Using Trigonometric Identities | 三角函数恒等式的运用

📚 Using Trigonometric Identities | 三角函数恒等式的运用

At A-Level, trigonometric identities are not just formulas to memorise. They are tools for rewriting expressions so that an equation becomes solvable, a proof becomes shorter, or an integral becomes possible. This guide covers the core identities and the strategic ways to use them.

在 A-Level 课程中，三角恒等式不只是需要记忆的公式，更是重写表达式的工具，使方程变得可解、证明变得更简洁，或使积分成为可能。本指南涵盖核心恒等式及其运用策略。

---

1. The Trigonometric Identity Toolkit | 三角函数恒等式工具箱

You should be able to recall the Pythagorean, reciprocal, quotient, compound angle, double angle and R-formula identities quickly. Edexcel provides many of these in the formulae booklet, but the exam rewards students who can select them without hesitation.

你应该能够快速回忆毕达哥拉斯恒等式、倒数恒等式、商数恒等式、复合角恒等式、二倍角恒等式和 R 公式。Edexcel 公式手册中提供了其中许多公式，但考试更青睐那些无需犹豫就能选出合适公式的学生。

The most useful identities can be grouped into a small toolkit. Always ask: can I replace one term with another to make the expression simpler?

最有用的恒等式可以归纳为一个小工具箱。始终问自己：我能否将某一项替换为另一项，使表达式更简单？

---

2. Pythagorean Identities and Their Rearrangements | 毕达哥拉斯恒等式及其变形

ntity sin²θ + cos²θ = 1. It is often used to replace one squared trig term with the other.

从基本恒等式 sin²θ + cos²θ = 1 出发。它常用于将一个平方三角项替换成另一个。

sin²θ + cos²θ = 1

Dividing the whole identity by cos²θ gives the tangent-secant form:

将整个恒等式除以 cos²θ，可得到正切-正割形式：

1 + tan²θ = sec²θ

Dividing by sin²θ gives the cotangent-cosecant form:

除以 sin²θ，可得到余切-余割形式：

1 + cot²θ = cosec²θ

cot²θ = cosec²θ − 1. These rearrangements are often the key step in proofs and equations.

记住这三种形式，包括变形版本，例如 tan²θ = sec²θ − 1 和 cot²θ = cosec²θ − 1。这些变形往往是证明题和方程题中的关键步骤。

---

3. Reciprocal and Quotient Identities | 倒数与商数恒等式

sθ is often the fastest way to turn a tangent expression into sines and cosines. The reciprocal identities connect cosecθ, secθ and cotθ to sinθ, cosθ and tanθ.

商数恒等式 tanθ = sinθ / cosθ 通常是把正切表达式转化为正弦与余弦的最快方法。倒数恒等式将 cosecθ、secθ、cotθ 与 sinθ、cosθ、tanθ 联系起来。

Identity / 恒等式	Equivalent form / 等价形式
tanθ = sinθ / cosθ	cotθ = cosθ / sinθ = 1 / tanθ
secθ = 1 / cosθ	cosecθ = 1 / sinθ

When proving identities, writing everything in terms of sinθ and cosθ often reveals cancellations that were hidden by tanθ, secθ or cotθ.

证明恒等式时，把一切写成 sinθ 和 cosθ 的形式，常常能揭示出被 tanθ、secθ 或 cotθ 掩盖的约分机会。

---

4. Compound Angle Identities | 复合角恒等式

Compound angle identities handle expressions such as sin(A + B) where the angle is a sum or difference. They are especially useful when exact angles like 15°, 75° or 105° appear, because those can be written as sums or differences of 30°, 45° and 60°.

复合角恒等式用于处理 sin(A + B) 这类角度为和或差的表达式。当出现 15°、75° 或 105° 等精确角时尤其有用，因为这些角可以写成 30°、45° 和 60° 的和或差。

sin(A ± B) = sinA cosB ± cosA sinB

cos(A ± B) = cosA cosB ∓ sinA sinB

tan(A ± B) = (tanA ± tanB) / (1 ∓ tanA tanB)

Notice the sign change in the cosine formula: cos(A + B) has a negative product term, while cos(A − B) has a positive product term.

注意余弦公式中的符号变化：cos(A + B) 的乘积项为负，而 cos(A − B) 的乘积项为正。

---

广告 / Sponsored

5. Double Angle Identities | 二倍角恒等式

e formulae with A = B = θ. They are extremely common in equations, proofs and integration.

二倍角恒等式是复合角公式在 A = B = θ 时的特例。它们在解方程、证明和积分中极为常见。

sin2θ = 2sinθ cosθ

cos2θ = cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ

tan2θ = 2tanθ / (1 − tan²θ)

The three forms of cos2θ are all the same. Choose the one that matches the other terms in the question: if you see sin²θ, use 1 − 2sin²θ; if you see cos²θ, use 2cos²θ − 1.

cos2θ 的三种形式完全相同。选择与题目中其他项匹配的那一种：如果看到 sin²θ，使用 1 − 2sin²θ；如果看到 cos²θ，使用 2cos²θ − 1。

---

6. Using Identities to Solve Equations | 用恒等式解方程

When an equation contains two different trig functions or mixed powers, use identities to rewrite it as a single function or as a quadratic in one function.

当方程含有两个不同的三角函数或混合幂时，使用恒等式将其改写为单一函数或某一函数的二次方程。

For example, consider:

例如，考虑：

3cos²θ − 2sinθ − 1 = 0

Replace cos²θ with 1 − sin²θ:

将 cos²θ 替换为 1 − sin²θ：

3(1 − sin²θ) − 2sinθ − 1 = 0

This simplifies to a quadratic in sinθ:

化简为关于 sinθ 的二次方程：

3sin²θ + 2sinθ − 2 = 0

Solving gives sinθ = (−1 + √7) / 3 ≈ 0.549 or sinθ = (−1 − √7) / 3 ≈ −1.215, which is outside the possible range for sine.

解得 sinθ = (−1 + √7) / 3 ≈ 0.549 或 sinθ = (−1 − √7) / 3 ≈ −1.215，后者超出了正弦的可能范围。

After solving the quadratic, use CAST or the sine graph to find all angles in the required interval.

解出二次方程后，使用 CAST 图或正弦图像求出给定区间内的所有角。

---

7. Proving Trigonometric Identities | 证明三角恒等式

For proof questions, start with the more complicated side and work towards the simpler side. Common moves include writing tanθ as sinθ / cosθ, using Pythagorean substitutions, and combining fractions over a common denominator.

对于证明题，从较复杂的一边开始，向较简单的一边推导。常用操作包括把 tanθ 写成 sinθ / cosθ、使用毕达哥拉斯替换，以及通分合并分式。

For example, to prove (sinθ + cosθ)² = 1 + sin2θ, expand the left side:

例如，要证明 (sinθ + cosθ)² = 1 + sin2θ，展开左边：

(sinθ + cosθ)² = sin²θ + 2sinθ cosθ + cos²θ = 1 + sin2θ

identity gives sin²θ + cos²θ = 1, and the double angle identity gives 2sinθ cosθ = sin2θ.

这里毕达哥拉斯恒等式给出 sin²θ + cos²θ = 1，二倍角恒等式给出 2sinθ cosθ = sin2θ。

---

8. The R-Formula and Transformations | R 公式与变形

ing maximum and minimum values and for solving equations with mixed sine and cosine terms.

形如 a sinθ ± b cosθ 的表达式可改写为 R sin(θ ± α) 或 R cos(θ ± α)。这对于求最大值和最小值以及解含有正余弦混合项的方程至关重要。

a sinθ + b cosθ = R sin(θ + α), R = √a² + b², tanα = b / a

a sinθ − b cosθ = R sin(θ − α), tanα = b / a

a cosθ + b sinθ = R cos(θ − α), tanα = b / a

a cosθ − b sinθ = R cos(θ + α), tanα = b / a

Since the amplitude is R, the expression has maximum value R and minimum value −R. You can find the angle at which the maximum or minimum

Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）