Inverse Trigonometric Functions | 反三角函数

📚 Inverse Trigonometric Functions | 反三角函数

Inverse trigonometric functions are the inverses of the trigonometric functions, defined by restricting the domain of the original functions to intervals where they are one-to-one. They are essential tools in calculus for integration, differentiation, and for solving equations involving trigonometric expressions. In IB Mathematics, a solid understanding of arcsin, arccos, and arctan – along with their derivatives, integrals, and key identities – is required for both the Analysis & Approaches and Applications & Interpretation courses at Higher Level. This article covers everything you need to master inverse trigonometric functions, with clear bilingual explanations and worked examples throughout.

反三角函数是三角函数的反函数，通过将原函数的定义域限制在使其一一对应的区间来定义。它们在微积分中是必不可少的工具，用于积分、微分以及解包含三角表达式的方程。在IB数学中，深刻理解arcsin、arccos和arctan——连同它们的导数、积分和重要恒等式——是分析与方法和应用与解释高水平课程的要求。本文涵盖掌握反三角函数所需的一切内容，并提供清晰的双语解释和示例。

1. What Are Inverse Trigonometric Functions? | 什么是反三角函数？

Trigonometric functions are periodic, so they are not injective over their entire real domains. To define an inverse, we restrict each function to an interval on which it is strictly monotonic. The inverse sine function, written as sin⁻¹ x or arcsin x, is defined by taking the sine function on [−π/2, π/2]. Similarly, cos⁻¹ x has domain [0, π] for the cosine function, and tan⁻¹ x uses (−π/2, π/2) for the tangent function. It is crucial to remember that sin⁻¹ x does not mean 1/sin x; the superscript '−1' denotes the inverse function, not a reciprocal.

三角函数是周期性的，因此它们在整个实数域上并非单射。为了定义反函数，我们将每个函数限制在其严格单调的区间上。反正弦函数记作 sin⁻¹ x 或 arcsin x，是将正弦函数限制在 [−π/2, π/2] 上定义的。类似地，cos⁻¹ x 的定义基于余弦函数在 [0, π] 上，而 tan⁻¹ x 则使用正切函数在 (−π/2, π/2) 上。必须牢记，sin⁻¹ x 不表示 1/sin x；上标 ‘−1’ 表示反函数，而非倒数。

2. Domain and Range of Inverse Trigonometric Functions | 反三角函数的定义域与值域

The domains and ranges of the three principal inverse trigonometric functions are strictly determined by the restrictions we place on the original trig functions. The table below summarises the standard specifications that you will encounter in IB examinations.

三个主要反三角函数的定义域和值域由我们对原三角函数施加的限制严格确定。下表总结了你在IB考试中会碰到的标准规格。

Function	Domain	Range (Principal Values)
sin⁻¹ x (arcsin x)	[−1, 1]	[−π/2, π/2]
cos⁻¹ x (arccos x)	[−1, 1]	[0, π]
tan⁻¹ x (arctan x)	ℝ (all real numbers)	(−π/2, π/2)

对于反正弦和反余弦，定义域限制在 [−1, 1] 是因为正弦和余弦的输出绝对不会超出这个范围。反正弦的值域为 [−π/2, π/2]，这意味着角度始终落在包含原点的递增分支上。反余弦的值域为 [0, π]，提供了一个递减的主值分支。反正切的定义域为所有实数，其值域为开放的 (−π/2, π/2)，以两条水平渐近线为界。

3. Graphs and Principal Values | 图像与主值

1, 0), and it is also bounded. The graph of y = tan⁻¹ x is S-shaped, with horizontal asymptotes y = −π/2 as x → −∞ and y = π/2 as x → ∞; it passes through (0,0) and is strictly increasing.

y = sin⁻¹ x 的图像是将 y = sin x 限制在 [−π/2, π/2] 上的部分关于直线 y = x 的反射。它通过原点，从 (−1, −π/2) 单调递增到 (1, π/2)，并在端点处有竖直切线。y = cos⁻¹ x 的图像是递减的，从 (−1, π) 开始到 (1, 0) 结束，且同样有界。y = tan⁻¹ x 的图像呈S形，有两条水平渐近线：当 x → −∞ 时 y = −π/2，当 x → ∞ 时 y = π/2；它通过 (0,0) 且严格递增。

ted interval. For instance, sin⁻¹(sin(5π/6)) = π/6, not 5π/6, because 5π/6 is not in [−π/2, π/2].

在计算诸如 sin⁻¹(sin θ) 或 tan⁻¹(tan θ) 的表达式时，主值的概念至关重要。由于反函数只能返回其受限值域内的值，只要原角度落在受限区间之外，结果就必须被调整到主值分支上。例如，sin⁻¹(sin(5π/6)) = π/6，而不是 5π/6，因为 5π/6 不在 [−π/2, π/2] 内。

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4. Derivatives of Inverse Trigonometric Functions | 反三角函数的导数

The derivatives of the inverse trigonometric functions are standard results that must be memorised for IB examinations. They are often applied together with the chain rule. The three fundamental derivative formulas are shown below.

反三角函数的导数是必须在IB考试中熟记的标准结论。它们经常与链式法则结合使用。下面给出三个基本的导数公式。

d/dx (sin⁻¹ x) = 1 / √1 − x², for |x| < 1

d/dx (cos⁻¹ x) = −1 / √1 − x², for |x| < 1

d/dx (tan⁻¹ x) = 1 / (1 + x²), for all real x

这些公式表明，反正弦和反余弦的导数只差一个符号，且两者的定义域均为开区间 (−1, 1)。反正切的导数是一有理函数，形式简洁，在整个实数范围内有效。当自变量是x的可微函数时，必须乘以该函数的导数（链式法则）。

For example, to differentiate f(x) = sin⁻¹(2x), we set u = 2x and apply the chain rule: f '(x) = [1 / √(1 − (2x)²)] · 2 = 2 / √1 − 4x², provided |2x| < 1, i.e., |x| < ½. A similar approach works for composite functions involving tan⁻¹ or cos⁻¹.

例如，为求 f(x) = sin⁻¹(2x) 的导数，令 u = 2x 并应用链式法则：f '(x) = [1 / √(1 − (2x)²)] · 2 = 2 / √1 − 4x²，前提是 |2x| < 1，即 |x| < ½。类似的方法适用于涉及 tan⁻¹ 或 cos⁻¹ 的复合函数。

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5. Proofs of the Derivative Formulas | 导数公式的证明

rote mistakes. To prove the derivative of y = sin⁻¹ x, start by writing sin y = x, where y ∈ 1 / cos y. Because y is in the principal range, cos y ≥ 0, and using the identity cos y = √1 − sin² y = √1 − x². Hence, dy/dx = 1 / √1 − x². The proof for cos⁻¹ x is analogous, but note that for y ∈ [0, π], sin y ≥ 0, giving sin y = √1 − x², and the implicit differentiation leads to −sin y · dy/dx = 1, yielding dy/dx = −1 / √1 − x².

理解如何推导这些公式可以加深你的概念把握，并有助于避免机械记忆带来的错误。为证明 y = sin⁻¹ x 的导数，首先写出 sin y = x，其中 y ∈ [−π/2, π/2]。对 x 进行隐函数求导：cos y · dy/dx = 1，于是 dy/dx = 1 / cos y。由于 y 在主值范围内，cos y ≥ 0，并利用恒等式 cos y = √1 − sin² y = √1 − x²。因此，dy/dx = 1 / √1 − x²。cos⁻¹ x 的证明类似，但需要注意，对于 y ∈ [0, π]，sin y ≥ 0，从而有 sin y = √1 − x²，隐函数求导给出 −sin y · dy/dx = 1，得到 dy/dx = −1 / √1 − x²。

The derivative of tan⁻¹ x can be obtained by setting y = tan⁻¹ x ⇔ tan y = x, with y ∈ ten appear in IB exam papers, sometimes asking for the derivation of one formula directly.

tan⁻¹ x 的导数可通过设 y = tan⁻¹ x ⇔ tan y = x，其中 y ∈ (−π/2, π/2) 来求得。求导：sec² y · dy/dx = 1。利用恒等式 sec² y = 1 + tan² y = 1 + x²，得到 dy/dx = 1 / (1 + x²)。这些证明经常出现在IB试卷中，有时会直接要求推导某个公式。

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6. Integrals Yielding Inverse Trigonometric Functions | 导致反三角函数的积分

A direct consequence of the derivative formulas is a set of standard integrals that evaluate to inverse trig functions. These integrals are indispensable in techniques such as trigonometric substitution. The following table lists the most commonly encountered forms.

导数公式的直接推论是一组结果为反三角函数的标准积分。这些积分在诸如三角代换等技巧中不可或缺。下表列出了最常见的形式。

Integral	Result	Condition
∫ 1 / √a² − x² dx	sin⁻¹(x / a) + C	a > 0
∫ −1 / √a² − x² dx	cos⁻¹(x / a) + C	a > 0
∫ 1 / (a² + x²) dx	(1/a) tan⁻¹(x / a) + C	a ≠ 0
∫ 1 / (x √x² − a²) dx	(1/a) sec⁻¹(x / a) + C	x > a > 0

注意，虽然 ∫ −1/√a²−x² dx 可以直接写成 cos⁻¹(x/a) + C，但在实际运算中，常利用恒等式将其与 sin⁻¹ 联系起来，因为微分后两者仅差一个常数。对于包含 √x² ± a² 的积分，完成平方后常常可以用反正切或反正弦形式表示结果。务必记住，这些公式中常数 a 必须为正，以确保根号内表达式有意义。

For instance, ∫ dx / √9 − x² 2) tan⁻¹(x/2) + C. In definite integrals, remember to apply the limits in radians within the principal range.

例如，∫ dx / √9 − x² = sin⁻¹(x/3) + C，而 ∫ dx / (4 + x²) = (1/2) tan⁻¹(x/2) + C。在定积分中，记得将主值范围内的弧度代入上下限。

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7. Important Identities Involving Inverse Trigonometric Functions | 涉及反三角函数的重要恒等式

solve equations. The complementary angle identity sin⁻¹ x + cos⁻¹ x = π/2 holds for all x ∈ is tan⁻¹ x + tan⁻¹(1/x) = π/2 for x > 0. Note that when x < 0, the sum equals −π/2.

若干恒等式有助于简化表达式和求解方程。互余角恒等式 sin⁻¹ x + cos⁻¹ x = π/2 对所有 x ∈ [−1, 1] 成立。这可以通过令 α = sin⁻¹ x 并观察到 cos(π/2 − α) = sin α = x，因此 π/2 − α = cos⁻¹ x 来证明。另一个有用的关系是，当 x > 0 时，tan⁻¹ x + tan⁻¹(1/x) = π/2。注意当 x < 0 时，其和为 −π/2。

erse trig functions to one another are also valuable. For instance, arcsin x = arctan( x / √1 − x² ) for |x| term, depending on the product ab. These extended identities are often examined in more challenging IB problems.

将不同反三角函数联系起来的恒等式也很有价值。例如，对于 |x| < 1，有 arcsin x = arctan( x / √1 − x² )。此外，还存在反函数的倍角和加法公式，例如 tan⁻¹ a + tan⁻¹ b = tan⁻¹((a + b) / (1 − ab)) 加上调整项，调整项取决于乘积 ab。这些拓展恒等式常在较难的IB考题中出现。

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8. Solving Equations with Inverse Trigonometric Functions | 解含有反三角函数的方程

x, then 2x = sin α and x = cos α. Squaring and adding gives (2x)² + x² = sin²α + cos²α = 1 ⇒ 5x² = 1 ⇒ x = ±1/√5. But we must check which solution lies within the domains: sin⁻¹(2x) requires |2x| ≤ 1, and for cos⁻¹ x, x ∈ [−1,1]. Here x = −1/√5 ive argument for sin⁻¹, but the equality may still hold; substituting back confirms x = 1/√5 is the correct solution.

诸如 sin⁻¹(2x) = cos⁻¹ x 或 tan⁻¹(x) + tan⁻¹(3x) = π/2 的方程，可以通过应用上述恒等式来求解，也可以在确保表达式落在合适范围内之后，对方程两边取正弦、余弦或正切。对于 sin⁻¹(2x) = cos⁻¹ x，设 α = sin⁻¹(2x) = cos⁻¹ x，则 2x = sin α 且 x = cos α。平方相加得 (2x)² + x² = sin²α + cos²α = 1 ⇒ 5x² = 1 ⇒ x = ±1/√5。但我们必须检查哪个解落在定义域内：sin⁻¹(2x) 要求 |2x| ≤ 1，而 cos⁻¹ x 要求 x ∈ [−1,1]。此处 x = −1/√5 给出 sin⁻¹ 的负参数，但等式仍可能成立；代回验证后确认 x = 1/√5 是正确的解。

(x)(3x)] = tan(π/2) which is undefined, implying the denominator must be zero: 1 − 3x² = 0 ⇒ x = ±1/√3. With careful sign checking, we find x = 1/√3 satisfies the positive sum equalling π/2. This method is highly effective but always requires verifying that the angles lie in the principal range.

当方程涉及反正切之和时，常用的技巧是对两边取正切。对于 tan⁻¹ x + tan⁻¹(3x) = π/2，两边取正切得 [x + 3x] / [1 − (x)(3x)] = tan(π/2)，这是无定义的，这意味着分母必须为零：1 − 3x² = 0 ⇒ x = ±1/√3。经过仔细的符号检验，我们发现 x = 1/√3 满足其正和为 π/2。该方法非常高效，但总是需要验证角度是否在主值范围内。

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9. Applications in Geometry and Calculus | 在几何与微积分中的应用

Inverse trigonometric functions appear in many physical situations, such as computing angles of elevation, phase shifts, and arc lengths. In calculus, they allow us to integrate rational functions involving quadratics that cannot be factored. For example, to find the area of a circular segment or to compute the arc length of a sine curve, one frequently encounters integrals that evaluate to an inverse trig function.

反三角函数出现在许多物理情形中，如计算仰角、相位偏移和弧长。在微积分中，它们使我们能够积分那些包含不可分解二次式的有理函数。例如，要求弓形面积或

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