In industrial control and installation, trigonometric and hyperbolic functions are vital for analyzing signal waveforms, designing control systems, and calculating component responses. Mastery of the half-angle formula allows professionals to simplify complex expressions and solve equations involving angles, especially in alternating current (AC) circuit analysis and mechanical linkages. This chapter focuses on the half-angle formula, providing practical methods and worked examples relevant to real-life industrial and engineering tasks in Kenya.

2.1 Half-Angle Formula

The half-angle formulas are essential for expressing trigonometric functions of half an angle in terms of the cosine or sine of the original angle. These formulas simplify calculations in signal analysis, control system design, and mechanical component positioning.

2.1.1 The Half-Angle Formulas for Sine, Cosine, and Tangent

The half-angle formulas allow you to find the sine, cosine, and tangent of half an angle using the cosine of the original angle. These are crucial when dealing with phase shifts in AC signals, calculating resultant forces, or analyzing mechanical linkages.

The main half-angle formulas are:$$ \sin\frac{\theta}{2} = \pm\sqrt{\frac{1 - \cos\theta}{2}} $$$$ \cos\frac{\theta}{2} = \pm\sqrt{\frac{1 + \cos\theta}{2}} $$$$ \tan\frac{\theta}{2} = \pm\sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}} $$

The sign (\(+\) or \(-\)) depends on the quadrant in which \(\frac{\theta}{2}\) lies.

Worked Examples

Example 1:
A technician needs to find \(\sin\frac{\theta}{2}\) if \(\theta = 60^\circ\), for setting a sensor at the correct angle.

Given: \(\theta = 60^\circ\)

$$ \sin\frac{\theta}{2} = \sqrt{\frac{1 - \cos\theta}{2}} $$$$ \sin\frac{60^\circ}{2} = \sqrt{\frac{1 - \cos60^\circ}{2}} $$$$ \sin30^\circ = \sqrt{\frac{1 - 0.5}{2}} $$$$ \sin30^\circ = \sqrt{\frac{0.5}{2}} $$$$ \sin30^\circ = \sqrt{0.25} $$ Answer: \(0.5\)

Example 2:
Determine \(\cos\frac{\theta}{2}\) for \(\theta = 120^\circ\), a common calculation when adjusting actuator rods.

Given: \(\theta = 120^\circ\)

$$ \cos\frac{\theta}{2} = \sqrt{\frac{1 + \cos\theta}{2}} $$$$ \cos\frac{120^\circ}{2} = \sqrt{\frac{1 + \cos120^\circ}{2}} $$$$ \cos60^\circ = \sqrt{\frac{1 + (-0.5)}{2}} $$$$ \cos60^\circ = \sqrt{\frac{0.5}{2}} $$$$ \cos60^\circ = \sqrt{0.25} $$ Answer: \(0.5\)

Example 3:
Find \(\tan\frac{\theta}{2}\) for \(\theta = 150^\circ\), as required when calculating the angular position for a rotating platform.

Given: \(\theta = 150^\circ\)

$$ \tan\frac{\theta}{2} = \sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}} $$$$ \tan\frac{150^\circ}{2} = \sqrt{\frac{1 - \cos150^\circ}{1 + \cos150^\circ}} $$$$ \tan75^\circ = \sqrt{\frac{1 - (-0.8660)}{1 + (-0.8660)}} $$$$ \tan75^\circ = \sqrt{\frac{1.8660}{0.1340}} $$$$ \tan75^\circ = \sqrt{13.93} $$ Answer: \(3.735\)

Example 4:
A relay in a control panel must be set at an angle such that \(\theta = 210^\circ\). Find \(\sin\frac{\theta}{2}\).

Given: \(\theta = 210^\circ\)

$$ \sin\frac{\theta}{2} = -\sqrt{\frac{1 - \cos\theta}{2}} $$$$ \sin\frac{210^\circ}{2} = -\sqrt{\frac{1 - \cos210^\circ}{2}} $$$$ \sin105^\circ = -\sqrt{\frac{1 - (-0.8660)}{2}} $$$$ \sin105^\circ = -\sqrt{\frac{1.8660}{2}} $$$$ \sin105^\circ = -\sqrt{0.9330} $$ Answer: \(-0.966\)

Example 5:
Calculate \(\cos\frac{\theta}{2}\) for \(\theta = 315^\circ\), used in determining the orientation of a sensor in an industrial conveyor system.

Given: \(\theta = 315^\circ\)

$$ \cos\frac{\theta}{2} = -\sqrt{\frac{1 + \cos\theta}{2}} $$$$ \cos\frac{315^\circ}{2} = -\sqrt{\frac{1 + \cos315^\circ}{2}} $$$$ \cos157.5^\circ = -\sqrt{\frac{1 + (-0.924)}{2}} $$$$ \cos157.5^\circ = -\sqrt{\frac{0.076}{2}} $$$$ \cos157.5^\circ = -\sqrt{0.038} $$ Answer: \(-0.195\)

2.1.2 Application of Half-Angle Formulas in Industrial Control

Half-angle formulas are frequently used in industrial control and installation to determine phasor relationships, analyze alternating currents, and adjust machinery that operates with rotational or oscillatory motion. These formulas simplify the process of finding unknown angles and lengths in control arms, sensor mounts, and linkage systems.

Worked Examples

Example 1:
In a water treatment plant, a technician measures a phase difference of \(\theta = 90^\circ\) between two AC signals. Find \(\tan\frac{\theta}{2}\) to help set the phase relay.

Given: \(\theta = 90^\circ\)

$$ \tan\frac{\theta}{2} = \sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}} $$$$ \tan\frac{90^\circ}{2} = \sqrt{\frac{1 - 0}{1 + 0}} $$$$ \tan45^\circ = \sqrt{\frac{1}{1}} $$$$ \tan45^\circ = \sqrt{1} $$ Answer: \(1\)

Example 2:
A robotic arm in a university laboratory is set to rotate through \(\theta = 45^\circ\). Find \(\sin\frac{\theta}{2}\) to calculate the required position for the arm.

Given: \(\theta = 45^\circ\)

$$ \sin\frac{\theta}{2} = \sqrt{\frac{1 - \cos\theta}{2}} $$$$ \sin\frac{45^\circ}{2} = \sqrt{\frac{1 - 0.7071}{2}} $$$$ \sin22.5^\circ = \sqrt{\frac{0.2929}{2}} $$$$ \sin22.5^\circ = \sqrt{0.14645} $$ Answer: \(0.383\)

Example 3:
In a county government office building, a backup generator's flywheel is positioned at \(\theta = 270^\circ\). Find \(\cos\frac{\theta}{2}\).

Given: \(\theta = 270^\circ\)

$$ \cos\frac{\theta}{2} = -\sqrt{\frac{1 + \cos\theta}{2}} $$$$ \cos\frac{270^\circ}{2} = -\sqrt{\frac{1 + \cos270^\circ}{2}} $$$$ \cos135^\circ = -\sqrt{\frac{1 + (-0.7071)}{2}} $$$$ \cos135^\circ = -\sqrt{\frac{0.2929}{2}} $$$$ \cos135^\circ = -\sqrt{0.14645} $$ Answer: \(-0.383\)

Example 4:
A technician in a retail business is aligning a conveyor system at \(\theta = 180^\circ\). What is \(\tan\frac{\theta}{2}\)?

Given: \(\theta = 180^\circ\)

$$ \tan\frac{\theta}{2} = \sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}} $$$$ \tan90^\circ = \sqrt{\frac{1 - (-1)}{1 + (-1)}} $$$$ \tan90^\circ = \sqrt{\frac{2}{0}} $$

Since division by zero is undefined, \(\tan90^\circ\) is undefined. This shows the importance of checking the denominator in the formula.

Example 5:
A hospital technician needs to adjust a sensor to an angle \(\theta = 240^\circ\). Find \(\sin\frac{\theta}{2}\).

Given: \(\theta = 240^\circ\)

$$ \sin\frac{\theta}{2} = -\sqrt{\frac{1 - \cos\theta}{2}} $$$$ \sin120^\circ = -\sqrt{\frac{1 - (-0.5)}{2}} $$$$ \sin120^\circ = -\sqrt{\frac{1.5}{2}} $$$$ \sin120^\circ = -\sqrt{0.75} $$ Answer: \(-0.866\)

2.1.3 Using Half-Angle Formulas with Radians

In industrial contexts, angles are often measured in radians, especially in automation and robotics. The half-angle formulas are applied in the same way, with the input angle in radians.

Worked Examples

Example 1:
A PLC-controlled valve opens through \(\theta = \frac{\pi}{2}\) radians. Find \(\sin\frac{\theta}{2}\).

Given: \(\theta = \frac{\pi}{2}\) radians

$$ \sin\frac{\theta}{2} = \sqrt{\frac{1 - \cos\theta}{2}} $$$$ \sin\frac{\pi}{4} = \sqrt{\frac{1 - \cos\frac{\pi}{2}}{2}} $$$$ \sin\frac{\pi}{4} = \sqrt{\frac{1 - 0}{2}} $$$$ \sin\frac{\pi}{4} = \sqrt{0.5} $$ Answer: \(0.707\)

Example 2:
A motor shaft rotates through \(\theta = \frac{2\pi}{3}\) radians. Find \(\cos\frac{\theta}{2}\).

Given: \(\theta = \frac{2\pi}{3}\) radians

$$ \cos\frac{\theta}{2} = \sqrt{\frac{1 + \cos\theta}{2}} $$$$ \cos\frac{\pi}{3} = \sqrt{\frac{1 + \cos\frac{2\pi}{3}}{2}} $$$$ \cos\frac{\pi}{3} = \sqrt{\frac{1 + (-0.5)}{2}} $$$$ \cos\frac{\pi}{3} = \sqrt{0.25} $$ Answer: \(0.5\)

Example 3:
A technician sets an actuator to \(\theta = \frac{5\pi}{6}\) radians. Calculate \(\tan\frac{\theta}{2}\).

Given: \(\theta = \frac{5\pi}{6}\) radians

$$ \tan\frac{\theta}{2} = \sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}} $$$$ \tan\frac{5\pi}{12} = \sqrt{\frac{1 - \cos\frac{5\pi}{6}}{1 + \cos\frac{5\pi}{6}}} $$$$ \tan\frac{5\pi}{12} = \sqrt{\frac{1 - (-0.866)}{1 + (-0.866)}} $$$$ \tan\frac{5\pi}{12} = \sqrt{\frac{1.866}{0.134}} $$$$ \tan\frac{5\pi}{12} = \sqrt{13.93} $$ Answer: \(3.735\)

Example 4:
A conveyor belt completes an angle of \(\theta = \pi\) radians in its rotation. Find \(\sin\frac{\theta}{2}\).

Given: \(\theta = \pi\) radians

$$ \sin\frac{\theta}{2} = \sqrt{\frac{1 - \cos\theta}{2}} $$$$ \sin\frac{\pi}{2} = \sqrt{\frac{1 - (-1)}{2}} $$$$ \sin\frac{\pi}{2} = \sqrt{1} $$ Answer: \(1\)

Example 5:
A technician needs to find \(\cos\frac{\theta}{2}\) for \(\theta = \frac{7\pi}{4}\) radians.

Given: \(\theta = \frac{7\pi}{4}\) radians

$$ \cos\frac{\theta}{2} = -\sqrt{\frac{1 + \cos\theta}{2}} $$$$ \cos\frac{7\pi}{8} = -\sqrt{\frac{1 + \cos\frac{7\pi}{4}}{2}} $$$$ \cos\frac{7\pi}{8} = -\sqrt{\frac{1 + 0.707}{2}} $$$$ \cos\frac{7\pi}{8} = -\sqrt{0.8535} $$ Answer: \(-0.924\)

Chapter Summary

This chapter explored advanced trigonometric and hyperbolic concepts essential for civil engineering applications. It began with the half-angle and factor formulas, providing methods to simplify and solve complex trigonometric expressions. The key trigonometric functions were reviewed, followed by parametric equations that describe curves using angles as parameters. The chapter distinguished between relative and absolute measures and demonstrated how to perform accurate calculations using these concepts. The meaning and properties of hyperbolic functions were introduced, including their evaluation and core identities. Osborne's Rule was discussed as a tool for converting trigonometric identities into hyperbolic forms, and the solution of equations such as \(a\cosh x + b\sinh x = C\) was addressed. The concept of one-to-one relationships in functions was explained, leading to a study of inverse functions for both general and trigonometric cases, as well as their graphical representations. Finally, the chapter concluded with the introduction and analysis of inverse hyperbolic functions, solidifying their role in engineering mathematics.