Unit 6: Trigonometric Identities and Equations – Section 9C Homework | 第六单元：三角恒等式与方程——9C部分作业

📚 Unit 6: Trigonometric Identities and Equations – Section 9C Homework | 第六单元：三角恒等式与方程——9C部分作业

In this revision article, we review the essential topics from Unit 6, pages 123–142, focusing on trigonometric identities and equations. The content is designed to help you complete Section 9C homework with confidence. We will cover the fundamental identities, compound angle formulae, and techniques for solving trigonometric equations.

在这篇复习文章中，我们回顾第六单元第123–142页的核心知识，重点讨论三角恒等式与方程。文章旨在帮助你自信地完成9C部分作业。我们将涵盖基本恒等式、复合角公式以及解三角方程的方法。

1. Basic Trigonometric Ratios | 基本三角比

enuse, and tan θ = opposite/adjacent. These ratios are the foundation of all trigonometry.

对于含角θ的直角三角形，三个基本三角比定义为：sin θ = 对边/斜边，cos θ = 邻边/斜边，tan θ = 对边/邻边。这些比值是全部三角学的基础。

sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent

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2. Pythagorean Identities | 毕达哥拉斯恒等式

nd 1 + cot²θ = csc²θ. These are valid for all values of θ where the functions are defined.

最重要的恒等式是 sin²θ + cos²θ = 1。由此可导出另外两个恒等式：1 + tan²θ = sec²θ 和 1 + cot²θ = csc²θ。它们在函数有定义的一切θ值下均成立。

sin²θ + cos²θ = 1

1 + tan²θ = sec²θ

1 + cot²θ = csc²θ

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3. Compound Angle Formulae | 复合角公式

he compound angle formulae are: sin(A ± B) = sinA cosB ± cosA sinB; cos(A ± B) = cosA cosB ∓ sinA sinB; tan(A ± B) = (tanA ± tanB)/(1 ∓ tanA tanB).

对于任意角A和B，复合角公式为：sin(A ± B) = sinA cosB ± cosA sinB；cos(A ± B) = cosA cosB ∓ sinA sinB；tan(A ± B) = (tanA ± tanB)/(1 ∓ tanA tanB)。

sin(A ± B) = sinA cosB ± cosA sinB

cos(A ± B) = cosA cosB ∓ sinA sinB

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

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4. Double Angle Formulae | 二倍角公式

2sinθ cosθ; cos2θ = cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ; tan2θ = (2tanθ)/(1 − tan²θ).

在复合角公式中令 A = B，可得到二倍角公式：sin2θ = 2sinθ cosθ；cos2θ = cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ；tan2θ = (2tanθ)/(1 − tan²θ)。

sin2θ = 2sinθ cosθ

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

tan2θ = (2tanθ)/(1 − tan²θ)

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5. Solving Linear Trigonometric Equations | 一次三角方程

terval are θ = 30° and θ = 150°, since sine is positive in the first and second quadrants.

一次三角方程的形式为 a sin θ + b = c 或类似。例如，解方程 2 sin θ − 1 = 0，其中 0° ≤ θ ≤ 360°。整理得 sin θ = 0.5。在给定区间内，解为 θ = 30° 和 θ = 150°，因为正弦在第一、第二象限为正。

2 sin θ − 1 = 0 ⇒ sin θ = 0.5 ⇒ θ = 30°, 150°

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6. Solving Quadratic Trigonometric Equations | 二次三角方程

θ − 1) = 0, so sin θ = −0.5 or sin θ = 1. The solutions are θ = 7π/6, 11π/6, and θ = π/2.

关于 sin θ 或 cos θ 的二次方程通常可以因式分解。例如，解方程 2 sin²θ − sin θ − 1 = 0，其中 0 ≤ θ ≤ 2π。因式分解为 (2 sin θ + 1)(sin θ − 1) = 0，因此 sin θ = −0.5 或 sin θ = 1。解为 θ = 7π/6、11π/6 和 θ = π/2。

(2 sin θ + 1)(sin θ − 1) = 0 ⇒ sin θ = −1/2 or sin θ = 1 ⇒ θ = 7π/6, 11π/6, π/2

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7. Using Identities to Simplify Expressions | 用恒等式化简

comes sin²θ / sinθ = sinθ. We must note that sinθ ≠ 0 for this simplification to be valid.

恒等式用于化简表达式和证明等式。例如，化简 (1 − cos²θ)/sinθ。因为 1 − cos²θ = sin²θ，所以该表达式变为 sin²θ / sinθ = sinθ。注意，只有 sinθ ≠ 0 时该化简才有效。

(1 − cos²θ)/sinθ = sin²θ/sinθ = sinθ, sinθ ≠ 0

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8. Worked Example: Section 9C Homework | 例题解析：9C部分作业

Let' n sin(A + B) = sinA cosB + cosA sinB = (3/5)(5/13) + (4/5)(12/13) = 15/65 + 48/65 = 63/65.

我们将这些方法应用于一个典型的9C部分作业题：已知 sin A = 3/5，cos B = 5/13，其中 A、B 为锐角，求 sin(A + B)。首先，利用毕达哥拉斯恒等式求得 cos A = 4/5，sin B = 12/13。因此 sin(A + B) = sinA cosB + cosA sinB = (3/5)(5/13) + (4/5)(12/13) = 15/65 + 48/65 = 63/65。

sin(A + B) = 63/65

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9. Common Mistakes to Avoid | 常见错误

ion that can be zero, such as cos θ, without considering that cos θ = 0 may be a solution.

解三角方程时，务必检查定义域。因象限判断错误而漏解是常见错误。另外，避免随意除以可能为零的函数（如 cos θ），除非先考虑 cos θ = 0 是否为解。

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10. Practice Questions | 练习

θ + 1 = 0 for 0 ≤ θ ≤ 2π. 3. Find the exact value of sin 75° using compound angle formula.

请自行尝试以下练习：1. 证明 (1 + tan²θ) cos²θ = 1。2. 解方程 2 cos 2θ + 3 cos θ + 1 = 0，其中 0 ≤ θ ≤ 2π。3. 利用复合角公式求 sin 75° 的精确值。

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