Exercise 6H – Sec, Cosec and Cot Functions | 习题6H：正割、余割与余切函数

📚 Exercise 6H – Sec, Cosec and Cot Functions | 习题6H：正割、余割与余切函数

Exercise 6H in the AQA A-Level Pure Mathematics Year 2 textbook closes the chapter on trigonometric functions. It expects you to combine the three new functions – secant (sec), cosecant (cosec) and cotangent (cot) – with algebraic manipulation, identities and equation solving. This revision guide explains the essential theory and works through the style of questions you will encounter in that exercise.

在 AQA A-Level 纯数学 Year 2 教材中，习题 6H 是三角函数章节的收尾练习。它要求你能够将新引入的三个函数——正割（sec）、余割（cosec）和余切（cot）——与代数变形、恒等式及方程求解结合起来。本篇复习指南将讲解核心理论，并带你逐步完成该练习中会出现的典型题型。

1. Definitions and Domains | 定义与定义域

The three new functions are defined as reciprocal trigonometric ratios. For an angle θ, we have:

三个新函数定义为三角比的倒数。对于角 θ，我们有：

sec θ = 1 / cos θ, cosec θ = 1 / sin θ, cot θ = 1 / tan θ = cos θ / sin θ

You must be careful about the values of θ for which these functions are defined. Since division by zero is undefined, sec θ is undefined when cos θ = 0, that is θ = (2k+1)π/2. Similarly, cosec θ is undefined when sin θ = 0, namely θ = kπ. Finally, cot θ is undefined when sin θ = 0, because cot θ = cos θ / sin θ.

你必须注意使这些函数有定义的 θ 值。因为分母不能为零，所以当 cos θ = 0 时，sec θ 无定义，即 θ = (2k+1)π/2。同理，当 sin θ = 0 时，cosec θ 无定义，即 θ = kπ。最后，cot θ 因写成 cos θ / sin θ，所以当 sin θ = 0 时也无定义。

2. Core Pythagorean Identities | 核心毕达哥拉斯恒等式

From the fundamental identity sin²θ + cos²θ = 1, we can derive two important results involving the new functions.

由基本恒等式 sin²θ + cos²θ = 1，我们可以推出两个涉及新函数的重要结论。

First, divide sin²θ + cos²θ = 1 by cos²θ. This gives tan²θ + 1 = sec²θ. Second, divide the same identity by sin²θ. This gives 1 + cot²θ = cosec²θ. These two identities are essential for simplifying expressions and proving other results.

首先，将 sin²θ + cos²θ = 1 两边同时除以 cos²θ，得到 tan²θ + 1 = sec²θ。其次，将同一个基本恒等式除以 sin²θ，得到 1 + cot²θ = cosec²θ。这两个恒等式在化简和证明中至关重要。

1 + tan²θ = sec²θ

1 + tan²θ = sec²θ
1 + cot²θ = cosec²θ

1 + cot²θ = cosec²θ

You also need to remember the quotient identity: tan θ = sin θ / cos θ, and hence cot θ = cos θ / sin θ.

你还需要记住商数恒等式：tan θ = sin θ / cos θ，因此 cot θ = cos θ / sin θ。

3. Graphs of Sec, Cosec and Cot | sec、cosec 和 cot 的图像

Understanding the graphs of these functions helps you solve equations and interpret inequalities.

理解这三个函数的图像有助于你解方程和判断不等式。

The graph of y = sec θ is the reciprocal of the cosine graph. It has vertical asymptotes where cos θ = 0, i.e. at θ = (2k+1)π/2. Its range is y ≤ -1 or y ≥ 1, and it is periodic with period 2π.

y = sec θ 的图像是余弦图像的倒数。它在 cos θ = 0 处有垂直渐近线，即在 θ = (2k+1)π/2 处。其值域为 y ≤ -1 或 y ≥ 1，周期为 2π。

Similarly, y = cosec θ is the reciprocal of the sine graph. It has vertical asymptotes at θ = kπ, its range is y ≤ -1 or y ≥ 1, and its period is 2π.

类似地，y = cosec θ 是正弦图像的倒数。它在 θ = kπ 有垂直渐近线，值域为 y ≤ -1 或 y ≥ 1，周期为 2π。

The graph of y = cot θ = cos θ / sin θ has asymptotes at θ = kπ. Unlike sec and cosec, its range is all real numbers, and its period is π. The graph decreases from +∞ to -∞ across each region between asymptotes.

y = cot θ = cos θ / sin θ 的图像在 θ = kπ 有渐近线。与 sec 和 cosec 不同，它的值域为全体实数，周期为 π。在每个渐近线之间的区间内，图像从 +∞ 递减到 -∞。

4. Inverse Trigonometric Functions | 反三角函数

Exercise 6H often involves inverse trigonometric functions. For a quantity x, the principal value arcs are used so that the inverse is a single-valued function.

习题 6H 经常涉及反三角函数。为了使反函数成为单值函数，我们使用主值范围。

For sin⁻¹ x (or arcsin x), the domain is -1 ≤ x ≤ 1 and the range is -π/2 ≤ y ≤ π/2. For cos⁻¹ x (or arccos x), the domain is -1 ≤ x ≤ 1 and the range is 0 ≤ y ≤ π. For tan⁻¹ x (or arctan x), the domain is all real numbers and the range is -π/2 < y < π/2.

对于 sin⁻¹ x（或 arcsin x），定义域为 -1 ≤ x ≤ 1，值域为 -π/2 ≤ y ≤ π/2。对于 cos⁻¹ x（或 arccos x），定义域为 -1 ≤ x ≤ 1，值域为 0 ≤ y ≤ π。对于 tan⁻¹ x（或 arctan x），定义域为全体实数，值域为 -π/2 < y < π/2。

Remember that these inverse functions are not the same as the reciprocals: sin⁻¹ x ≠ 1 / sin x. Exam questions may ask you to evaluate expressions such as sin⁻¹(1/2) or to simplify composite functions like tan(cos⁻¹ x).

请记住，反函数与倒数不是同一个概念：sin⁻¹ x ≠ 1 / sin x。考试题可能要求你计算 sin⁻¹(1/2) 的值，或者化简复合函数如 tan(cos⁻¹ x)。

5. Solving Equations with Sec, Cosec and Cot | 解含 sec、cosec 和 cot 的方程

Exercise 6H contains equations where the new functions appear. The standard strategy is to rewrite them in terms of sin and cos, or to use the Pythagorean identities to form a quadratic equation.

习题 6H 中包含含有新函数的方程。常规策略是将其改写为 sin 和 cos 的形式，或利用毕达哥拉斯恒等式构造二次方程。

Worked Example 1: Solve sec θ = 2 for 0 ≤ θ < 2π.

例 1：解方程 sec θ = 2，其中 0 ≤ θ < 2π。

Since sec θ = 1 / cos θ, the equation is equivalent to 1 / cos θ = 2, so cos θ = 1/2. The solutions in the given range are θ = π/3 and θ = 5π/3.

因为 sec θ = 1 / cos θ，原方程等价于 1 / cos θ = 2，所以 cos θ = 1/2。在给定范围内，解为 θ = π/3 和 θ = 5π/3。

Worked Example 2: Solve cot θ = -√3 for 0 ≤ θ < 2π.

例 2：解方程 cot θ = -√3，其中 0 ≤ θ < 2π。

cot θ = -√3 means cos θ / sin θ = -√3, which is equivalent to tan θ = -1/√3. The reference angle is π/6. Since tan is negative in the second and fourth quadrants, we obtain θ = π – π/6 = 5π/6 and θ = 2π – π/6 = 11π/6.

cot θ = -√3 即 cos θ / sin θ = -√3，等价于 tan θ = -1/√3。参考角为 π/6。因为 tan 在第二和第四象限为负，所以得到 θ = π – π/6 = 5π/6 和 θ = 2π – π/6 = 11π/6。

Worked Example 3: Solve 2 sec²θ + tan θ = 5 for 0 ≤ θ < 2π.

例 3：解方程 2 sec²θ + tan θ = 5，其中 0 ≤ θ < 2π。

Using 1 + tan²θ = sec²θ, the equation becomes 2(1 + tan²θ) + tan θ = 5. Simplify to 2 tan²θ + tan θ – 3 = 0. Factorise: (2 tan θ + 3)(tan θ – 1) = 0. Hence tan θ = -3/2 or tan θ = 1. Using a calculator, tan θ = -3/2 gives θ = 2.159 (to 3 d.p.) and θ = 5.301; tan θ = 1 gives θ = π/4 and θ = 5π/4.

利用 1 + tan²θ = sec²θ，原方程变为 2(1 + tan²θ) + tan θ = 5。化简得 2 tan²θ + tan θ – 3 = 0。因式分解：(2 tan θ + 3)(tan θ – 1) = 0。因此 tan θ = -3/2 或 tan θ = 1。用计算器计算，tan θ = -3/2 时 θ ≈ 2.159 和 5.301；tan θ = 1 时 θ = π/4 和 5π/4。

6. Proving Trig Identities | 证明三角恒等式

A common question type in Exercise 6H asks you to prove identities involving sec, cosec and cot. You should start from the more complicated side and use the identities from Section 2 to reduce it to the other side.

习题 6H 中的常见题型是证明含 sec、cosec 和 cot 的恒等式。你应从较复杂的一边开始，利用第 2 节的恒等式将其化为另一边。

Worked Example 4: Prove that (sec θ + tan θ)(sec θ – tan θ) = 1.

例 4：证明 (sec θ + tan θ)(sec θ – tan θ) = 1。

Expand the left-hand side: sec²θ – tan²θ. From 1 + tan²θ = sec²θ, we have sec²θ – tan²θ = 1. Hence the identity is proved.

展开左边：sec²θ – tan²θ。由 1 + tan²θ = sec²θ 可得 sec²θ – tan²θ = 1，故恒等式成立。

Worked Example 5: Prove that (1 + tan²θ) / (1 + cot²θ) = tan²θ.

例 5：证明 (1 + tan²θ) / (1 + cot²θ) = tan²θ。

Use the identities: 1 + tan²θ = sec²θ and 1 + cot²θ = cosec²θ. The left-hand side becomes sec²θ / cosec²θ = (1/cos²θ) / (1/sin²θ) = sin²θ / cos²θ = tan²θ.

利用恒等式：1 + tan²θ = sec²θ，1 + cot²θ = cosec²θ。左边变为 sec²θ / cosec²θ = (1/cos²θ) / (1/sin²θ) = sin²θ / cos²θ = tan²θ。

Worked Example 6: Prove that 1 / (1 + sin θ) + 1 / (1 – sin θ) = 2 sec²θ.

例 6：证明 1 / (1 + sin θ) + 1 / (1 – sin θ) = 2 sec²θ。

Combine the fractions on the left: ((1 – sin θ) + (1 + sin θ)) / ((1 + sin θ)(1 – sin θ)) = 2 / (1 – sin²θ) = 2 / cos²θ = 2 sec²θ.

将左边的分数合并：((1 – sin θ) + (1 + sin θ)) / ((1 + sin θ)(1 – sin θ)) = 2 / (1 – sin²θ) = 2 / cos²θ = 2 sec²θ。

7. Exam Pitfalls | 考试易错点

When working through Exercise 6H, be careful with the following common mistakes.

在完成习题 6H 时，请注意以下常见错误。

Forgetting that sec, cosec and cot are undefined for certain angles. Always check the domain before finalising your answer.

忘记 sec、cosec 和 cot 在某些角度处无定义。在确定最终答案前，务必检查定义域。
Confusing the inverse functions sin⁻¹ x with the reciprocal 1 / sin x. They are entirely different.

混淆反函数 sin⁻¹ x 与倒数 1 / sin x。它们是完全不同的概念。
When solving equations like tan θ = k, remembering to add multiples of π to find all solutions in the range, not just using the calculator’s principal value.

在解 tan θ = k 这类方程时，需加上 π 的整数倍以得到给定范围内的全部解，而不能只使用计算器显示的主值。
When proving identities, not working on both sides at the same time; it is better to transform one side until it equals the other.

证明恒等式时，不要同时处理两边；最好只变形一边，直到它等于另一边。

8. Practice Questions in the Style of Exercise 6H | 6H 风格练习

Try the following questions before checking the solutions. They are representative of the level in Exercise 6H.

先尝试完成以下题目，再核对答案。它们代表了习题 6H 的难度。

Question 1: Solve cosec θ = 2 for 0 ≤ θ < 2π.

题目 1：解方程 cosec θ = 2，其中 0 ≤ θ < 2π。

Question 2: Prove that (sec θ + cosec θ)² = sec²θ + 2 sec θ cosec θ + cosec²θ and then simplify the middle term.

题目 2：证明 (sec θ + cosec θ)² = sec²θ + 2 sec θ cosec θ + cosec²θ，并化简中间项。

Question 3: Evaluate sin⁻¹(√3 / 2) + tan⁻¹(1), giving your answer in terms of π.

题目 3：计算 sin⁻¹(√3 / 2) + tan⁻¹(1)，用 π 表示结果。

Question 4: Solve sec θ – 2 cos θ = 0 for 0 ≤ θ < 2π.

题目 4：解方程 sec θ – 2 cos θ = 0，其中 0 ≤ θ < 2π。

Solutions:

解答：

1. cosec θ = 2 ⇒ sin θ = 1/2 ⇒ θ = π/6 or 5π/6.

1. cosec θ = 2 ⇒ sin θ = 1/2 ⇒ θ = π/6 或 5π/6。

2. The expansion is standard. The middle term simplifies to 2 / (sin θ cos θ) = 4 cosec 2θ or 2 sec θ cosec θ.

2. 展开是标准的。中间项化简为 2 / (sin θ cos θ) = 4 cosec 2θ 或 2 sec θ cosec θ。

3. sin⁻¹(√3 / 2) = π/3 and tan⁻¹(1) = π/4, so the sum is 7π/12.

3. sin⁻¹(√3 / 2) = π/3，tan⁻¹(1) = π/4，所以和为 7π/12。

4. sec θ – 2 cos θ = 0 ⇒ 1/cos θ – 2 cos θ = 0 ⇒ 1 – 2 cos²θ = 0 ⇒ cos²θ = 1/2 ⇒ cos θ = ±1/√2. The solutions are θ = π/4, 3π/4, 5π/4, 7π/4.

4. sec θ – 2 cos θ = 0 ⇒ 1/cos θ – 2 cos θ = 0 ⇒ 1 – 2 cos²θ = 0 ⇒ cos²θ = 1/2 ⇒ cos θ = ±1/√2。解为 θ = π/4, 3π/4, 5π/4, 7π/4。

9. Integration and Differentiation Links | 与积分和微分的联系

Although Exercise 6H focuses on algebra and graphs, the same functions appear in later chapters on differentiation and integration. You may need to recall the derivatives: d/dx(sec x) = sec x tan x, d/dx(cosec x) = -cosec x cot x, d/dx(cot x) = -cosec²x. The corresponding integrals are also useful.

尽管习题 6H 侧重代数和图像，这些函数也会出现在后续的微分与积分章节。你需要记住导数公式：d/dx(sec x) = sec x tan x，d/dx(cosec x) = -cosec x cot x，d/dx(cot x) = -cosec²x。对应的积分公式同样重要。

AQA exam papers often ask you to integrate these functions, so mastering the identities now will build a strong foundation for later work.

AQA 考试卷经常要求你对这些函数进行积分，因此现在掌握这些恒等式将为后续学习打下坚实的基础。

10. Summary | 小结

Exercise 6H consolidates the key ideas of the trigonometric functions chapter. You should know the definitions of sec, cosec and cot, be able to sketch their graphs, use the Pythagorean identities to prove new identities, and solve equations involving these functions accurately.

习题 6H 巩固了三角函数章节的核心概念。你需要掌握 sec、cosec 和 cot 的定义，能够画出它们的图像，使用毕达哥拉斯恒等式证明新的恒等式，并准确求解含有这些函数的方程。

Practising algebraically from both sides of an identity, checking domains, and handling inverse functions will help you earn full marks in this topic. Good luck with your revision.

练习恒等式两边变形、检查定义域以及处理反函数，将帮助你在这一主题上获得满分。祝你复习顺利。

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Exercise 6H – Sec, Cosec and Cot Functions | 习题6H：正割、余割与余切函数