2.2 Trigonometry — Identities, Triangles & Spherical Trig

2.2 Trigonometry — Identities, Triangles & Spherical Trig

Trigonometry is indispensable in electrical engineering for solving plane geometric problems, surveying power line routes, and analyzing alternating current (AC) voltage and current waveforms. Spherical trigonometry extends these concepts to three-dimensional spherical surfaces, critical for terrestrial navigation, power grid satellite positioning, and global geographic calculations.

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1. Fundamental Trigonometric Identities

Pythagorean Identities

$\sin^2 \theta + \cos^2 \theta = 1$

$1 + \tan^2 \theta = \sec^2 \theta$

$1 + \cot^2 \theta = \csc^2 \theta$

Angle Addition & Subtraction Formulas

$\sin(\alpha \pm \beta) = \sin \alpha \cos \beta \pm \cos \alpha \sin \beta$

$\cos(\alpha \pm \beta) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta$

$\tan(\alpha \pm \beta) = \frac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta}$

Double-Angle & Half-Angle Formulas

Power Reduction Formulas in AC Circuits: $\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}, \quad \cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$ These identities allow electrical engineers to compute instantaneous power and average power in AC circuits.

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2. Oblique Plane Triangle Solutions

For any plane triangle with sides $a, b, c$ opposite to angles $A, B, C$:

Law of Sines

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

where $R$ is the radius of the circumscribed circle (circumradius).

Law of Cosines

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

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

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

Triangle Area Formulas

1. Side-Angle-Side (SAS): $\text{Area} = \frac{1}{2} ab \sin C = \frac{1}{2} bc \sin A = \frac{1}{2} ac \sin B$
2. Heron's Formula (Three Sides - SSS): $\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}$ where the semi-perimeter $s = \frac{a + b + c}{2}$.
3. In terms of Circumradius ($R$) and Inradius ($r$): $\text{Area} = \frac{abc}{4R} = r \cdot s$

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3. Spherical Trigonometry Fundamentals

A spherical triangle is formed on the surface of a sphere by three intersecting arcs of great circles.

Key Properties of Spherical Triangles

• Sides $a, b, c$ are measured as central angles (in degrees or radians), not linear distances.
• Interior angles $A, B, C$ satisfy: $180^\circ < A + B + C < 540^\circ$
• Spherical Excess ($E$): The amount by which the sum of interior angles exceeds $180^\circ$: $E = A + B + C - 180^\circ$
• Area of a Spherical Triangle: $\text{Area} = \frac{\pi R^2 E^\circ}{180^\circ} = R^2 E_{\text{rad}}$ where $R$ is the sphere's radius and $E^\circ$ is in degrees.

Fundamental Spherical Laws

• Spherical Law of Sines: $\frac{\sin a}{\sin A} = \frac{\sin b}{\sin B} = \frac{\sin c}{\sin C}$
• Spherical Law of Cosines for Sides: $\cos a = \cos b \cos c + \sin b \sin c \cos A$
• Spherical Law of Cosines for Angles: $\cos A = -\cos B \cos C + \sin B \sin C \cos a$

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4. Napier's Rules for Right Spherical Triangles

For a right spherical triangle where $C = 90^\circ$, John Napier devised a simple rule utilizing a "circle of five parts":

The Five Parts (Arranged in Circle)

1. Side $a$
2. Side $b$
3. Complement of Angle $A$ (denoted $\text{co-}A = 90^\circ - A$)
4. Complement of Hypotenuse $c$ (denoted $\text{co-}c = 90^\circ - c$)
5. Complement of Angle $B$ (denoted $\text{co-}B = 90^\circ - B$)

The Two Fundamental Rules

• Rule 1 (SIN-COOP): The sine of any middle part is equal to the product of the cosines of the opposite parts. $\sin(\text{middle part}) = \cos(\text{opposite}_1) \cdot \cos(\text{opposite}_2)$
• Rule 2 (SIN-TAD): The sine of any middle part is equal to the product of the tangents of the adjacent parts. $\sin(\text{middle part}) = \tan(\text{adjacent}_1) \cdot \tan(\text{adjacent}_2)$

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Solved Practice Examples

Example 1: Oblique Triangle & Area via Heron's Formula

Problem: A transmission tower guy-wire anchor layout forms a triangle with side lengths $a = 7\ \text{m}$, $b = 8\ \text{m}$, and $c = 9\ \text{m}$. Find the area of the triangular plot and the largest interior angle.

Solution:

1. Compute semi-perimeter $s$: $s = \frac{7 + 8 + 9}{2} = 12\ \text{m}$
2. Calculate area using Heron's formula: $\text{Area} = \sqrt{12(12 - 7)(12 - 8)(12 - 9)} = \sqrt{12 \times 5 \times 4 \times 3} = \sqrt{720} \approx 26.8328\ \text{m}^2$
3. Find largest angle $C$ (opposite longest side $c = 9\ \text{m}$) using Law of Cosines: $c^2 = a^2 + b^2 - 2ab \cos C \implies 9^2 = 7^2 + 8^2 - 2(7)(8) \cos C$ $81 = 49 + 64 - 112 \cos C \implies 81 = 113 - 112 \cos C$ $112 \cos C = 32 \implies \cos C = \frac{32}{112} = \frac{2}{7} \approx 0.2857$ $C = \arccos(0.2857) \approx 73.40^\circ$

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Example 2: Napier's Rule Application

Problem: In a right spherical triangle with $C = 90^\circ$, given side $a = 45^\circ$ and side $b = 60^\circ$, find the hypotenuse arc $c$.

Solution:

1. Select middle part $\text{co-}c = 90^\circ - c$. Its opposite parts on Napier's circle are sides $a$ and $b$.
2. Apply SIN-COOP rule: $\sin(\text{co-}c) = \cos(a) \cdot \cos(b)$
3. Recall $\sin(90^\circ - c) = \cos c$: $\cos c = \cos(45^\circ) \cdot \cos(60^\circ) = \left(\frac{\sqrt{2}}{2}\right) \cdot \left(\frac{1}{2}\right) = \frac{\sqrt{2}}{4} \approx 0.35355$
4. Solve for $c$: $c = \arccos(0.35355) \approx 69.30^\circ$