链式法则与三角函数
In the A-Level curriculum, trigonometric functions rarely appear in isolation. They are typically composed with other functions, requiring the chain rule. If y = sin[f(x)], then dy/dx = f ‘(x) · cos[f(x)].
在 A-Level 课程中，三角函数很少单独出现。它们通常与其他函数复合，这时需要使用链式法则。若 y = sin[f(x)]，则 dy/dx = f ‘(x) · cos[f(x)]。
Worked Example 1: Differentiate y = sin(3x² + 2x). Let u = 3x² + 2x, so y = sin u. Then du/dx = 6x + 2 and dy/du = cos u. Therefore dy/dx = (6x + 2) · cos(3x² + 2x).
例题 1： 求 y = sin(3x² + 2x) 的导数。令 u = 3x² + 2x，则 y = sin u。于是 du/dx = 6x + 2，dy/du = cos u。因此 dy/dx = (6x + 2) · cos(3x² + 2x)。
Worked Example 2: Differentiate y = cos⁵ x. This requires both the chain rule and the power rule. Writing y = (cos x)⁵, let u = cos x, so y = u⁵. Then dy/du = 5u⁴ and du/dx = −sin x. Hence dy/dx = 5(cos x)⁴ · (−sin x) = −5 sin x · cos⁴ x.
例题 2： 求 y = cos⁵ x 的导数。这需要同时运用链式法则和幂法则。将 y 写作 (cos x)⁵，令 u = cos x，则 y = u⁵。那么 dy/du = 5u⁴，du/dx = −sin x。因此 dy/dx = 5(cos x)⁴ · (−sin x) = −5 sin x · cos⁴ x。
When differentiating tan or sec of a function, remember to multiply by the derivative of the inner function: d/dx [tan(g(x))] = g ‘(x) · sec²(g(x)).
在求 tan 或 sec 复合函数的导数时，记得乘以内层函数的导数：d/dx [tan(g(x))] = g ‘(x) · sec²(g(x))。
5. Implicit Differentiation with Trigonometric Functions | 三角函数的隐函数求导
When y is defined implicitly as a function of x through a trigonometric equation, we differentiate both sides with respect to x, treating y as a function of x and using the chain rule. The derivative of sin y with respect to x is cos y · (dy/dx).
当 y 通过一个三角方程隐式地定义为 x 的函数时，我们需要对等式两边关于 x 求导，将 y 视为 x 的函数并使用链式法则。sin y 关于 x 的导数为 cos y · (dy/dx)。
Worked Example: Find dy/dx given that sin y = x².
例题： 已知 sin y = x²，求 dy/dx。
Differentiating both sides with respect to x:
对等式两边关于 x 求导：
cos y · (dy/dx) = 2x
Therefore dy/dx = 2x / cos y. Using the identity cos y = √(1 − sin² y) = √(1 − x⁴), we can express the answer purely in terms of x:
因此 dy/dx = 2x / cos y。利用恒等式 cos y = √(1 − sin² y) = √(1 − x⁴)，我们可以将答案完全用 x 表示：
dy/dx = 2x / √(1 − x⁴)
This technique is particularly useful for finding the derivatives of inverse trigonometric functions, which are discussed in the next section.
这一技巧在求反三角函数的导数时尤其有用，我们将在下一节讨论。
6. Derivatives of Inverse Trigonometric Functions | 反三角函数的导数
Inverse trigonometric functions arise frequently in integration problems and differential equations. Their derivatives are algebraic expressions, which surprises many students at first. Let y = arcsin x, meaning sin y = x. Differentiating implicitly:
反三角函数在积分问题和微分方程中频繁出现。它们的导数是代数表达式，许多同学第一次看到时会感到惊讶。设 y = arcsin x，即 sin y = x。隐式求导得到：
cos y · (dy/dx) = 1 → dy/dx = 1/cos y = 1/√(1 − x²)
The other key results are summarised below:
其他重要结果总结如下：
| Function 函数 | Derivative 导数 | Domain 定义域 |
| arcsin x | 1/√(1 − x²) | |x| < 1 |
| arccos x | −1/√(1 − x²) | |x| < 1 |
| arctan x | 1/(1 + x²) | all real x |
For composite inverse trigonometric functions, apply the chain rule. For example, if y = arctan(2x), then dy/dx = 2/(1 + 4x²).
对于复合反三角函数，应用链式法则。例如，若 y = arctan(2x)，则 dy/dx = 2/(1 + 4x²)。
7. Second Derivatives and Oscillatory Motion | 二阶导数与振荡运动
The second derivative of a trigonometric function reveals important information about concavity and acceleration. For y = sin x, we have y’ = cos x and y” = −sin x = −y. This remarkable property — that the second derivative of sine is the negative of the original function — is the mathematical foundation of simple harmonic motion.
三角函数的二阶导数揭示了关于凹凸性和加速度的重要信息。对于 y = sin x，有 y’ = cos x 和 y” = −sin x = −y。这个显著的性质——正弦的二阶导数是原函数的负值——是简谐运动的数学基础。
Worked Example: A particle moves such that its displacement is x = 5 sin(3t) cm. Find its acceleration.
例题： 一个质点的位移为 x = 5 sin(3t) cm。求其加速度。
Velocity: v = dx/dt = 15 cos(3t) cm/s. Acceleration: a = dv/dt = −45 sin(3t) cm/s².
速度：v = dx/dt = 15 cos(3t) cm/s。加速度：a = dv/dt = −45 sin(3t) cm/s²。
Observe that acceleration is proportional to displacement but opposite in direction. This is the defining characteristic of SHM. Understanding this connection between differentiation and physics demonstrates the power of calculus.
注意加速度与位移成正比但方向相反。这是简谐运动的本质特征。理解微分与物理之间的这种联系，展现了微积分的强大力量。
8. Small-Angle Approximations and Their Derivatives | 小角度近似及其导数
For small angles measured in radians, we have the approximations sin θ ≈ θ, cos θ ≈ 1, and tan θ ≈ θ. These approximations are derived from the Taylor series expansions of the trigonometric functions and connect naturally to the derivatives discussed earlier.
对于以弧度为单位的小角度，有近似公式 sin θ ≈ θ，cos θ ≈ 1，tan θ ≈ θ。这些近似公式来源于三角函数的泰勒级数展开，并且与之前讨论的导数自然相关。
Worked Example: Show that for small x, the derivative of sin x is approximately equal to 1.
例题： 证明对于小 x，sin x 的导数近似等于 1。
Since sin x ≈ x for small x, and d/dx (sin x) = cos x ≈ 1 for small x, the result follows immediately. This is consistent because the slope of y = x is exactly 1 everywhere.
因为对于小 x 有 sin x ≈ x，而 d/dx (sin x) = cos x，且对于小 x 有 cos x ≈ 1，所以结论立即成立。这与直线 y = x 的斜率处处等于 1 是一致的。
In physical applications, the pendulum equation θ” + (g/L)·sin θ = 0 is often simplified to θ” + (g/L)·θ = 0 for small oscillations, leveraging sin θ ≈ θ. This linearisation makes the differential equation much easier to solve.
在物理应用中，单摆方程 θ” + (g/L)·sin θ = 0 在小振动条件下通常利用 sin θ ≈ θ 简化为 θ” + (g/L)·θ = 0。这种线性化处理使得微分方程的求解大为简化。
9. Common Mistakes and How to Avoid Them | 常见错误及避免方法
Students often lose marks on trigonometric differentiation due to a handful of recurring errors. The most common include forgetting the negative sign when differentiating cosine, mixing up the derivatives of sec x (sec x tan x) and cosec x (−cosec x cot x), and omitting the chain-rule factor when differentiating composite functions.
同学们在三角函数的求导中经常因为几个反复出现的错误而丢分。最常见的包括：求余弦导数时忘记负号；混淆 sec x（导数 sec x tan x）和 cosec x（导数 −cosec x cot x）的导数；以及在求复合函数导数时遗漏了链式法则的因子。
- Mistake 1: Writing d/dx (cos x) = sin x. Correction: Always remember the negative: d/dx (cos x) = −sin x.
- 错误 1： 写下 d/dx (cos x) = sin x。纠正：始终记得负号：d/dx (cos x) = −sin x。
- Mistake 2: Forgetting the inner derivative in sin(2x). Correction: The answer must include the factor 2: d/dx [sin(2x)] = 2cos(2x).
- 错误 2： 在 sin(2x) 中忘记内层导数。纠正：答案必须包含因子 2：d/dx [sin(2x)] = 2cos(2x)。
- Mistake 3: Using degrees instead of radians in derivative formulas. Correction: All calculus formulas for trig functions assume radian measure.
- 错误 3： 在导数公式中使用角度制而非弧度制。纠正：所有三角函数的微积分公式都默认使用弧度制。
A reliable strategy is to check your answer by differentiating a second time or by evaluating the derivative at a specific point numerically. For instance, at x = 0, the derivative of sin x should equal 1, and the derivative of cos x should equal 0.
一个可靠的策略是：对答案再求一次导，或者在某个特定点数值化地验证导数。例如，在 x = 0 处，sin x 的导数应该等于 1，cos x 的导数应该等于 0。
10. Mixed Practice Problems | 综合练习
Apply everything you have learned to the following questions. Attempt each one independently before checking the answers provided.
将所学的知识应用到以下问题中。请先独立尝试每一题，再对照答案。
Question 1: Differentiate y = tan(4x).
练习题 1： 求 y = tan(4x) 的导数。
Answer: dy/dx = 4sec²(4x).
Question 2: Differentiate y = x² · sin x.
练习题 2： 求 y = x² · sin x 的导数。
Answer: Using the product rule, dy/dx = 2x sin x + x² cos x.
答案： 应用乘积法则，dy/dx = 2x sin x + x² cos x。
Question 3: Given y = sin x / cos x, confirm that dy/dx = sec² x.
练习题 3： 已知 y = sin x / cos x，确认 dy/dx = sec² x。
Answer: This confirms our earlier derivation that tan x differentiates to sec² x.
Question 4: Find the equation of the tangent to y = sin x at x = π/4.
练习题 4： 求曲线 y = sin x 在 x = π/4 处的切线方程。
Answer: At x = π/4, y = √2/2 and the gradient is cos(π/4) = √2/2. The tangent equation is y − √2/2 = (√2/2)(x − π/4).
答案： 在 x = π/4 处，y = √2/2，斜率为 cos(π/4) = √2/2。切线方程为 y − √2/2 = (√2/2)(x − π/4)。
Question 5: Differentiate y = ln(sec x + tan x).
练习题 5： 求 y = ln(sec x + tan x) 的导数。
Answer: dy/dx = sec x. This is a classic result worth memorising — the derivative of ln(sec x + tan x) is simply sec x.
答案： dy/dx = sec x。这是一个值得记忆的经典结论——ln(sec x + tan x) 的导数恰好是 sec x。
Mastering trigonometric differentiation requires both memorisation of the standard formulas and consistent practice with the chain rule, product rule, and quotient rule. Regular practice with mixed problems will build fluency and confidence, ensuring you are well-prepared for examination questions on this topic. Remember: radians, not degrees, are the currency of calculus.
掌握三角函数的求导既需要熟记标准公式，也需要通过持续的练习来熟练运用链式法则、乘积法则和商法则。定期进行混合练习将帮助你建立流畅度和信心，确保你在考试中充分准备好应对该主题的问题。切记：弧度制，而非角度制，才是微积分世界的通用语言。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导