📚 AQA Pure Maths Examples 6K: Trigonometric Equations, Identities and Modelling | AQA纯数学例题6K：三角方程、恒等式与建模
Exercise 6K in the AQA Pure Mathematics course gathers the final set of examples on trigonometric equations, identities and simple modelling. Students are expected to move from finding one reference angle to listing all solutions in a given interval, and to choose the correct identity when an equation mixes sin, cos and tan.
AQA纯数学课程中的练习6K通常汇集了关于三角方程、恒等式与简单建模的最后一组例题。学生需要从求出一个参考角进阶到列出给定区间内的所有解，并在方程混合出现 sin、cos 和 tan 时正确选择恒等式。
1. Where Exercise 6K Sits in AQA Pure | 6K练习在AQA纯数学中的位置
Within the AQA Pure Year 1 specification, the trigonometry block covers graphs of sine, cosine and tangent, exact values for special angles, the two core identities, and solving equations in degrees or radians. Exercise 6K is usually a consolidation set: it mixes these skills and often adds a modelling context.
在AQA纯数学第一学年的考试大纲中，三角学部分包括正弦、余弦和正切的图像，特殊角的精确值，两个核心恒等式，以及以度或弧度为单位解方程。练习6K通常是综合性练习：它混合了这些技能，并经常加入建模背景。
2. Key Facts and Exact Values | 关键事实与精确值
You must be able to quote exact values for 30°, 45° and 60°. These come from the two special right-angled triangles. The table below summarises the values you will need constantly in 6K.
你必须能够直接写出30°、45°和60°的精确值。这些值来自两个特殊直角三角形。下表总结了你在6K中会反复用到的值。
| Angle θ | sin θ | cos θ | tan θ |
|---|---|---|---|
| 30° = π/6 | 1/2 | √3/2 | 1/√3 = √3/3 |
| 45° = π/4 | 1/√2 = √2/2 | 1/√2 = √2/2 | 1 |
| 60° = π/3 | √3/2 | 1/2 | √3 |
w both degrees and radians, but you must match the interval notation: if the question uses radians, write all final answers in radians.
对于弧度，请记住 180° = π，因此 360° = 2π。许多AQA 6K题目允许使用度或弧度，但你必须与区间记号保持一致：如果题目使用弧度，所有最终答案都要写成弧度。
3. The Two Core Identities | 两个核心恒等式
h are valid for all real θ where the functions are defined, and both are essential when an equation contains more than one trigonometric function.
AQA要求你熟练使用两个恒等式。第一个是 tan θ = sin θ / cos θ。第二个是 sin² θ + cos² θ = 1。这两个恒等式在函数有定义的所有实数θ上都成立，当方程中包含多个三角函数时它们都必不可少。
sin² θ + cos² θ = 1
tan θ = sin θ / cos θ
θ, or cos² θ with 1 − sin² θ. This is the most common algebraic step in Exercise 6K questions.
使用毕达哥拉斯恒等式 sin² θ + cos² θ = 1，可以将 sin² θ 替换为 1 − cos² θ，或将 cos² θ 替换为 1 − sin² θ。这是练习6K题目中最常见的代数步骤。
4. Solving sin x = k in [0°, 360°] | 在[0°, 360°]内解 sin x = k
° − α. If k is negative, sine is negative in the third and fourth quadrants; solve sin x = |k| first to get α, then the solutions are 180° + α and 360° − α.
要求解 sin x = k，首先求出主值 α = sin⁻¹(k)。如果 k 为正，正弦在第一和第二象限为正，因此解为 α 和 180° − α。如果 k 为负，正弦在第三和第四象限为负；先解 sin x = |k| 得到 α，然后解为 180° + α 和 360° − α。
Example: Solve sin x = 1/2 for 0° ≤ x ≤ 360°.
示例：在 0° ≤ x ≤ 360° 内解 sin x = 1/2。
pal value is α = 30°. Since sine is positive, the solutions are 30° and 180° − 30° = 150°.
主值为 α = 30°。因为正弦为正，解为 30° 和 180° − 30° = 150°。
x = 30°, 150°
5. Solving cos x = k and tan x = k | 解 cos x = k 与 tan x = k
s positive in the first and fourth quadrants, so the solutions are α and 360° − α. If k is negative, cosine is negative in the second and third quadrants, giving 180° − α and 180° + α.
对于 cos x = k，若 k 为正，余弦在第一和第四象限为正，因此解为 α 和 360° − α。若 k 为负，余弦在第二和第三象限为负，解为 180° − α 和 180° + α。
an⁻¹(k). Tangent has period 180°, so add 180° to α to find every solution in [0°, 360°]. When k is positive, tan is positive in the first and third quadrants; when k is negative, the same formula works because tan⁻¹ gives a negative principal value.
对于 tan x = k，基本解为 α = tan⁻¹(k)。正切函数的周期为180°，因此将 α 加上180°即可找到 [0°, 360°] 内的所有解。当 k 为正时，正切在第一和第三象限为正；当 k 为负时，同样公式仍然适用，因为 tan⁻¹ 会给出负的主值。
Example: Solve cos x = −1/2 for 0° ≤ x ≤ 360°.
示例：在 0° ≤ x ≤ 360° 内解 cos x = −1/2。
60°. Since cosine is negative, the solutions are 180° − 60° = 120° and 180° + 60° = 240°.
先解 cos x = 1/2，得到 α = 60°。因为余弦为负，解为 180° − 60° = 120° 和 180° + 60° = 240°。
x = 120°, 240°
6. CAST Diagrams and Multiple Solutions | CAST图与多解
A CAST diagram labels the four quadrants to show which function is positive: C for cosine in the fourth quadrant, A for all in the first quadrant, S for sine in the second quadrant, and T for tangent in the third quadrant. Always start from the principal value α and use symmetry about the axes to write all solutions in the required interval.
CAST图标注了四个象限，以显示哪些函数为正：第四象限是 C（余弦），第一象限是 A（全部），第二象限是 S（正弦），第三象限是 T（正切）。始终从主值 α 出发，利用关于坐标轴的对称性写出所需区间内的所有解。
For sine and cosine, add or subtract α from 180° or 360° depending on the quadrant. For tangent, simply add multiples of 180°. In radians, replace 180° with π and 360° with 2π.
对于正弦和余弦，根据象限将 α 从180°或360°中相加或相减。对于正切，只需加上180°的整数倍。在弧度制中，将180°替换为 π，将360°替换为 2π。
7. Equations That Need tan x = sin x / cos x | 需要 tan x = sin x / cos x 的方程
check that cos x = 0 is not already a solution before dividing, or consider it separately.
当方程同时包含 sin x 和 cos x 且没有平方项时，通常最简单的方法是将每一项都除以 cos x，从而构造出 tan x。在除以 cos x 之前，必须检查 cos x = 0 是否已经是解，或者单独考虑这种情况。
Example: Solve 2 sin x = 3 cos x for 0° ≤ x ≤ 360°.
示例：在 0° ≤ x ≤ 360° 内解 2 sin x = 3 cos x。
Divide through by cos x, assuming cos x ≠ 0:
除以 cos x，假设 cos x ≠ 0：
2 tan x = 3
tan x = 3/2
α = tan⁻¹(3/2) ≈ 56.3°. Since tangent has period 180°, the solutions are 56.3° and 236.3°.
主值为 α = tan⁻¹(3/2) ≈ 56.3°。因为正切的周期为180°，解为 56.3° 和 236.3°。
8. Equations That Need sin² x + cos² x = 1 | 需要 sin² x + cos² x = 1 的方程
If an equation mixes squared trigonometric functions, replace either sin² x or cos² x using the Pythagorean identity. The aim is to produce a quadratic in one trigonometric function only.
如果方程中混合了平方三角函数，请使用毕达哥拉斯恒等式替换 sin² x 或 cos² x。目标是得到一个仅含一个三角函数的二次方程。
Example: Solve 2 cos² x + sin x = 1 for 0° ≤ x ≤ 360°.
示例：在 0° ≤ x ≤ 360° 内解 2 cos² x + sin x = 1。
Replace cos² x with 1 − sin² x:
将 cos² x 替换为 1 − sin² x：
2(1 − sin² x) + sin x = 1
2 − 2 sin² x + sin x = 1
2 sin² x − sin x − 1 = 0
Factorise the quadratic:
将二次式因式分解：
(2 sin x + 1)(sin x − 1) = 0
r sin x = 1. In [0°, 360°], sin x = 1 gives x = 90°; sin x = −1/2 gives x = 210° and 330°.
因此 sin x = −1/2 或 sin x = 1。在 [0°, 360°] 内，sin x = 1 给出 x = 90°；sin x = −1/2 给出 x = 210° 和 330°。
x = 90°, 210°, 330°
9. Transformed Arguments: sin(2x) and cos(x − 30°) | 变换角度：sin(2x) 与 cos(x − 30°)
When the angle is transformed, solve the basic equation for the whole angle first, then rearrange to find x. Be careful to expand the range for the transformed angle before listing solutions.
当角度发生变换时，首先对整体角度求解基本方程，然后再整理求出 x。列解前要小心扩展变换角的范围。
Example: Solve sin(2x) = 1/2 for 0° ≤ x ≤ 360°.
示例：在 0° ≤ x ≤ 360° 内解 sin
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