📚 Using sec x, cosec x and cot x | 使用 sec x、cosec x 和 cot x
Reciprocal trigonometric functions—secant, cosecant, and cotangent—extend your toolkit beyond sine, cosine, and tangent. In A-Level Mathematics, you will need to manipulate these functions fluently, sketch their graphs, simplify expressions, solve equations, and differentiate and integrate them. Mastering sec x, cosec x, and cot x is essential for success in the Edexcel Pure Mathematics syllabus.
倒数三角函数——正割、余割和余切——将你的工具包拓展到正弦、余弦和正切之外。在 A-Level 数学中，你需要熟练地操作这些函数，绘制它们的图像，化简表达式，解方程，并对它们进行微分和积分。掌握 sec x、cosec x 和 cot x 是 Edexcel 纯数学大纲成功的关键。
1. Definitions and Basic Properties | 定义与基本性质
tan x = cos x / sin x. These definitions hold for all real x except where the denominator is zero.
函数 sec x、cosec x 和 cot x 分别定义为余弦、正弦和正切的倒数：sec x = 1 / cos x，cosec x = 1 / sin x，cot x = 1 / tan x = cos x / sin x。这些定义在分母不为零的所有实数 x 上成立。
The domain of sec x excludes odd multiples of π/2: x ≠ (2n+1)π/2, n ∈ ℤ, because cos x = 0 there. Its range is (−∞, −1] ∪ [1, ∞). Sec x is an even function, periodic with period 2π.
sec x 的定义域排除 π/2 的奇数倍：x ≠ (2n+1)π/2, n ∈ ℤ，因为此时 cos x = 0。它的值域是 (−∞, −1] ∪ [1, ∞)。sec x 是偶函数，周期为 2π。
Similarly, cosec x has domain x ≠ nπ, with range (−∞, −1] ∪ [1, ∞). It is an odd function, also of period 2π. Cot x has domain x ≠ nπ, range ℝ, and is an odd function with period π.
类似地，cosec x 的定义域为 x ≠ nπ，值域为 (−∞, −1] ∪ [1, ∞)。它是奇函数，周期也为 2π。cot x 的定义域为 x ≠ nπ，值域为 ℝ，是奇函数，周期为 π。
2. Graphs of sec x, cosec x and cot x | sec x、cosec x 和 cot x 的图像
The graph of y = sec x is characterised by U‑ when sec x ≤ −1. Vertical asymptotes occur at x = π/2 + nπ. The curve never crosses the x‑axis and has minimum points at (2nπ, 1) and maximum points at ((2n+1)π, −1).
y = sec x 的图像由一系列 U 形分支组成，当 sec x ≥ 1 时开口向上，当 sec x ≤ −1 时开口向下。垂直渐近线位于 x = π/2 + nπ，曲线永不穿过 x 轴，极小值点出现在 (2nπ, 1)，极大值点在 ((2n+1)π, −1)。
sec x ≥ 1 and downwards for cosec x ≤ −1. Its minimum points are at (π/2 + 2nπ, 1) and maxima at (3π/2 + 2nπ, −1).
y = cosec x 的图像类似于 sec x 经 π/2 平移后的倒像。垂直渐近线在 x = nπ，当 cosec x ≥ 1 时开口向上，cosec x ≤ −1 时开口向下。极小值点在 (π/2 + 2nπ, 1)，极大值点在 (3π/2 + 2nπ, −1)。
y to negative infinity between consecutive vertical asymptotes at x = nπ. It crosses the x‑axis at odd multiples of π/2 and has period π. There are no maximum or minimum points.
对于 y = cot x，图像在相邻的垂直渐近线 x = nπ 之间从正无穷递减到负无穷。它在 π/2 的奇数倍处穿过 x 轴，周期为 π，没有极大或极小值点。
A summary table can help recall key features:
下面表格有助记忆关键特征：
| Function | Domain | Range | Period | Asymptotes |
|---|---|---|---|---|
| sec x | x ≠ π/2 + nπ | (−∞, −1] ∪ [1, ∞) | 2π | x = π/2 + nπ |
| cosec x | x ≠ nπ | (−∞, −1] ∪ [1, ∞) | 2π | x = nπ |
| cot x | x ≠ nπ | ℝ | π | x = nπ |
3. Simplifying Expressions | 化简表达式
A common task is to rewrite expressions involving sec, cosec and cot in terms of sin and cos, then apply identities. For instance, simplify (sin x)(cot x + tan x):
常见任务是将涉及 sec、cosec 和 cot 的表达式重写为 sin 和 cos 形式，然后应用恒等式。例如，化简 (sin x)(cot x + tan x)：
= sin x * [(cos² x + sin² x)/(sin x cos x)] = sin x * [1/(sin x cos x)] = 1/cos x = sec x.
将 cot x 写为 cos x / sin x，tan x 写为 sin x / cos x。则 = sin x * [(cos² x + sin² x)/(sin x cos x)] = sin x * [1/(sin x cos x)] = 1/cos x = sec x。
ec² x gives immediately 1. Always look for opportunities to use the squared relationships.
另一个例子：化简 (cosec² x − cot² x)。使用毕达哥拉斯恒等式 1 + cot² x = cosec² x 直接得到 1。始终寻找使用平方关系的机会。
4. Proving Trigonometric Identities | 证明三角恒等式
ing to sin and cos and factoring. Prove that (cosec x − cot x)² = (1 − cos x)/(1 + cos x).
证明恒等式通常需要转换为 sin 和 cos 并进行因式分解。证明 (cosec x − cot x)² = (1 − cos x)/(1 + cos x)。
ction simplifies to (1 − cos x)²/[(1 − cos x)(1 + cos x)] = (1 − cos x)/(1 + cos x) = RHS.
左边：(1/sin x − cos x/sin x)² = ((1 − cos x)/sin x)² = (1 − cos x)²/sin² x。因为 sin² x = 1 − cos² x = (1 − cos x)(1 + cos x)，分式化简为 (1 − cos x)²/[(1 − cos x)(1 + cos x)] = (1 − cos x)/(1 + cos x) = 右边。
in x cos x) = cosec x sec x. Always state the domain restrictions when proving identities.
另一个常见的证明：证明 cot x + tan x = cosec x sec x。左边： x/sin x + sin x/cos x = (cos² x + sin² x)/(sin x cos x) = 1/(sin x cos x) = cosec x sec x。在证明恒等式时，始终注明定义域限制。
5. Solving Equations Involving sec, cosec, cot | 解含 sec、cosec、cot 的方程
using identities to form a quadratic. Example: Solve 2 sec² x = 5 + tan x for 0 ≤ x < 2π.
含倒数三角函数的方程可通过重写为 tan、sin 或 cos 形式，然后利用恒等式构造二次方程来求解。例如：在 0 ≤ x < 2π 内解 2 sec² x = 5 + tan x。
x − tan x − 3 = 0. Factorise: (2 tan x − 3)(tan x + 1) = 0. So tan x = 3/2 or tan x = −1.
使用 sec² x = 1 + tan² x：2(1 + tan² x) = 5 + tan x → 2 + 2 tan² x = 5 + tan x → 2 tan² x − tan x − 3 = 0。因式分解：(2 tan x − 3)(tan x + 1) = 0，所以 tan x = 3/2 或 tan x = −1。
n x = 3/2, the principal value is x = tan⁻¹(3/2) ≈ 0.9828 rad. The general solution is x =
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