IB Mathematics: Negative Angle and Complementary Angle Formulas | IB数学：负角与余角公式

📚 IB Mathematics: Negative Angle and Complementary Angle Formulas | IB数学：负角与余角公式

Trigonometric identities are a core part of IB Mathematics. Among them, the negative angle identities and the complementary angle (cofunction) identities are particularly useful for simplifying expressions, solving equations, and proving other identities.

三角恒等式是 IB 数学的核心内容。其中，负角恒等式与余角（余函数）恒等式在化简表达式、解方程以及证明其它恒等式时特别有用。

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1. Negative Angle Identities | 负角恒等式

The negative angle identities describe the behaviour of trigonometric functions when the input angle is replaced by its negative, −θ.

负角恒等式描述的是当输入角替换为其相反数 −θ 时，三角函数的行为。

For any real angle θ, the following identities hold:

对于任意实数角 θ，以下恒等式成立：

sin(−θ) = −sin θ

cos(−θ) = cos θ

tan(−θ) = −tan θ

The reciprocal functions follow the same pattern:

倒数函数也有相同的规律：

• csc(−θ) = −csc θ
• sec(−θ) = sec θ
• cot(−θ) = −cot θ

These identities show that sine, tangent, cosecant, and cotangent are odd functions, while cosine and secant are even functions.

这些恒等式表明，正弦、正切、余割和余切是奇函数，而余弦和正割是偶函数。

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2. Complementary Angle Identities | 余角恒等式

Two angles are complementary if their measures add to π/2 rad (90°). The complementary angle identities, also called cofunction identities, relate each trigonometric function to its cofunction at the complementary angle.

如果两个角的和为 π/2 rad（90°），则它们互为余角。余角恒等式，也称余函数恒等式，将每个三角函数与其在余角处的余函数联系起来。

The basic cofunction identities are:

基本的余函数恒等式如下：

sin(π/2 − θ) = cos θ
cos(π/2 − θ) = sin θ
tan(π/2 − θ) = cot θ

Similarly, for the reciprocal functions:

对于倒数函数，也有类似关系：

• sec(π/2 − θ) = csc θ
• csc(π/2 − θ) = sec θ
• cot(π/2 − θ) = tan θ

In degrees, these become, for example, sin(90° − θ) = cos θ.

在角度制下，例如 sin(90° − θ) = cos θ。

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3. Unit Circle Derivation | 单位圆推导

The unit circle provides a clear visual derivation of the negative angle identities. If a point on the unit circle corresponds to angle θ, its coordinates are (cos θ, sin θ).

单位圆为负角恒等式提供了清晰的几何推导。如果单位圆上的一个点对应角 θ，那么它的坐标是 (cos θ, sin θ)。

so its coordinates become (cos θ, −sin θ). Therefore cos(−θ) = cos θ and sin(−θ) = −sin θ.

对于角 −θ，该点关于 x 轴对称，因此坐标变为 (cos θ, −sin θ)。所以 cos(−θ) = cos θ，sin(−θ) = −sin θ。

tes, giving the point (sin θ, cos θ). Hence sin(π/2 − θ) = cos θ and cos(π/2 − θ) = sin θ.

对于余角 π/2 − θ，该点关于直线 y = x 对称。这会交换横纵坐标，得到点 (sin θ, cos θ)。因此 sin(π/2 − θ) = cos θ，cos(π/2 − θ) = sin θ。

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4. Right-Triangle Derivation | 直角三角形推导

ght triangle, the two acute angles α and β satisfy α + β = π/2, so they are complementary.

在直角三角形中，两个锐角 α 和 β 满足 α + β = π/2，因此它们互为余角。

= adjacent side to α / hypotenuse = sin α. Since β = π/2 − α, we get sin α = cos(π/2 − α).

设 α 为一个锐角，则 sin α = 对边 / 斜边。其余角 β 的对边恰好是 α 的邻边，所以 cos β = 邻边 / 斜边 = α 的邻边 / 斜边 = sin α。因为 β = π/2 − α，所以 sin α = cos(π/2 − α)。

This derivation is intuitive, but it only works for acute angles. The unit circle or algebraic proofs are needed to justify the identities for all angles in radians or degrees.

这种推导非常直观，但只适用于锐角。要证明对任意角度（弧度制或角度制）均成立，需要借助单位圆或代数证明。

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5. Even and Odd Function Perspective | 偶函数与奇函数视角

Recall that a function f is even if f(−x) = f(x) and odd if f(−x) = −f(x).

回顾一下：如果 f(−x) = f(x)，则 f 是偶函数；如果 f(−x) = −f(x)，则 f 是奇函数。

From the graphs of the trigonometric functions, cosine is even and sine is odd. This immediately gives the negative angle identities for sin and cos. Tangent is odd because it is the ratio of sine to cosine, and an odd function divided by an even function is odd.

从三角函数图像可以看出，余弦是偶函数，正弦是奇函数。这直接给出了 sin 和 cos 的负角恒等式。正切是奇函数，因为它是正弦与余弦之比，而奇函数除以偶函数仍为奇函数。

cos θ. This is because the sine curve is closely related to the cosine curve through a phase shift. The identity expresses that relationship in a compact form.

余角恒等式可以看作水平平移关系：sin(π/2 − θ) = cos θ。这是因为正弦曲线与余弦曲线通过相位平移紧密相关。该恒等式以紧凑的形式体现了这种关系。

Understanding the parity of the functions helps you predict the sign without re-deriving.

理解函数的奇偶性有助于你无需重新推导即可判断符号。

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6. Simplifying Expressions | 化简表达式

These identities are powerful tools for simplifying trigonometric expressions.

这些恒等式是化简三角函数表达式的有力工具。

Example 1: Simplify sin(π/2 − x) cos(−x).

例 1：化简 sin(π/2 − x) cos(−x)。

angle identity, cos(−x) = cos x. Therefore the expression becomes cos x · cos x = cos² x.

利用余角恒等式，sin(π/2 − x) = cos x。利用负角恒等式，cos(−x) = cos x。因此原式变为 cos x · cos x = cos

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