6.1 Sine Theorem, Cosine Theorem, and Triangle Area Formulas

6.1 Sine Theorem, Cosine Theorem, and Triangle Area Formulas

In the National College Entrance Examination (Gaokao) Mathematics syllabus, the study of triangles serves as the primary bridge linking pure trigonometry with plane geometry and vector algebra. A foundational understanding of triangle quantitative relations relies on two core theorems: the Sine Theorem and the Cosine Theorem, complemented by flexible Area Formulas and side-angle transformation principles.

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1. Rigorous Statement and Proofs of Core Theorems

Let $\triangle ABC$ be an arbitrary triangle with interior angles $A, B, C$ and respective opposite side lengths $a, b, c$. Let $R$ denote the radius of its circumscribed circle (circumradius) and $r$ denote the radius of its inscribed circle (inradius).

The Sine Theorem (正弦定理)

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$

Geometric Derivation (Circumscribed Circle Method)

Consider $\triangle ABC$ inscribed in circle $O$ with radius $R$. Construct diameter $BD$ passing through center $O$ and endpoint $B$, intersecting circle $O$ at $D$. Connect $CD$.

1. In right triangle $\triangle BCD$, the angle $\angle BCD = 90^\circ$ by Thales' theorem.
2. By the inscribed angle theorem, angles subtended by the same arc $BC$ are equal: $\angle BDC = \angle BAC = A$ (when $A$ is acute). Thus, $\sin A = \sin(\angle BDC) = \frac{BC}{BD} = \frac{a}{2R}$.
3. Rearranging yields $\frac{a}{\sin A} = 2R$. If $A = 90^\circ$, $\sin A = 1$ and $a = 2R$, holding trivially. If $A$ is obtuse, $\angle BDC = 180^\circ - A$, so $\sin(\angle BDC) = \sin(180^\circ - A) = \sin A = \frac{a}{2R}$.
4. By symmetry across all vertices, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.

The Cosine Theorem (余弦定理)

$a^2 = b^2 + c^2 - 2bc \cos A$ $b^2 = a^2 + c^2 - 2ac \cos B$ $c^2 = a^2 + b^2 - 2ab \cos C$

Equivalently, the angle forms are: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}, \quad \cos B = \frac{a^2 + c^2 - b^2}{2ac}, \quad \cos C = \frac{a^2 + b^2 - c^2}{2ab}$

Vector Proof

Define vectors along the sides of $\triangle ABC$: $\vec{BC} = \vec{AC} - \vec{AB}$. Taking the vector dot product of $\vec{BC}$ with itself: $\vec{BC} \cdot \vec{BC} = (\vec{AC} - \vec{AB}) \cdot (\vec{AC} - \vec{AB})$ $|\vec{BC}|^2 = |\vec{AC}|^2 + |\vec{AB}|^2 - 2 (\vec{AC} \cdot \vec{AB})$ Since $|\vec{BC}| = a$, $|\vec{AC}| = b$, $|\vec{AB}| = c$, and $\vec{AC} \cdot \vec{AB} = |\vec{AC}||\vec{AB}| \cos A = bc \cos A$, substitution directly yields: $a^2 = b^2 + c^2 - 2bc \cos A$ This proof holds universally for acute, right, and obtuse angles without separate case analysis.

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2. Triangle Area Formulas and Geometric Radii

The area $S$ of $\triangle ABC$ can be expressed in several equivalent formulations depending on given parameters:

1. Standard Sine Area Formula: $S = \frac{1}{2} ab \sin C = \frac{1}{2} bc \sin A = \frac{1}{2} ac \sin B$
2. Circumradius Formula: Substituting $\sin C = \frac{c}{2R}$ into $S = \frac{1}{2}ab \sin C$ yields: $S = \frac{abc}{4R} \implies R = \frac{abc}{4S}$
3. Inradius Formula: Decomposing $\triangle ABC$ into three subtriangles $\triangle OBC, \triangle OCA, \triangle OAB$ sharing the incenter $O$ as common vertex with height $r$: $S = \frac{1}{2}ar + \frac{1}{2}br + \frac{1}{2}cr = \frac{a+b+c}{2} r = pr \implies r = \frac{2S}{a+b+c}$
4. Heron's Formula (海伦公式): Let $p = \frac{a+b+c}{2}$ be the semi-perimeter. Then: $S = \sqrt{p(p-a)(p-b)(p-c)}$

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3. Systematic Side-Angle Conversion Strategies (边角互化法则)

Solving trigonometric identity equations in triangles requires converting side relations into angle relations (or vice-versa):

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4. Worked Gaokao Exam Examples

Example 1 (Gaokao Identity Transformation & Area Analysis)

In $\triangle ABC$, the side lengths opposite to angles $A, B, C$ are $a, b, c$ respectively. Given that: $(2a - c)\cos B = b \cos C$

1. Find the value of angle $B$.
2. If $b = \sqrt{3}$ and $a + c = 3$, find the area $S$ of $\triangle ABC$.

Solution:

Part 1: Convert the side lengths to sines using the Sine Theorem ($a = 2R\sin A, b = 2R\sin B, c = 2R\sin C$): $(2\sin A - \sin C)\cos B = \sin B \cos C$ Expand the left-hand side: $2\sin A \cos B - \sin C \cos B = \sin B \cos C$ Move $\sin C \cos B$ to the right-hand side: $2\sin A \cos B = \sin B \cos C + \sin C \cos B$ By the sine addition identity, $\sin B \cos C + \sin C \cos B = \sin(B+C)$. Since $A + B + C = \pi$, we have $\sin(B+C) = \sin(\pi - A) = \sin A$. Therefore: $2\sin A \cos B = \sin A$ Because $A \in (0, \pi)$, $\sin A \neq 0$. Dividing both sides by $\sin A$ yields: $\cos B = \frac{1}{2}$ Since $B \in (0, \pi)$, we conclude $B = \frac{\pi}{3}$ ($60^\circ$).

Part 2: Applying the Cosine Theorem for angle $B$: $b^2 = a^2 + c^2 - 2ac \cos B$ Substitute $b = \sqrt{3}$ and $B = \frac{\pi}{3}$ ($\cos B = \frac{1}{2}$): $3 = a^2 + c^2 - ac$ Rewrite $a^2 + c^2$ as $(a+c)^2 - 2ac$: $3 = (a+c)^2 - 3ac$ Given $a + c = 3$, substitute this value: $3 = 3^2 - 3ac \implies 3 = 9 - 3ac \implies 3ac = 6 \implies ac = 2$ Now calculate area $S$: $S = \frac{1}{2} ac \sin B = \frac{1}{2} \cdot 2 \cdot \sin\frac{\pi}{3} = 1 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$

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Example 2 (Circumradius and Inequality Bound)

In $\triangle ABC$, $a^2 + b^2 - c^2 = \sqrt{3}ab$. The circumradius of $\triangle ABC$ is $R = 2$.

1. Find side length $c$.
2. Determine the maximum possible area of $\triangle ABC$.

Solution:

Part 1: By the Cosine Theorem: $\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{\sqrt{3}ab}{2ab} = \frac{\sqrt{3}}{2}$ Since $C \in (0, \pi)$, we get $C = \frac{\pi}{6}$ ($30^\circ$). By the Sine Theorem: $c = 2R \sin C = 2 \cdot 2 \cdot \sin\frac{\pi}{6} = 4 \cdot \frac{1}{2} = 2$

Part 2: From $a^2 + b^2 - c^2 = \sqrt{3}ab$ and $c = 2$, we have $a^2 + b^2 - 4 = \sqrt{3}ab$, so: $a^2 + b^2 = 4 + \sqrt{3}ab$ By the AM-GM inequality, $a^2 + b^2 \ge 2ab$. Thus: $2ab \le 4 + \sqrt{3}ab \implies (2 - \sqrt{3})ab \le 4$ Dividing by $2 - \sqrt{3}$ and rationalizing the denominator: $ab \le \frac{4}{2 - \sqrt{3}} = 4(2 + \sqrt{3}) = 8 + 4\sqrt{3}$ Equality holds when $a = b = \sqrt{4 + 2\sqrt{3}} = 1 + \sqrt{3}$. Therefore, maximum area $S_{max}$ is: $S_{max} = \frac{1}{2} ab_{max} \sin C = \frac{1}{2} (8 + 4\sqrt{3}) \sin\frac{\pi}{6} = \frac{1}{4} (8 + 4\sqrt{3}) = 2 + \sqrt{3}$