📚 Differentiating sin x and cos x | sin x与cos x的求导法则
Differentiation is one of the most powerful tools in A-Level Mathematics, and trigonometric functions form an essential part of the syllabus. In this article, we will explore the derivative rules for sin x and cos x, understand where they come from, and see how to apply them in a variety of exam-style problems.
微分是A-Level数学中最强大的工具之一，而三角函数更是考纲中不可或缺的一部分。在这篇文章中，我们将深入探讨sin x与cos x的求导法则，理解它们的来源，并学习如何在各类考试题型中灵活运用。
1. The Derivative of sin x | sin x的导数
For any angle x measured in radians, the derivative of sin x with respect to x is cos x. That is:
d/dx (sin x) = cos x
This result is beautifully simple, but it is essential to remember that x must be in radians for this formula to hold. If x were in degrees, an extra conversion factor of π/180 would appear, which is why A-Level exams always assume radians unless stated otherwise.
对于以弧度为单位的角度x，sin x关于x的导数为cos x，即：
d/dx (sin x) = cos x
这一结果简洁而优美，但必须牢记：仅当x以弧度为单位时该公式才成立。如果x以度为单位，则会出现额外的转换因子π/180，因此A-Level考试中除非特别说明，一律默认使用弧度制。
2. The Derivative of cos x | cos x的导数
Similarly, the derivative of cos x with respect to x is -sin x:
d/dx (cos x) = -sin x
at x = 0, which is consistent with the graph of cos x falling from its maximum value of 1.
类似地，cos x关于x的导数为-sin x：
d/dx (cos x) = -sin x
请注意这里的负号。它源于当x增大时cos x先减小这一事实。这意味着曲线y = cos x在x = 0处的切线斜率为负，这与cos x从最大值1下降的图形完全吻合。
3. Deriving the Rules from First Principles | 从第一原理推导法则
To truly understand these formulas, it is valuable to see how they emerge from the definition of the derivative. The derivative of a function f(x) is defined as:
f'(x) = lim₍ₕ→₀₎ [f(x+h) - f(x)] / h
sin x, we use the trigonometric identity sin(A+B) = sin A cos B + cos A sin B. This gives:
d/dx (sin x) = lim₍ₕ→₀₎ [sin x cos h + cos x sin h - sin x] / h
Rearranging terms and taking the limit, we obtain:
d/dx (sin x) = sin x · lim₍ₕ→₀₎ (cos h - 1)/h + cos x · lim₍ₕ→₀₎ (sin h)/h
ocess for cos x, using cos(A+B) = cos A cos B - sin A sin B, yields d/dx (cos x) = -sin x.
为真正理解这些公式，值得看看它们如何从导数的定义中产生。函数f(x)的导数定义为：
f'(x) = lim₍ₕ→₀₎ [f(x+h) - f(x)] / h
将此应用于f(x) = sin x，使用三角恒等式sin(A+B) = sin A cos B + cos A sin B，可得：
d/dx (sin x) = lim₍ₕ→₀₎ [sin x cos h + cos x sin h - sin x] / h
整理各项并取极限，得到：
d/dx (sin x) = sin x · lim₍ₕ→₀₎ (cos h - 1)/h + cos x · lim₍ₕ→₀₎ (sin h)/h
利用标准极限lim₍ₕ→₀₎ (sin h)/h = 1和lim₍ₕ→₀₎ (cos h - 1)/h = 0，我们得到d/dx (sin x) = cos x。对cos x应用类似的过程，使用cos(A+B) = cos A cos B - sin A sin B，可得到d/dx (cos x) = -sin x。
4. Key Limits You Must Know | 必须掌握的关键极限
There are two standard limits that appear repeatedly in the derivation of trigonometric derivatives. You may be asked to use them or, in more advanced questions, to prove them using geometric arguments or the squeeze theorem.
有两个标准极限在三角函数求导的推导中反复出现。你可能会被要求直接使用它们，也可能在更高阶的题目中被要求用几何论证或夹逼定理来证明它们。
lim₍ₓ→₀₎ (sin x)/x = 1 — This is the most important trigonometric limit.
lim₍ₓ→₀₎ (cos x - 1)/x = 0 — This follows from the identity cos x - 1 = -2sin²(x/2).
lim₍ₓ→₀₎ (sin x)/x = 1 — 这是最重要的三角极限。
lim₍ₓ→₀₎ (cos x - 1)/x = 0 — 由恒等式cos x - 1 = -2sin²(x/2)得出。
5. Differentiating sin kx and cos kx | 对sin kx与cos kx求导
Using the chain rule, we can extend our formulas to handle multiples of x. If k is a constant, then:
d/dx [sin(kx)] = k cos(kx)
d/dx [cos(kx)] = -k sin(kx)
which equals k. For example, d/dx [sin(3x)] = 3 cos(3x), and d/dx [cos(5x)] = -5 sin(5x).
利用链式法则，我们可以将公式推广到处理x的倍数。若k为常数，则：
d/dx [sin(kx)] = k cos(kx)
d/dx [cos(kx)] = -k sin(kx)
因子k的出现是因为内层函数kx的导数为k。例如，d/dx [sin(3x)] = 3 cos(3x)，而d/dx [cos(5x)] = -5 sin(5x)。
6. The Chain Rule and Composite Functions | 链式法则与复合函数
a more complex expression, we rely on the chain rule. If y = sin(u) where u = g(x), then:
dy/dx = cos(u) · du/dx
/dx = -sin(u) · du/dx. This is one of the most frequently tested skills in A-Level papers.
当sin或cos作用于更复杂的表达式（即复合函数）时，我们需要借助链式法则。若y = sin(u)，其中u = g(x)，则：
dy/dx = cos(u) · du/dx
同理，若y = cos(u)，则dy/dx = -sin(u) · du/dx。这是A-Level考试中最常考的题型之一。
7. Worked Example: Differentiating x²sin x | 示例：对x²sin x求导
function y = x²sin x. This is a product of two functions, so we must use the product rule:
dy/dx = 2x · sin x + x² · cos x
Here, the product rule states that if y = uv, then dy/dx = u'v + uv'. Taking u = x² and v = sin x, we get u' = 2x and v' whether the product rule, quotient rule, or chain rule is required before differentiating.
考虑函数y = x²sin x。这是两个函数的乘积，因此需要用到乘法法则：
dy/dx = 2x · sin x + x² · cos x
乘法法则说明：若y = uv，则dy/dx = u'v + uv'。令u = x²，v = sin x，则u' = 2x，v' = cos x。将其代入公式即得最终答案。务必在求导前判断并选择正确的法则（乘法法则、商法则或链式法则）。
8. Worked Example: Differentiating sin(2x + 1) | 示例：对sin(2x + 1)求导
Let us differentiate y = sin(2x + 1). Here, the inner function is u = 2x + 1, so u' = 2. Using the chain rule:
dy/dx = 2 cos(2x + 1)
. Notice that the derivative of the inner function (2x + 1) appears as a multiplicative factor in the final result.
我们来求y = sin(2x + 1)的导数。这里内层函数为u = 2x + 1，因此u' = 2。使用链式法则：
dy/dx = 2 cos(2x + 1)
同样的逻辑适用于cos(2x + 1)，其导数为dy/dx = -2 sin(2x + 1)。注意内层函数(2x + 1)的导数作为乘法因子出现在最终结果中。
9. Tangents and Normals | 切线与法线
Once we know how to differentiate trigonometric functions, we can find the equation of tangents and normals to curves. The slope of the tangent at a given point is simply the value of the derivative at that point, and the slope of the normal is the negative reciprocal of the tangent's slope.
掌握了三角函数的求导方法后，我们就能求曲线的切线方程和法线方程。切线的斜率就是函数在该点导数的值，而法线斜率则是切线斜率的负倒数。
Example: Find the equation of the tangent to y = sin x at x = π/3.
| Step 1: Find the derivative | y' = cos x |
| Step 2: Evaluate at x = π/3 | y'(π/3) = cos(π/3) = 1/2 |
| Step 3: Find y-coordinate | y = sin(π/3) = √3/2 |
| Step 4: Equation of tangent | y - √3/2 = (1/2)(x - π/3) |
示例：求y = sin x在x = π/3处的切线方程。
| 步骤1：求导数 | y' = cos x |
| 步骤2：在x = π/3处求值 | y'(π/3) = cos(π/3) = 1/2 |
| 步骤3：求y坐标 | y = sin(π/3) = √3/2 |
| 步骤4：切线方程 | y - √3/2 = (1/2)(x - π/3) |
10. Stationary Points of Trigonometric Functions | 三角函数的驻点
ue -1. Similarly, for y = cos x, stationary points occur where sin x = 0, i.e., at x = nπ.
为了求驻点（曲线水平的位置），我们令导数等于零。由于sin x的导数为cos x，y = sin x的驻点出现在cos x = 0处，即x = π/2 + nπ（n为任意整数）。在这些点，sin x取最大值1或最小值-1。类似地，对于y = cos x，驻点出现在sin x = 0处，即x = nπ。
11. Common Exam Mistakes | 常见考试错误
Students frequently lose marks on trigonometric differentiation questions for reasons that are entirely avoidable. Being aware of these pitfalls is the first step to avoiding them.
学生在三角求导题目中丢分的原因往往是完全可以避免的。了解这些陷阱是避免它们的第一步。
Forgetting the minus sign: The derivative of cos x is -sin x, not sin x. This simple sign error is one of the most common mistakes.
Using degrees instead of radians: The standard formulas only work in radians. Make sure your calculator is in radian mode.
Forgetting the chain rule factor: When differentiating sin(kx), you must multiply by k. Forgetting this factor leads to wrong answers.
Confusing the product rule and chain rule: Always check the structure of the function first — is it a product, a quotient, or a composite?
忘记负号：cos x的导数为-sin x，而非sin x。这个简单的符号错误是最常见的失分原因。
使用度数而非弧度：标准公式仅在弧度制下成立。请确保计算器处于弧度模式。
忘记链式法则因子：对sin(kx)求导时，必须乘以k。遗漏该因子将导致错误答案。
混淆乘法法则与链式法则：务必先分析函数的结构——它是乘积、商还是复合函数？
12. Summary and Final Tips | 总结与备考建议
From these, combined with the chain rule, product rule, and quotient rule, you can differentiate any expression involving sin x and cos x. Practice is essential — al attention to questions that combine trigonometric functions with algebraic expressions.
核心公式是d/dx (sin x) = cos x和d/dx (cos x) = -sin x。以它们为基础，结合链式法则、乘法法则和商法则，你就能对任何包含sin x与cos x的表达式求导。练习至关重要——请尽量多做历年真题，尤其关注将三角函数与代数表达式结合起来的题目。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图