📚 IB Mathematics: Trigonometric Equations and Identities | IB数学：三角方程与恒等式
The unit circle is the foundation of every trigonometric function. For any angle θ measured from the positive x-axis, the point where the terminal side intersects the circle has coordinates (cos θ, sin θ). This visual model defines the sine, cosine and tangent functions, and it explains why each function behaves differently in the four quadrants.
单位圆是所有三角函数的基础。从x轴正方向起测的任意角θ，其终边与单位圆的交点坐标为(cosθ, sinθ)。这个直观模型定义了正弦、余弦和正切，也解释了各函数在四个象限中为何呈现不同的正负号。
1. The Unit Circle and Fundamental Definitions | 单位圆与基本定义
Consider a point P(x, y) on the unit circle. The angle θ is measured counterclockwise from the positive x-axis. The definitions are:
设单位圆上有一点P(x, y)，角度θ从x轴正方向逆时针度量。基本定义如下：
sin θ = y, cos θ = x, tan θ = y/x (x ≠ 0)
Because the radius is 1, the coordinates automatically satisfy x² + y² = 1. This gives the most important identity in trigonometry: sin²θ + cos²θ = 1.
因为半径为1，所以坐标自然满足x² + y² = 1。由此得到三角学中最重要的恒等式：sin²θ + cos²θ = 1。
The signs of sin, cos and tan depend on the quadrant. In the first quadrant all are positive; in the second quadrant only sine is positive; in the third only tangent is positive; in the fourth only cosine is positive. This CAST pattern helps when solving equations.
sin、cos和tan的正负取决于象限：第一象限全正，第二象限仅sin正，第三象限仅tan正，第四象限仅cos正。这种CAST正负分布模式有助于解方程。
2. Reciprocal and Pythagorean Identities | 倒数恒等式与毕达哥拉斯恒等式
The reciprocal functions are defined as the reciprocals of the primary functions. Alongside the tangent identity, they form a set of identities that can simplify almost any trigonometric expression.
倒数函数定义为三个基本函数的倒数。配合正切恒等式，它们构成了可化简几乎所有三角表达式的恒等式组。
| csc θ = 1 / sin θ | sec θ = 1 / cos θ | cot θ = 1 / tan θ |
| tan θ = sin θ / cos θ | cot θ = cos θ / sin θ | sin²θ + cos²θ = 1 |
From the Pythagorean identity, two alternative forms can be derived by dividing by cos²θ or sin²θ. These are especially useful in IB Paper 2 for simplifying fractions.
由毕达哥拉斯恒等式，通过除以cos²θ或sin²θ可得到两个常用变形，这些在IB笔试二中尤其适用于化简分式。
1 + tan²θ = sec²θ, 1 + cot²θ = csc²θ
For example, to simplify 1/(1 + tan²θ), replace the denominator by sec²θ to obtain cos²θ. Always look for these hidden quadratic relationships before attempting other substitutions.
例如，化简1/(1 + tan²θ)时，将分母换成sec²θ，立即得到cos²θ。求解时，先寻找这类隐藏的二次齐次关系，往往快于尝试其他代换。
3. Compound Angle Identities | 复合角恒等式
When two angles A and B are added or subtracted, the sine and cosine do not distribute. The compound angle identities give the correct expansion and are essential for solving equations that contain expressions such as sin(x + y).
两个角A和B相加或相减时，sin和cos不能简单分配。复合角恒等式给出了正确的展开式，对于处理如sin(x + y)这类表达式至关重要。
sin(A ± B) = sin A cos B ± cos A sin B
cos(A ± B) = cos A cos B ∓ sin A sin B
tan(A ± B) = (tan A ± tan B) / (1 ∓ tan A tan B)
These identities are useful for proving other results. For example, if you need to solve cos(x + 60°) = 0.5, you could expand the left side, but it is simpler to recognize that x + 60° is a single reference angle.
这些恒等式常用于证明其他结论。例如，解cos(x + 60°) = 0.5时，可以展开左边，但更简便的方法是视x + 60°为一个整体参照角。
A common IB exam task is to write an expression in the form R sin(θ + α). This technique is a direct application of the compound angle formula for sine, and it transforms a linear combination into a single sine function.
IB考试常见任务是把某个表达式写成R sin(θ + α)的形式。此技巧直接应用sin的复合角公式，会将线性组合归并为单一正弦函数。
4. Double Angle Identities | 二倍角恒等式
Setting A = B in the compound angle identities produces the double angle identities. These are frequently used in IB to rewrite powers, eliminate products, or reduce the degree of an equation.
在复合角恒等式中令A = B，便得到二倍角恒等式。IB中常用它们改写幂次、消去乘积或降低方程的次数。
sin 2A = 2 sin A cos A
cos 2A = cos²A – sin²A = 2 cos²A – 1 = 1 – 2 sin²A
tan 2A = 2 tan A / (1 – tan²A)
The three forms of cos 2A are equivalent. You should choose the form that matches the given expression. For instance, when integrating in calculus, cos²A = (1 + cos 2A)/2 and sin²A = (1 – cos 2A)/2 are crucial for power reduction.
cos 2A的三种形式彼此等价。选用哪种形式应取决于所给表达式。例如在微积分积分时，cos²A = (1 + cos 2A)/2与sin²A = (1 – cos 2A)/2是降幂的关键变形。
Double angle identities also help solve equations such as sin 2θ = cos θ. Replacing sin 2θ by 2 sinθ cosθ allows factoring: cosθ(2 sinθ – 1) = 0.
二倍角公式也有助于解方程，例如sin 2θ = cos θ。将sin 2θ替换为2 sinθ cosθ后可因式分解：cosθ(2 sinθ – 1) = 0。
5. Solving Basic Trigonometric Equations | 解基本三角方程
Solving a basic equation like sin θ = a or cos θ = a requires two steps: find the acute reference angle, and then determine all angles in the required interval using the ASTC sign diagram.
解形如sin θ = a或cos θ = a的基本方程需两步：先求锐角参照角，再利用ASTC象限符号图确定给定区间内的全部角。
For sin θ = 0.5 on the interval 0° ≤ θ < 360°, the reference angle is 30°. Since sine is positive in quadrants I and II, the solutions are θ = 30° and θ = 150°.
对于区间0° ≤ θ < 360°内的sin θ = 0.5，参照角是30°。由于正弦在第一、二象限为正，所以解为θ = 30°和θ = 150°。
For cos θ = 0.5, cosine is positive in quadrants I and IV, giving θ = 60° and θ = 300°. For tan θ = 1, tangent is positive in quadrants I and III, giving θ = 45° and θ = 225°.
对于cos θ = 0.5，余弦在第一、四象限为正，解为θ = 60°和θ = 300°。对于tan θ = 1，正切在第一、三象限为正，解为θ = 45°和θ = 225°。
Watch out for negative values. For sin θ = -0.5, the reference angle is still 30° but the solutions lie in quadrants III and IV: θ = 210° and θ = 330°.
注意负值情形。对于sin θ = -0.5，参照角仍是30°，但解落在第三、四象限：θ = 210°和θ = 330°。
6. Factoring and Quadratic-Type Equations | 因式分解法与二次型方程
Many trigonometric equations are actually quadratic in one function. Replace the function with a variable such as u, solve the quadratic, and then solve the corresponding simple trigonometric equations.
许多三角方程实质上是关于某个三角函数的二次方程。用一个变量u代替该函数，解二次方程，再解对应的简单三角方程即可。
Example: solve 2sin²θ + 3sinθ + 1 = 0 for 0 ≤ θ < 2π. Let u = sinθ. Then 2u² + 3u + 1 = 0 factors as (2u + 1)(u + 1) = 0, so u = -1/2 or u = -1.
例：在0 ≤ θ < 2π内解2sin²θ + 3sinθ + 1 = 0。令u = sinθ，则2u² + 3u + 1 = 0分解为(2u + 1)(u + 1) = 0，故u = -1/2或u = -1。
For u = -1/2, the reference angle is π/6. Sine is negative in quadrants III and IV, so θ = 7π/6 or 11π/6. For u = -1, θ = 3π/2.
当u = -1/2时，参照角为π/6。正弦为负的象限是第三、四象限，故θ = 7π/6或11π/6。当u = -1时，θ = 3π/2。
Always check whether a factored second term exists. An equation like sinθ cosθ – sinθ = 0 can be factored as sinθ(cosθ – 1) = 0, producing solutions for both factors separately.
始终检查是否有可分解的第二项。方程sinθ cosθ – sinθ = 0可分解为sinθ(cosθ – 1) = 0，需对两个因子分别求解放置的解。
7. General Solutions and Periodicity | 通解与周期性
Since sine and cosine have period 360° or 2π, and tangent has period 180° or π, equations usually have infinitely many solutions. A general solution captures all possible values.
由于正弦、余弦的周期为360°或2π，而正切的周期为180°或π，方程通常有无穷多个解。通解可以概括所有可能取值。
For sin θ = a, if the principal solution is θ₀, then the general solution is θ = n·360° + θ₀ or θ = n·360° + 180° – θ₀, depending on the quadrant. In radians this becomes θ = 2nπ + θ₀ or θ = 2nπ + π – θ₀.
对sin θ = a，若主解为θ₀，则通解为θ = n·360° + θ₀或θ = n·360° + 180° – θ₀，具体由象限决定。弧度制下写成θ = 2nπ + θ₀或θ = 2nπ + π – θ₀。
For cos θ = a, the general solution is θ = n·360° ± θ₀. For tan θ = a, the general solution is θ = n·180° + θ₀.
对cos θ = a，通解为θ = n·360° ± θ₀。对tan θ = a，通解为θ = n·180° + θ₀。
Example: solve tan θ = √3. The reference angle is 60°, so the general solution is θ = n·180° + 60°. For a restricted interval, substitute integer values of n.
例：解tan θ = √3。参照角为60°，故通解为θ = n·180° + 60°。若限定区间，只需代入不同的整数n。
8. Auxiliary Angle Method | 辅助角方法
The expression a sin θ + b cos θ can be written as a single sine or cosine function using the auxiliary angle method. This is a frequent question in IB analysis and approaches.
表达式a sin θ + b cos θ可通过辅助角方法改写为单个正弦或余弦函数。这是IB分析与方法中常见的题型。
a sin θ + b cos θ = R sin(θ + α), R = √(a² + b²), α = arctan(b/a)
When matching coefficients, expand R sin(θ + α) as R(sinθ cosα + cosθ sinα). Thus R cosα = a and R sinα = b. The constant R is the amplitude, and α is the phase shift.
匹配系数时，将R sin(θ + α)展开为R(sinθ cosα + cosθ sinα)。因此R cosα = a且R sinα = b。常数R是振幅，α是相位移动。
For example, 3 sinθ + 4 cosθ becomes 5 sin(θ + 53.13°). This form makes it easy to find maximum and minimum values or to solve equations.
例如，3 sinθ + 4 cosθ可化为5 sin(θ + 53.13°)。这种形式便于求最大值、最小值或解方程。
To solve 3 sinθ + 4 cosθ = 4, rewrite as 5 sin(θ + 53.13°) = 4, so sin(θ + 53.13°) = 0.8. Then use the standard method for a basic sine equation.
解3 sinθ + 4 cosθ = 4时，改写为5 sin(θ + 53.13°) = 4，即sin(θ + 53.13°) = 0.8，再按基本正弦方程的方法求解。
9. Proving Trigonometric Identities | 证明三角恒等式
Identity proofs require transforming one side of the equation into the other. The key is to work with the more complicated side, convert all functions to sines and cosines, and factor or cancel carefully.
恒等式证明要求把等式的一边变形为另一边。关键是从较复杂的一边入手，把所有函数化为正弦和余弦，再谨慎因式分解或约分。
For example, prove that cos³θ/(1 – sinθ) = 1 + sinθ. Multiply numerator and denominator by (1 + sinθ) to obtain cos²θ(1 + sinθ)/(1 – sin²θ). Since 1 – sin²θ = cos²θ, the expression reduces to 1 + sinθ.
例如，证明cos³θ/(1 – sinθ) = 1 + sinθ。将分子分母同乘(1 + sinθ)，得到cos²θ(1 + sinθ)/(1 – sin²θ)。因为1 – sin²θ = cos²θ，原式化简为1 + sinθ。
Always state the non-permissible values. If a denominator becomes zero, the identity is undefined. In IB, writing these restrictions earns separate marks.
始终说明非允许值。若分母为零，恒等式无定义。在IB考试中，写出这些限制条件可单独得分。
Another common proof uses the double angle identity: cotθ + tanθ = cscθ secθ. Write cotθ = cosθ/sinθ and tanθ = sinθ/cosθ, then combine fractions: (cos²θ + sin²θ)/(sinθ cosθ) = 1/(sinθ cosθ) = cscθ secθ.
另一个常见证明是cotθ + tanθ = cscθ secθ。令cotθ = cosθ/sinθ、tanθ = sinθ/cosθ，再合并分式：(cos²θ + sin²θ)/(sinθ cosθ) = 1/(sinθ cosθ) = cscθ secθ。
10. Common Pitfalls and Exam Strategy | 常见易错点与应试策略
The most frequent error is forgetting that sinθ = a has two primary solutions in one cycle, not one. Another is mixing up the sign of the reference angle when extending to general solutions.
最常见的错误是忘记一个周期内sinθ = a有两个主解，而不是一个。另一个常见错误是在推广到通解时混淆参照角的正负号。
When squaring both sides of a trigonometric equation, ensure you do not introduce extraneous roots. For example, solving sinθ + cosθ = 1 by squaring requires checking each candidate solution.
对三角方程两边同时平方时，要小心产生增根。例如，用平方的方法解sinθ + cosθ = 1时，必须检验每一个候选解。
| Do: state the interval, use exact values, check all solutions. | Don’t: drop the ± sign, ignore the quadrant, or cancel by sinθ without considering sinθ = 0. |
In Paper 1, show each algebraic step clearly. In Paper 2, use your calculator to verify solutions and to solve equations that cannot be solved analytically, such as cos2θ = θ.
在笔试一中，要清晰写出每一步代数变形。在笔试二中，用计算器验证解，并求解无法解析求解的方程，例如cos2θ = θ。
Finally, memorise the standard table of exact values for 0°, 30°, 45°, 60° and 90°. This speeds up many solutions and gives you confidence in part (b), which often asks you to use part (a) to solve an equation.
最后，牢记0°、30°、45°、60°、90°的标准特殊角精确值表。这能加快解题速度，并帮助你完成第二小问——它通常要求利用第一小问的结果求解方程。
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