📚 Edexcel Maths Unit Test: Trigonometric Equations & Identities | 爱德思数学单元测试卷：三角方程与恒等式
This unit test focuses on trigonometric equations and identities, a core topic in Edexcel A Level Mathematics. You will find a selection of typical exam-style questions covering solving simple trig equations, proving identities, applying R-formulae, and sketching trigonometric graphs. Work through each question carefully and check your solutions against the worked answers provided. This revision resource mimics the structure of a timed unit test to help you identify any gaps in understanding before the real assessment.
本单元测试围绕三角函数方程与恒等式展开，这是爱德思 A Level 数学的核心主题。你将领略到一系列典型的考试风格题目，涵盖解简单三角方程、证明恒等式、应用 R 公式以及绘制三角函数图像。请认真完成每一道题，并对照详细解答进行核对。这份复习资料模拟限时单元测试的结构，帮助你在正式评估前发现理解上的薄弱环节。
1. Solving sin 2θ = ½ for 0° ≤ θ ≤ 360° | 求解 sin 2θ = ½ 在 0° 到 360° 范围内
Solve sin 2θ = ½, giving all solutions for θ in the interval 0° ≤ θ ≤ 360°.
解方程 sin 2θ = ½，求出 θ 在区间 0° 到 360° 内的所有解。
is positive in the first and second quadrants, the general solutions for 2θ are 30°, 180° – All four values lie within 0° ≤ θ ≤ 360°, so the solution set is θ = 15°, 75°, 195°, 255°.
首先，求出 sin 2θ = ½ 的参考角。主解为 2θ = 30°。由于正弦在第一和第二象限为正，2θ 的通解为 30°、180° – 30° = 150°，然后加上 360° 的整数倍：30° + 360° = 390°，150° + 360° = 510° 等。接着每项除以 2 得到 θ：15°、75°、195°、255°。这四个值均落在 0° ≤ θ ≤ 360° 内，因此解集为 θ = 15°, 75°, 195°, 255°。
2. Proving a Trigonometric Identity | 证明三角恒等式
Prove that (1 – cos²θ) / sin θ ≡ sin θ. State any restrictions on θ.
证明恒等式 (1 – cos²θ) / sin θ ≡ sin θ。说明 θ 的限制条件。
Use the Pythagorean identity sin²θ + cos²θ = 1, which implies 1 – holds for all θ except θ = nπ (or 180°n), where n is an integer. This completes the proof.
利用勾股恒等式 sin²θ + cos²θ = 1，可得 1 – cos²θ = sin²θ。代入左边：sin²θ / sin θ = sin θ，前提是 sin θ ≠ 0。因此，该恒等式对所有 θ 成立，但 θ ≠ nπ（即 180°n），n 为整数。证明完毕。
3. Solving cos(x + 30°) = √3/2 | 解 cos(x + 30°) = √3/2
Solve cos(x + 30°) = √3/2 for 0° ≤ x ≤ 360°.
解方程 cos(x + 30°) = √3/2，x 的区间为 0° 到 360°。
Recognise that √3/2 corresponds to cos 30°. Thus, x + 30° = ±30° + 360°k. Solve for x: x = 0° + 360°k or x = –60° + 360°k. Within the given range, k = 0 gives x = 0° and x = –60° (reject). k = 1 gives x = 360° (which is allowed as 0° ≤ x ≤ 360°) and x = 300°. k = –1 would give x = –420°, outside the range. Hence, the final solutions are x = 0°, 300°, and 360°.
识别 √3/2 对应 cos 30°。于是，x + 30° = ±30° + 360°k。解出 x：x = 0° + 360°k 或 x = –60° + 360°k。在给定范围内，k = 0 时 x = 0° 和 x = –60°（舍去）。k = 1 时得 x = 360°（允许，因为包含端点）和 x = 300°。k = –1 会得到 –420°，超出范围。因此，最终解为 x = 0°, 300°, 360°。
4. Solving tan²θ – 1 = 0 for 0 ≤ θ ≤ 2π | 解 tan²θ – 1 = 0，θ ∈ [0, 2π]
Find all solutions to tan²θ – 1 = 0 in the interval 0 ≤ θ ≤ 2π, giving answers in radians.
求方程 tan²θ – 1 = 0 在区间 0 ≤ θ ≤ 2π 内的所有解，答案以弧度表示。
Factor the equation as (tan θ – 1)(tan θ + 1) = 0, so tan θ = 1 or tan θ = – he principal solution is θ = π/4, and the period of tan is π, so θ = π/4 + nπ. For tan θ = – /4, 5π/4; from 3π/4 + nπ → 3π/4, 7π/4. Therefore, solutions are θ = π/4, 3π/4, 5π/4, 7π/4.
将方程分解为 (tan θ – 1)(tan θ + 1) = 0，因此 tan θ = 1 或 tan θ = –1。对于 tan θ = 1，主解为 θ = π/4，正切函数周期为 π，故 θ = π/4 + nπ。对于 tan θ = –1，θ = 3π/4 + nπ。在 [0, 2π] 内取值：由 π/4 + nπ 得 π/4、5π/4；由 3π/4 + nπ 得 3π/4、7π/4。因此，解为 θ = π/4, 3π/4, 5π/4, 7π/4。
5. Using R sin(θ + α) to Solve 3 sin θ + 4 cos θ = 2 | 利用 R sin(θ + α) 解 3 sin θ + 4 cos θ = 2
α), where R > 0 and 0° < α < 90°, and hence solve 3 sin θ + 4 cos θ = 2 for 0° ≤ θ ≤ 360°.
将 3 sin θ + 4 cos θ 写成 R sin(θ + α) 的形式，其中 R > 0 且 0° < α < 90°，并由此解方程 3 sin θ + 4 cos θ = 2，区间 0° ≤ θ ≤ 360°。
= R sin θ cos α + R cos θ sin α. Compare coefficients: R cos α = 3, R sin α = 4. Then R = √3² + 4² = 5, and tan α = 4/3 ⇒ .13°) = 0.4. Find the principal angle: sin⁻¹ 0.4 ≈ 23.58°. Thus, θ + 53.13° ≈ 23.58°, 180° – 23.58° = 156.42°, and adding 360° for further cycles. Solve for θ: θ ≈ 23.58° – 53.13° = –29.55° (reject) and θ ≈ 156.42° – 53.13° = 103.29°. Including +360° shifts: 23.58° + 360° = 383.58° ⇒ θ ≈ 330.45°, and 156.42° + 360° = 516.42° ⇒ θ ≈ 463.29° (out of range). So solutions are θ ≈ 103.3° and 330.5° (1 d.p.).
展开 R sin(θ + α) = R sin θ cos α + R cos θ sin α。比较系数得 R cos α = 3，R sin α = 4。则 R = √3² + 4² = 5，tan α = 4/3 ⇒ α ≈ 53.13°。方程化为 5 sin(θ + 53.13°) = 2，即 sin(θ + 53.13°) = 0.4。求主角：sin⁻¹ 0.4 ≈ 23.58°。于是 θ + 53.13° ≈ 23.58°, 180° – 23.58° = 156.42°，以及加 360° 的周期。解出 θ：θ ≈ 23.58° – 53.13° = –29.55°（舍去），θ ≈ 156.42° – 53.13° = 103.29°。考虑加 360° 的情形：23.58° + 360° = 383.58° ⇒ θ ≈ 330.45°，156.42° + 360° = 516.42° ⇒ θ ≈ 463.29°（超限）。所以解为 θ ≈ 103.3° 和 330.5°（保留一位小数）。
6. Sketching y = 2 sin(θ – 30°) and Transformations | 绘制 y = 2 sin(θ – 30°) 的图像与变换
Sketch the graph of y = 2 sin(θ – ning points. Describe the sequence of transformations that maps y = sin θ onto this graph.
画出 y = 2 sin(θ – 30°) 在 0° ≤ θ ≤ 360° 的图像，标出所有截距和拐点。描述将 y = sin θ 映射到该图像的一系列变换。
Start with y = sin θ. A horizontal translation of 30° to the right gives y = sin(θ – 30°). Then a vertical stretch by factor 2 produces y = 2 sin(θ – 60°, and is shifted 30° right. Key points: when θ = 30°, sin(0) = 0; at 120°, sin(90°) = 1 ⇒ y = 2 (max); at 210°, sin(180°) = 0; at 300°, sin(270°) = –1 ⇒ y = – cale) returns to 2. Within given domain, intercepts: (30°,0), (210°,0), also at end points? Check θ = 0°: y = 2 sin(–30°) = –1, intercept at (0,–1); θ = 360°: y = 2 sin(330°) = –1, so (360°,–1). Turning points: (120°,2) and (300°,–2). Plot these and sketch a smooth sine curve.
从 y = sin θ 出发。向右平移 30° 得到 y = sin(θ – 30°)。然后垂直拉伸因子 2 得到 y = 2 sin(θ – 30°)。图像振幅为 2，周期为 360°，右移 30°。关键点：θ = 30° 时，sin(0) = 0；θ = 120° 时，sin(90°) = 1 ⇒ y = 2（极大值）；θ = 210° 时，sin(180°) = 0；θ = 300° 时，sin(270°) = –1 ⇒ y = –2（极小值）；θ = 390°（超范围）回到 2。给定域内，截距：(30°,0) 和 (210°,0)，还有端点？检验 θ = 0°：y = 2 sin(–30°) = –1，截距 (0,–1)；θ = 360°：y = 2 sin(330°) = –1，点 (360°,–1)。极值点：(120°,2) 和 (300°,–2)。描点并画出光滑正弦曲线。
7. Solving a Quadratic in Trigonometric Form | 解三角形式的二次方程
Solve 2 cos²θ – 3 cos θ + 1 = 0 for 0° ≤ θ ≤ 360°.
解方程 2 cos²θ – 3 cos θ + 1 = 0，区间 0° ≤ θ ≤ 360°。
Treat the equation as a quadratic in cos θ. Factorise: (2 cos θ – 1)(cos θ – 0° (both boundaries included). Thus, the complete solution set is θ = 0°, 60°, 300°, 360°.
把方程看作关于 cos θ 的二次式。分解因式：(2 cos θ – 1)(cos θ – 1) = 0。因此，cos θ = ½ 或 cos θ = 1。对于 cos θ = ½，0° 到 360° 内的解为 θ = 60° 和 θ = 300°。对于 cos θ = 1，唯一解为 θ = 0° 和 θ = 360°（均包含端点）。所以完整解集为 θ = 0°, 60°, 300°, 360°。
8. Applying Double-Angle Formula | 应用倍角公式
Given that sin θ = ⅗ and θ is acute, find the exact value of sin 2θ and cos 2θ.
已知 sin θ = ⅗ 且 θ 为锐角，求 sin 2θ 和 cos 2θ 的精确值。
Since θ is acute, cos θ is positive. Use cos θ = √1 – sin²θ = √(1 – (⅗)²) = √1 – 9/25 = √16/25 = ⅘. Now apply double-angle formulas: sin 2θ = 2 sin θ cos θ = 2 × ⅗ × ⅘ = 24/25. cos 2θ = cos²θ – sin²θ = (⅘)² – (⅗)² = 16/25 – 9/25 = 7/25. These are the exact values.
因为 θ 是锐角，cos θ 为正。利用 cos θ = √1 – sin²θ = √(1 – (⅗)²) = √1 – 9/25 = √16/25 = ⅘。然后应用倍角公式：sin 2θ = 2 sin θ cos θ = 2 × ⅗ × ⅘ = 24/25。cos 2θ = cos²θ – sin²θ = (⅘)² – (⅗)² = 16/25 – 9/25 = 7/25。这就是精确值。
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