4.3 Solutions of Triangles (Sine/Cosine Laws) & Trigonometric Equations

Solutions of Triangles (Sine/Cosine Laws) & Trigonometric Equations

In the Intermediate Mathematics Part I syllabus, triangle trigonometry and the theory of trigonometric equations represent the culmination of analytical trigonometry. ECAT questions on this topic ask candidates to compute triangle elements, solve heights-and-distances problems, and solve trigonometric equations quickly without a calculator. Mastering the geometric identities connecting triangle circles with side lengths, combined with standard solution formulas for general trigonometric equations, guarantees a significant competitive advantage.

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1. The Law of Sines and the Circumscribed Circle

In any triangle $ABC$, let $a, b, c$ denote the lengths of the sides opposite to angles $A, B, C$ respectively. Let $R$ denote the radius of the circumscribed circle (the circumcircle) that passes through all three vertices $A, B, C$.

The Extended Law of Sines

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

Geometric Proof Sketch

Construct a diameter $BD = 2R$ of the circumcircle. Draw chord $CD$. Because the angle subtended by a diameter at the circumference is a right angle, $\angle BCD = 90^\circ$. Furthermore, $\angle BDC = \angle BAC = A$ because angles subtended by the same arc $BC$ are equal. In the right-angled triangle $BCD$:

$\sin(\angle BDC) = \frac{BC}{BD} \implies \sin A = \frac{a}{2R} \implies \frac{a}{\sin A} = 2R$

Repeating the construction for chords $AC$ and $AB$ establishes the complete equality.

Triangle Applications of the Law of Sines

1. Case 1: AAS or ASA (Two angles and one side known): Compute the third angle using $A + B + C = 180^\circ$, then apply $\frac{a}{\sin A} = \frac{b}{\sin B}$ to determine remaining sides directly.
2. Case 2: The Ambiguous Case (SSA): Given two sides $a, b$ and a non-included angle $A$:
   • Let $h = b\sin A$ be the perpendicular altitude from vertex $C$ to side $c$.
   • If $A < 90^\circ$:
     • $a < h$: Side $a$ is too short to reach the base line $\implies$ No triangle exists.
     • $a = h$: Side $a$ forms a right triangle with base $\implies$ Exactly one triangle.
     • $h < a < b$: Side $a$ can intersect the base line in two distinct points $\implies$ Two triangles (Ambiguous Case).
     • $a \ge b$: Side $a$ intersects the base line only on one side of ray $AB$ $\implies$ Exactly one triangle.
   • If $A \ge 90^\circ$:
     • $a \le b$: No triangle can exist.
     • $a > b$: Exactly one obtuse or right triangle exists.

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2. The Law of Cosines and the Law of Tangents

When two sides and their included angle (SAS) or all three sides (SSS) are specified, the Law of Sines cannot be initiated without additional information. The Law of Cosines provides the necessary generalized formulation of the Pythagorean theorem.

Formulation of the Law of Cosines

$a^2 = b^2 + c^2 - 2bc\cos A$

$b^2 = c^2 + a^2 - 2ca\cos B$

$c^2 = a^2 + b^2 - 2ab\cos C$

Solving for the cosine of the angles yields:

$\cos A = \frac{b^2 + c^2 - a^2}{2bc}, \quad \cos B = \frac{c^2 + a^2 - b^2}{2ca}, \quad \cos C = \frac{a^2 + b^2 - c^2}{2ab}$

Classification of Triangles by the Law of Cosines

Let $a$ be the longest side of a triangle ($a \ge b, a \ge c$):

• If $a^2 < b^2 + c^2 \iff \cos A > 0 \implies$ The triangle is acute-angled.
• If $a^2 = b^2 + c^2 \iff \cos A = 0 \implies$ The triangle is right-angled ($A = 90^\circ$).
• If $a^2 > b^2 + c^2 \iff \cos A < 0 \implies$ The triangle is obtuse-angled ($A > 90^\circ$).

The Law of Tangents

$\frac{a - b}{a + b} = \frac{\tan\left(\frac{A - B}{2}\right)}{\tan\left(\frac{A + B}{2}\right)}$

Because $\frac{A + B}{2} = 90^\circ - \frac{C}{2}$, we have $\tan\left(\frac{A+B}{2}\right) = \cot\left(\frac{C}{2}\right)$, yielding:

$\tan\left(\frac{A - B}{2}\right) = \frac{a - b}{a + b}\cot\left(\frac{C}{2}\right)$

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3. Half-Angle Formulas in Terms of Semi-Perimeter

Let the semi-perimeter of the triangle be:

$s = \frac{a + b + c}{2}$

Using the cosine identity $\cos A = 1 - 2\sin^2\left(\frac{A}{2}\right) = 2\cos^2\left(\frac{A}{2}\right) - 1$ in conjunction with the Law of Cosines yields the classic half-angle formulas:

1. Sine of Half-Angles

$\sin\left(\frac{A}{2}\right) = \sqrt{\frac{(s-b)(s-c)}{bc}}$

$\sin\left(\frac{B}{2}\right) = \sqrt{\frac{(s-c)(s-a)}{ca}}, \quad \sin\left(\frac{C}{2}\right) = \sqrt{\frac{(s-a)(s-b)}{ab}}$

2. Cosine of Half-Angles

$\cos\left(\frac{A}{2}\right) = \sqrt{\frac{s(s-a)}{bc}}$

$\cos\left(\frac{B}{2}\right) = \sqrt{\frac{s(s-b)}{ca}}, \quad \cos\left(\frac{C}{2}\right) = \sqrt{\frac{s(s-c)}{ab}}$

3. Tangent of Half-Angles

$\tan\left(\frac{A}{2}\right) = \frac{\sin(A/2)}{\cos(A/2)} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} = \frac{\Delta}{s(s-a)} = \frac{(s-b)(s-c)}{\Delta}$

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4. Area of a Triangle ($\Delta$)

The area of a triangle, denoted by $\Delta$, can be computed under multiple geometric representations depending on available data:

1. Two Sides and Included Angle (SAS): $\Delta = \frac{1}{2}ab\sin C = \frac{1}{2}bc\sin A = \frac{1}{2}ca\sin B$
2. One Side and Two Angles (ASA/AAS): Using $b = \frac{a\sin B}{\sin A}$: $\Delta = \frac{a^2\sin B\sin C}{2\sin A} = \frac{b^2\sin A\sin C}{2\sin B} = \frac{c^2\sin A\sin B}{2\sin C}$
3. Three Sides Known (SSS - Heron's Formula): $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$
4. Radii Relationships: $\Delta = r s = \frac{abc}{4R} = \sqrt{r \cdot r_1 \cdot r_2 \cdot r_3}$

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5. Associated Circles and Their Radii ($R, r, r_1, r_2, r_3$)

Every triangle possesses four fundamental tangent circles: one circumscribed circle, one inscribed circle, and three escribed circles (excircles).

1. The Circumcircle and Circumradius ($R$)

The circle passing through all three vertices. Its center is the circumcenter (intersection of perpendicular bisectors of sides):

$R = \frac{a}{2\sin A} = \frac{b}{2\sin B} = \frac{c}{2\sin C} = \frac{abc}{4\Delta}$

2. The Incircle and Inradius ($r$)

The circle touching all three sides internally. Its center is the incenter (intersection of internal angle bisectors):

$r = \frac{\Delta}{s} = (s-a)\tan\left(\frac{A}{2}\right) = 4R\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)$

3. The Escribed Circles and Exradii ($r_1, r_2, r_3$)

Circles tangent to one side externally and to the extensions of the other two sides. The excenter $I_1$ opposite to vertex $A$ is the intersection of the internal bisector of angle $A$ and the external bisectors of angles $B$ and $C$:

$r_1 = \frac{\Delta}{s-a} = s\tan\left(\frac{A}{2}\right) = 4R\sin\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)$

$r_2 = \frac{\Delta}{s-b} = s\tan\left(\frac{B}{2}\right) = 4R\cos\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)$

$r_3 = \frac{\Delta}{s-c} = s\tan\left(\frac{C}{2}\right) = 4R\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)$

Fundamental Radii Identities Table

ECAT Proof Shortcut (Equilateral Triangle): For an equilateral triangle with side $a$, $s = \frac{3a}{2}$, $\Delta = \frac{\sqrt{3}}{4}a^2$, and $s - a = \frac{a}{2}$:

• $r = \frac{\Delta}{s} = \frac{\sqrt{3}a^2 / 4}{3a / 2} = \frac{a}{2\sqrt{3}}$
• $R = \frac{a}{2\sin 60^\circ} = \frac{a}{\sqrt{3}} = 2r$
• $r_1 = \frac{\Delta}{s - a} = \frac{\sqrt{3}a^2 / 4}{a / 2} = \frac{\sqrt{3}a}{2} = 3r$ Hence: $R : r : r_1 = 2r : r : 3r = 2 : 1 : 3$ identically.

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6. General Solutions of Trigonometric Equations

A trigonometric equation is an equation containing one or more trigonometric functions of variable angles that is valid only for specific values of the variable (distinguishing it from an identity, which is true for all permissible angles).

Principal Solutions vs. General Solutions

• Principal Solutions: Solutions lying in the primary standard interval $[0, 2\pi)$.
• General Solutions: Because circular functions are periodic, if $\theta_0$ is a solution, an infinite family of solutions exists. The expression representing all possible solutions across $\mathbb{R}$ parameterized by an integer $n \in \mathbb{Z}$ is the general solution.

The Standard General Solution Formulas

Let $\alpha$ be the reference angle (the principal inverse value) satisfying the basic equation:

1. Sine Equation: $\sin\theta = k \quad (|k| \le 1) \implies \theta = n\pi + (-1)^n \alpha, \quad n \in \mathbb{Z}$ (For even $n = 2m$, $\theta = 2m\pi + \alpha$; for odd $n = 2m+1$, $\theta = (2m+1)\pi - \alpha = 2m\pi + (\pi - \alpha)$).
2. Cosine Equation: $\cos\theta = k \quad (|k| \le 1) \implies \theta = 2n\pi \pm \alpha, \quad n \in \mathbb{Z}$
3. Tangent Equation: $\tan\theta = k \quad (k \in \mathbb{R}) \implies \theta = n\pi + \alpha, \quad n \in \mathbb{Z}$
4. Squared Trigonometric Equations: If an equation reduces to squared form $\sin^2\theta = \sin^2\alpha$, $\cos^2\theta = \cos^2\alpha$, or $\tan^2\theta = \tan^2\alpha$: $\theta = n\pi \pm \alpha, \quad n \in \mathbb{Z}$

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7. Advanced Equation Solving Techniques & The Auxiliary Angle Method

The Auxiliary Angle Method for $a\cos\theta + b\sin\theta = c$

Linear combinations of sine and cosine are reduced to a single phase-shifted sinusoid by dividing the entire equation by the magnitude $\sqrt{a^2 + b^2}$:

$\frac{a}{\sqrt{a^2 + b^2}}\cos\theta + \frac{b}{\sqrt{a^2 + b^2}}\sin\theta = \frac{c}{\sqrt{a^2 + b^2}}$

Let $\cos\phi = \frac{a}{\sqrt{a^2 + b^2}}$ and $\sin\phi = \frac{b}{\sqrt{a^2 + b^2}}$ where $\tan\phi = \frac{b}{a}$. Then:

$\cos\theta\cos\phi + \sin\theta\sin\phi = \frac{c}{\sqrt{a^2 + b^2}} \implies \cos(\theta - \phi) = \frac{c}{\sqrt{a^2 + b^2}}$

Solvability Condition: Because $|\cos(\theta - \phi)| \le 1$, real solutions exist if and only if: $|c| \le \sqrt{a^2 + b^2} \iff c^2 \le a^2 + b^2$ If $|c| > \sqrt{a^2 + b^2}$, the equation has no real solutions.

Systematic Protocol for Solutions in $[0, 2\pi)$

1. Step 1: Solve algebraically for the primary trigonometric ratio (e.g., $\sin\theta, \cos\theta, \tan\theta$).
2. Step 2: Compute the acute reference angle $\alpha$ using the positive absolute value: $\alpha = \arcsin(|k|)$ or $\arccos(|k|)$.
3. Step 3: Identify the target quadrants where the trigonometric function has the specified sign ($+$ or $-$).
4. Step 4: Construct the terminal angles in $[0, 2\pi)$:
   • Quadrant I: $\theta = \alpha$
   • Quadrant II: $\theta = \pi - \alpha$
   • Quadrant III: $\theta = \pi + \alpha$
   • Quadrant IV: $\theta = 2\pi - \alpha$
5. Step 5: Discard any extraneous roots introduced by squaring or values that cause division by zero in original tangent or secant terms.

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8. Heights and Distances (Angles of Elevation and Depression)

The angle of elevation is measured upward from the horizontal line of sight to an object above the observer; the angle of depression is measured downward from the horizontal to an object below. Because the two horizontal lines are parallel, the angle of depression from $A$ to $B$ equals the angle of elevation from $B$ to $A$ (alternate angles).

Example 1 (one observation): From a point 40 m from the foot of a tower, the angle of elevation of the top is $60^\circ$. Height $h = 40\tan 60^\circ = 40\sqrt{3} \approx 69.3$ m.

Example 2 (two observations on a line): The angle of elevation of the top of a tower changes from $30^\circ$ to $60^\circ$ as an observer walks 20 m towards it. Let $d$ be the final distance from the foot:

$h = d\tan 60^\circ = (d + 20)\tan 30^\circ \implies d\sqrt{3} = \frac{d + 20}{\sqrt{3}} \implies 3d = d + 20 \implies d = 10 \text{ m}$

so $h = 10\sqrt{3} \approx 17.3$ m. In general, for elevations $\alpha < \beta$ measured from two points $s$ apart on a straight line towards the foot:

$h = \frac{s\tan\alpha\tan\beta}{\tan\beta - \tan\alpha}$

Example 3 (depression): From the top of a 50 m cliff, the angle of depression of a boat is $30^\circ$. Horizontal distance $= 50\cot 30^\circ = 50\sqrt{3} \approx 86.6$ m.

When the observation points are not in line with the foot of the object, draw the triangle formed by the two observers and the object and use the law of sines to find the needed side before applying $\tan$.