📚 Derivatives of sin x and cos x: Derivation and Applications | sin x 与 cos x 的导数推导与应用
The derivatives of sine and cosine are foundational in calculus. They appear in physics, engineering, and every branch of applied mathematics. In this article, we will derive these derivatives from first principles using the limit definition, explore the key trigonometric limits involved, and then apply the results to chain rule problems, second derivatives, and real-world models.
正弦函数和余弦函数的导数是微积分的基石。它们在物理、工程以及应用数学的各个分支中频繁出现。本文将利用极限定义从第一性原理推导这两个导数，探讨所涉及的关键三角极限，随后将结果应用于链式法则、二阶导数以及真实世界模型。
1. The Limit Definition of a Derivative | 导数的极限定义
For any function f(x), the derivative f'(x) is defined as the limit of the difference quotient as h approaches 0:
对于任意函数 f(x)，导数 f'(x) 定义为差商在 h 趋近于 0 时的极限：
f'(x) = limh→0 [f(x+h) − f(x)] / h
This definition gives the instantaneous rate of change of f at x. To differentiate sin x and cos x, we substitute them directly into this formula and evaluate the resulting limits.
这个定义给出了 f 在 x 处的瞬时变化率。为了对 sin x 和 cos x 求导，我们直接将它们代入此公式，并计算所得极限。
2. Deriving the Derivative of sin x | 推导 sin x 的导数
Let f(x) = sin x. Using the definition:
设 f(x) = sin x。利用定义：
sin′(x) = limh→0 [sin(x+h) − sin x] / h
Apply the sine addition formula: sin(x+h) = sin x cos h + cos x sin h. Then the numerator becomes sin x (cos h − 1) + cos x sin h. Rearranging:
应用正弦加法公式：sin(x+h) = sin x cos h + cos x sin h。则分子变为 sin x (cos h − 1) + cos x sin h。整理得：
sin′(x) = sin x · limh→0 (cos h − 1)/h + cos x · limh→0 (sin h)/h
We now need two well-known limits. If we already know that limh→0 (sin h)/h = 1 and limh→0 (cos h − 1)/h = 0, then sin′(x) = sin x · 0 + cos x · 1 = cos x.
现在我们需要两个著名极限。若已知 limh→0 (sin h)/h = 1 和 limh→0 (cos h − 1)/h = 0，则 sin′(x) = sin x · 0 + cos x · 1 = cos x。
d/dx (sin x) = cos x
3. Deriving the Derivative of cos x | 推导 cos x 的导数
Let f(x) = cos x. The limit definition gives:
设 f(x) = cos x。极限定义给出：
cos′(x) = limh→0 [cos(x+h) − cos x] / h
Use the cosine addition formula: cos(x+h) = cos x cos h − sin x sin h. Then the numerator is cos x (cos h − 1) − sin x sin h. Therefore:
使用余弦加法公式：cos(x+h) = cos x cos h − sin x sin h。则分子为 cos x (cos h − 1) − sin x sin h。因此：
cos′(x) = cos x · limh→0 (cos h − 1)/h − sin x · limh→0 (sin h)/h
Substituting the same two limits gives cos′(x) = cos x · 0 − sin x · 1 = −sin x.
代入同样的两个极限，得到 cos′(x) = cos x · 0 − sin x · 1 = −sin x。
d/dx (cos x) = −sin x
4. Proving the Key Trigonometric Limits | 证明关键三角极限
The two limits above are not given by chance. They are consequences of the geometry of the unit circle. Let θ be measured in radians. Consider the unit circle and the areas of a sector and two triangles.
上述两个极限并非凭空产生。它们是单位圆几何性质的推论。设 θ 以弧度为单位。考虑单位圆中一个扇形和两个三角形的面积。
For θ > 0 small, the area of the sector is θ/2. The area of the smaller triangle is (sin θ cos θ)/2, and the area of the larger triangle is (tan θ)/2. Comparing areas gives:
对于较小的 θ > 0，扇形面积为 θ/2。较小三角形面积为 (sin θ cos θ)/2，较大三角形面积为 (tan θ)/2。比较面积得：
sin θ cos θ ≤ θ ≤ tan θ
Dividing by sin θ > 0 and taking reciprocals, then using the squeeze theorem, yields limθ→0 (sin θ)/θ = 1.
除以 sin θ > 0 并取倒数，再利用夹逼定理，得到 limθ→0 (sin θ)/θ = 1。
For the second limit, multiply numerator and denominator by (cos θ + 1):
对于第二个极限，将分子分母同乘以 (cos θ + 1)：
(cos θ − 1)/θ = −(sin² θ) / [θ(cos θ + 1)] = −(sin θ/θ) · sin θ / (cos θ + 1)
As θ → 0, the first factor tends to 1 and the second factor tends to 0/2 = 0, so the whole limit is 0.
当 θ → 0 时，第一个因子趋于 1，第二个因子趋于 0/2 = 0，因此整个极限为 0。
5. Geometric Intuition: The Unit Circle | 几何直观：单位圆
Visually, the derivative of sin x being cos x means that the slope of the sine curve at any point equals the cosine of that point. At x = 0, the sine curve is rising steeply, and cos 0 = 1. At x = π/2, the slope is 0, matching cos (π/2) = 0.
从视觉上看，sin x 的导数为 cos x 意味着正弦曲线上任意一点的斜率等于该点的余弦值。在 x = 0 处，正弦曲线陡峭上升，而 cos 0 = 1。在 x = π/2 处，斜率为 0，与 cos (π/2) = 0 相符。
Similarly, the derivative of cos x is −sin x. At x = 0, the cosine curve has a horizontal tangent, so the slope is 0, and −sin 0 = 0. Near x = π/2, the cosine curve is falling, and −sin (π/2) = −1.
类似地，cos x 的导数为 −sin x。在 x = 0 处，余弦曲线有水平切线，所以斜率为 0，而 −sin 0 = 0。在 x = π/2 附近，余弦曲线下降，而 −sin (π/2) = −1。
Remember that radian measure is essential. If x were in degrees, these simple formulas would acquire extra constants such as π/180.
请记住弧度制是必需的。如果 x 以度为单位，这些简单公式将带有额外的常数，例如 π/180。
6. Application: Chain Rule | 应用：链式法则
When the sine or cosine function is applied to a function u(x), the chain rule gives:
当正弦或余弦函数作用于某个函数 u(x) 时，链式法则给出：
d/dx [sin u] = cos u · u′ , d/dx [cos u] = −sin u · u′
For example, differentiate y = sin (3x² + 2). Here u = 3x² + 2, so u′ = 6x. Thus:
例如，求 y = sin (3x² + 2) 的导数。这里 u = 3x² + 2，因此 u′ = 6x。于是：
dy/dx = 6x cos (3x² + 2)
Another example: y = cos (eˣ). Let u = eˣ, so u′ = eˣ. Hence dy/dx = −eˣ sin (eˣ).
另一个例子：y = cos (eˣ)。设 u = eˣ，所以 u′ = eˣ。因此 dy/dx = −eˣ sin (eˣ)。
7. Application: Second Derivatives and Oscillators | 应用：二阶导数与振荡器
Taking derivatives repeatedly reveals a beautiful cycle:
反复求导会呈现一个优美循环：
-
y = sin x → y′ = cos x → y″ = −sin x → y‴ = −cos x → y⁗ = sin x
y = sin x → y′ = cos x → y″ = −sin x → y‴ = −cos x → y⁗ = sin x
Thus the second derivative of sin x is −sin x, and the same holds for cos x. Both functions satisfy the differential equation y″ = −y. This equation is the mathematical heart of simple harmonic motion.
因此 sin x 的二阶导数为 −sin x，cos x 同样如此。两个函数都满足微分方程 y″ = −y。这个方程是简谐运动的数学核心。
8. Application: Simple Harmonic Motion in Physics | 应用：物理中的简谐运动
In mechanics, a particle moving on a spring with no friction has position x(t) satisfying x″ = −ω² x. The solutions are x(t) = A sin (ωt + φ) or x(t) = A cos (ωt + φ).
在力学中，无摩擦弹簧上质点的位置 x(t) 满足 x″ = −ω² x。其解为 x(t) = A sin (ωt + φ) 或 x(t) = A cos (ωt + φ)。
Using the chain rule, the velocity is:
利用链式法则，速度为：
v(t) = x′(t) = Aω cos (ωt + φ) or v(t) = −Aω sin (ωt + φ)
The acceleration is then x″ = −ω²x, which confirms the harmonic nature. This shows why trigonometric derivatives are central to oscillatory systems.
加速度则为 x″ = −ω²x，这证实了谐振的本质。这展示了三角导数为何成为振荡系统的核心。
9. Application: Related Rates and Triangles | 应用：相关变化率与三角形
Consider a ladder sliding down a wall. If the angle θ between the ladder and the floor changes at a known rate, the height of the top of the ladder can be expressed as h = L sin θ. Differentiating with respect to time:
考虑一把梯子从墙上滑落。如果梯子与地面之间的夹角 θ 以已知速率变化，梯子顶端的高度可表示为 h = L sin θ。对时间求导：
dh/dt = L cos θ · dθ/dt
This application of the derivative of sin x relates the angular speed to the vertical speed. Similar calculations appear in navigation, robotics, and radar tracking.
这个 sin x 导数的应用将角速度与竖直速度联系起来。类似的计算出现在导航、机器人和雷达追踪中。
10. Common Mistakes and Study Tips | 常见错误与学习技巧
| Mistake | 错误 | Correction | 纠正 |
|---|---|
| Forgetting the negative sign in cos′ x | Always write d/dx (cos x) = −sin x |
| Using degree mode in calculus | Use radians; otherwise multiply by π/180 |
| Missing the inner derivative in chain rule | Multiply by u′ whenever differentiating sin u or cos u |
To memorize the derivatives, say: “sine goes to cosine, cosine goes to negative sine.” Practice differentiating nested expressions to build fluency.
为记忆这些导数，可以默念：“正弦变余弦，余弦变负正弦。”通过练习复合函数求导来提升熟练度。
11. Derivation Using the Complex Exponential | 利用复数指数推导
For advanced students, Euler’s formula eⁱˣ = cos x + i sin x provides a compact alternative. Differentiating both sides with respect to x gives i eⁱˣ = cos′ x + i sin′ x. But i eⁱˣ = i cos x − sin x, so equating real and imaginary parts:
对于进阶学生，欧拉公式 eⁱˣ = cos x + i sin x 提供了紧凑的替代方法。两边对 x 求导得 i eⁱˣ = cos′ x + i sin′ x。而 i eⁱˣ = i cos x − sin x，因此比较实部和虚部：
cos′ x = −sin x , sin′ x = cos x
This method is elegant but relies on complex analysis. The limit derivation remains the primary proof in introductory calculus.
这种方法很优雅，但依赖于复分析。在入门微积分中，极限推导仍是主要证明。
12. Summary | 总结
We have derived the derivatives of sin x and cos x using the limit definition, proved the essential trigonometric limits, and explored applications including the chain rule, second derivatives, and simple harmonic motion.
我们利用极限定义推导了 sin x 和 cos x 的导数，证明了必要的三角极限，并探讨了链式法则、二阶导数以及简谐运动等应用。
d/dx (sin x) = cos x , d/dx (cos x) = −sin x
These two simple formulas unlock a vast range of calculus problems. Understanding their derivation ensures you can handle them confidently in any examination.
这两个简单公式解锁了广泛的微积分问题。理解其推导过程，能确保你在任何考试中自信地驾驭它们。
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