📚 Differentiating sin x and cos x | 正弦与余弦函数的微分
In A-Level Edexcel Mathematics, differentiating trigonometric functions is a core skill that appears across pure mathematics, mechanics and applied contexts. This article focuses on the two most fundamental results: the derivatives of sin x and cos x. You need to know the standard results, understand where they come from, and be able to apply the chain rule, product rule and quotient rule confidently. Everything else in trigonometric calculus - the derivative of tan x, the second derivative cycle, stationary points on trig curves and simple harmonic motion - is built on these two lines.
在 A-Level Edexcel 数学中，对三角函数求导是一项核心技能，贯穿纯数学、力学和应用情境。本文重点讨论两个最基本的结果：sin x 和 cos x 的导数。你需要掌握标准结论，理解它们的来源，并能够熟练应用链式法则、乘积法则和商法则。三角微积分中的其他内容——tan x 的导数、二阶导数的循环规律、三角曲线上的驻点以及简谐运动——全都建立在这两行结论之上。
1. The Key Results | 核心结论
The two standard derivatives you must memorise are given below. They are the foundation for nearly all differentiation involving trigonometric functions at A-Level, and they should be recalled instantly, without hesitation, in the exam.
以下是两个必须牢记的标准导数结果。它们是 A-Level 中几乎所有涉及三角函数的微分问题的基础，在考试中应当能够毫不犹豫地立刻写出。
d/dx (sin x) = cos x
d/dx (cos x) = −sin x
at differentiating sine moves you forward in the cycle sin → cos → −sin → −cos, and differentiating cosine moves you forward in the same cycle. The minus sign is the single most common source of lost marks in this topic.
规律很简单，却很容易记错：正弦的导数是正的余弦，而余弦的导数是负的正弦。注意，对正弦求导就是沿着 sin → cos → −sin → −cos 这个循环向前走一步，对余弦求导同样沿着这个循环向前走一步。负号是这个专题中最常见的失分点。
- Sine: differentiate once to get cosine, keeping the positive sign.
正弦：求一次导得到余弦，符号保持为正。 - Cosine: differentiate once to get negative sine - never write positive sine.
余弦：求一次导得到负的正弦，绝不能写成正的正弦。 - Both results: valid only when the angle is measured in radians.
两个结论：只在角度以弧度为单位时成立。
2. Why Radian Measure Matters | 为什么必须使用弧度制
At A-Level, trigonometric differentiation is always assumed to be in radians unless a question explicitly states otherwise. The reason comes from the small-angle limit that underpins the first-principles proof. This limit is the engine of the whole topic, so it deserves careful attention.
在 A-Level 考试中，除非题目明确说明，否则三角函数的微分均默认使用弧度制。原因在于支撑第一原理证明的小角度极限。这个极限是整个专题的发动机，值得仔细关注。
lim (h → 0) sin h / h = 1
lim (h → 0) (cos h − 1) / h = 0
This limit only equals 1 when h is measured in radians. If h were in degrees, the limit would involve an extra factor of π/180, and the derivative of sin x would become (π/180) cos x. Exam questions assume radians, so do not add this factor unless told otherwise. In practical terms: always check that your calculator is in radian mode before evaluating any trigonometric expression in a calculus question, and leave answers in terms of π rather than converting to degrees.
这个极限只有在 h 以弧度表示时才等于 1。如果 h 以度数表示，极限就会含有一个额外的因子 π/180，sin x 的导数也会变成 (π/180) cos x。考试题目默认使用弧度制，因此除非题目另作说明，否则不要加上这个因子。实际操作上：在做微积分题时，先检查计算器是否处于弧度模式，并且答案尽量用 π 表示，而不是换算成度数。
- Radians: the default unit in all calculus work, with the derivative of sin x equal to cos x.
弧度制：所有微积分运算的默认单位，此时 sin x 的导数为 cos x。 - Degrees: require the extra factor π/180, which is never expected at A-Level unless stated.
角度制：需要额外的因子 π/180，A-Level 中除非明确说明，否则不会这样考查。 - Small-angle approximations: sin x ≈ x and cos x ≈ 1 − x²/2 for small x in radians.
小角度近似：当 x 很小且用弧度表示时，sin x ≈ x，cos x ≈ 1 − x²/2。
3. Derivative of sin x from First Principles | 从第一原理推导 sin x 的导数
By definition, the derivative is the limit of the gradient of a chord as the chord shrinks to zero length. Applying this definition to sin x, and using the compound-angle identity, gives a proof that examiners sometimes ask you to reproduce in structure.
按照定义，导数就是弦长趋近于零时弦斜率的极限。把这个定义应用于 sin x，并借助复合角恒等式，就得到了一种推导过程，考官有时会要求你复现其结构。
d/dx (sin x) = lim (h → 0) [ sin(x + h) − sin x ] / h
becomes sin x cos h + cos x sin h − sin x. Grouping the terms that contain sin x together gives the next line, which is the key step of the proof.
使用加法公式 sin(x + h) = sin x cos h + cos x sin h，分子变为 sin x cos h + cos x sin h − sin x。把含有 sin x 的项合并在一起，就得到下一行，这是整个证明的关键一步。
= lim (h → 0) [ sin x (cos h − 1) / h + cos x (sin h / h) ]
= cos x. The two standard limits quoted in Section 2 are exactly what make this step work.
现在分别对两个括号取极限。第一个括号趋于 sin x × 0 = 0，第二个括号趋于 cos x × 1 = cos x。正是第 2 节引用的两个标准极限让这一步成立。
d/dx (sin x) = 0 + cos x = cos x
Two features of this proof are worth remembering for written questions: the compound-angle expansion is essential, and the result depends critically on the radian limit. If you are asked to prove the result from first principles, show both limit statements explicitly rather than simply asserting them.
这个证明有两点值得为笔答题记住：复合角展开是必不可少的，而结果关键依赖于弧度制下的极限。如果被要求从第一原理证明该结论，要把两个极限明确写出来，而不是直接断言。
4. Derivative of cos x from First Principles | 从第一原理推导 cos x 的导数
sin x sin h. The structure of the algebra is almost identical, which is why the two proofs are usually learned together.
同样的方法对余弦也适用，只需使用恒等式 cos(x + h) = cos x cos h − sin x sin h。代数结构几乎完全相同，这也是两个证明通常一起学习的原因。
d/dx (cos x) = lim (h → 0) [ cos x (cos h − 1) / h − sin x (sin h / h) ]
in x. The negative sign arises naturally from the minus sign inside the compound-angle identity - it is not an arbitrary convention to be memorised.
逐项取极限得到 cos x × 0 − sin x × 1，因此导数为 −sin x。负号自然来自复合角恒等式内部的减号，它并不是一个可以随意记住的约定。
d/dx (cos x) = −sin x
cos(π/2 − x) = −sin x. This alternative derivation is a useful way to test whether your memory of the sign is correct.
还有一个很好的快捷检验方法值得掌握。由于 cos x = sin(π/2 − x)，用链式法则并代入内层导数 −1，得到 d/dx (cos x) = −cos(π/2 − x) = −sin x。这个替代推导是检验你对符号记忆是否正确的好办法。
5. Constant Multiples and Simple Linear Combinations | 常数倍与简单线性组合
Because differentiation is linear, constants pass straight through the operator. This means the two standard results extend immediately to sums and constant multiples of sin x and cos x without any extra work.
由于求导是线性的，常数可以直接穿过算子。这意味着两个标准结果可以立即推广到 sin x 与 cos x 的和以及它们的常数倍，而不需要做额外的工作。
x, because the constant −2 multiplies the derivative −sin x, giving +2 sin x. Careless sign handling here costs easy marks.
例如，若 y = 5 sin x，则 dy/dx = 5 cos x。若 y = −2 cos x，则 dy/dx = 2 sin x，因为常数 −2 乘以导数 −sin x，得到 +2 sin x。这里的符号处理不慎会白白丢分。
y = 5 sin x → dy/dx = 5 cos x
y = −2 cos x → dy/dx = 2 sin x
Each term is differentiated separately and the coefficients are carried through unchanged.
同样的思路适用于更长的组合。若 y = 3 sin x + 4 cos x，则 dy/dx = 3 cos x − 4 sin x。每一项分别求导，系数原封不动地保留。
| Function | 函数 | Derivative | 导数 |
| y = 5 sin x | dy/dx = 5 cos x |
| y = −2 cos x | dy/dx = 2 sin x |
| y = 3 sin x + 4 cos x | dy/dx = 3 cos x − 4 sin x |
| y = 7 − 6 cos x | dy/dx = 6 sin x |
6. The Chain Rule with Composite Trigonometric Functions | 复合三角函数的链式法则
Most exam questions do not use plain sin x or cos x. They use composite arguments such as sin 3x, cos(2x − π/4) or sin(x²), and here the chain rule is unavoidable. The rule is the same as for any other function: differentiate the outer function, then multiply by the derivative of the inner function.
大多数考题并不使用简单的 sin x 或 cos x，而是使用复合变量，例如 sin 3x、cos(2x − π/4) 或 sin(x²)，这时链式法则不可或缺。规则与其他函数完全相同：先对外层函数求导，再乘以内层函数的导数。
d/dx [ sin(ax + b) ] = a cos(ax + b)
d/dx [ cos(ax + b) ] = −a sin(ax + b)
- y = sin 3x: dy/dx = 3 cos 3x, because the inner derivative is 3.
y = sin 3x：dy/dx = 3 cos 3x，因为内层导数为 3。 - y = cos(2x − π/4): dy/dx = −2 sin(2x − π/4).
y = cos(2x − π/4)：dy/dx = −2 sin(2x − π/4)。 - y = sin(x²): dy/dx = 2x cos(x²), because the inner derivative is 2x.
y = sin(x²)：dy/dx = 2x cos(x²)，因为内层导数为 2x。 - y = cos³x, meaning (cos x)³: dy/dx = 3 cos²x × (−sin x) = −3 cos²x sin x.
y = cos³x，即 (cos x)³：dy/dx = 3 cos²x × (−sin x) = −3 cos²x sin x。 - y = sin²(3x): dy/dx = 2 sin 3x × 3 cos 3x = 6 sin 3x cos 3x = 3 sin 6x.
y = sin²(3x)：dy/dx = 2 sin 3x × 3 cos 3x = 6 sin 3x cos 3x = 3 sin 6x。
Note the notation trap. sin²x means (sin x)², so it is the square of the sine - not the sine of x². The derivative of sin²x is 2 sin x cos x, which equals sin 2x, whereas the derivative of sin(x²) is 2x cos(x²). Reading the bracket carefully is essential.
注意符号陷阱。sin²x 表示 (sin x)²，即正弦的平方，而不是 x² 的正弦。sin²x 的导数是 2 sin x cos x，即 sin 2x；而 sin(x²) 的导数是 2x cos(x²)。仔细读括号至关重要。
7. Product Rule and Quotient Rule | 乘积法则与商法则
When sin x or cos x is multiplied or divided by another function, you need the product rule or the quotient rule. It is worth checking whether a chain-rule substitution or a trigonometric identity would be quicker before committing to a long algebraic expansion.
当 sin x 或 cos x 与另一个函数相乘或相除时，就需要用到乘积法则或商法则。在开始冗长的代数展开之前，值得先想一想是否存在更快的链式法则代换或三角恒等式。
Product rule: if y = u·v, then dy/dx = u·dv/dx + v·du/dx
Quotient rule: if y = u/v, then dy/dx = ( v·du/dx − u·dv/dx ) / v²
- y = x sin x: dy/dx = sin x + x cos x, with the second term carrying the cosine.
y = x sin x：dy/dx = sin x + x cos x，第二项带有余弦。 - y = x² cos x: dy/dx = 2x cos x − x² sin x, keeping the negative sign in the second term.
y = x² cos x：dy/dx = 2x cos x − x² sin x，第二项保留负号。 - y = x cos x: dy/dx = cos x − x sin x.
y = x cos x：dy/dx = cos x − x sin x。 - y = sin x / x: dy/dx = ( x cos x − sin x ) / x², with the square in the denominator.
y = sin x / x：dy/dx = ( x cos x − sin x ) / x²，分母为平方。 - y = sin x cos x: dy/dx = cos²x − sin²x = cos 2x, a result that also follows from writing y = ½ sin 2x.
y = sin x cos x：dy/dx = cos²x − sin²x = cos 2x，若先写成 y = ½ sin 2x 也能得到同样结果。
The quotient rule is also the standard route to the derivative of tan x, discussed in the next section. A frequent error is to reverse the order of the two terms in the numerator of the quotient rule; remember that the term with the derivative of the numerator comes first and carries a minus sign in front of the second term.
商法则也是求 tan x 导数的标准途径，将在下一节讨论。一个常见错误是把商法则分子中两项的顺序颠倒；请记住，含分子导数的那一项在前，第二项前面带负号。
8. Extending to tan x and Other Consequences | 推广到 tan x 及其它推论
Once the derivatives of sine and cosine
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