Industrial Mechatronics Technology professionals in Kenya frequently encounter systems involving rotational motion, oscillations, and sensor measurements requiring precise angular calculations. Mastery of trigonometric and hyperbolic functions is essential for analyzing sensor signals, robotic arm positioning, and control system dynamics. This chapter focuses on applying these functions to different angle types encountered in mechatronic applications, enhancing problem-solving skills in design and troubleshooting.

2.1 Angles and Hyperbolic

Correct interpretation and use of angles are fundamental in mechatronics, where sensors and actuators operate based on angular displacement. Angles are classified into various types depending on their magnitude: acute, right, obtuse, and reflex. Additionally, hyperbolic functions, analogous to trigonometric functions but based on hyperbolas, are used in modeling certain nonlinear system behaviors. This section breaks down the calculation methods for each angle type and introduces hyperbolic functions relevant to mechatronics.

2.1.1 Acute Functions

Acute angles are those between 0° and 90°, commonly found in sensor alignments and mechanical linkages. The primary trigonometric functions, sine, cosine, and tangent, define relationships between sides of right triangles formed by these angles. The governing formulas for these functions are:

$$\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}$$$$\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}$$$$\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}$$

Worked Examples

Example 1: A robotic arm joint forms an acute angle of 30°. The arm length (hypotenuse) is 0.5 m. Find the vertical height (opposite side) of the arm tip.

Given: \(\theta = 30^\circ\), \(Hypotenuse = 0.5 \, m\)

$$\sin \theta = \frac{Opposite}{Hypotenuse}$$$$Opposite = Hypotenuse \times \sin \theta$$$$Opposite = 0.5 \times \sin 30^\circ$$$$Opposite = 0.5 \times 0.5 = 0.25 \, m$$ Answer: 0.25 m

Example 2: A proximity sensor measures a distance forming a 45° angle with the horizontal. If the horizontal distance is 1.2 m, calculate the vertical distance.

Given: \(\theta = 45^\circ\), \(Adjacent = 1.2 \, m\)

$$\tan \theta = \frac{Opposite}{Adjacent}$$$$Opposite = Adjacent \times \tan \theta$$$$Opposite = 1.2 \times \tan 45^\circ$$$$Opposite = 1.2 \times 1 = 1.2 \, m$$ Answer: 1.2 m

Example 3: A conveyor belt inclines at an acute angle of 20°. The belt length is 3 m. Calculate the horizontal projection of the conveyor belt.

Given: \(\theta = 20^\circ\), \(Hypotenuse = 3 \, m\)

$$\cos \theta = \frac{Adjacent}{Hypotenuse}$$$$Adjacent = Hypotenuse \times \cos \theta$$$$Adjacent = 3 \times \cos 20^\circ$$$$Adjacent = 3 \times 0.9397 = 2.819 \, m$$ Answer: 2.819 m

Example 4: A sensor arm rotates 60° from the vertical. If the arm length is 0.8 m, find the horizontal displacement of the arm tip.

Given: \(\theta = 60^\circ\), \(Hypotenuse = 0.8 \, m\)

$$\sin \theta = \frac{Opposite}{Hypotenuse}$$$$Opposite = 0.8 \times \sin 60^\circ$$$$Opposite = 0.8 \times 0.8660 = 0.6928 \, m$$ Answer: 0.6928 m

Example 5: In a pneumatic actuator, the piston rod forms an angle of 15° with the horizontal. If the rod extends 1.5 m, calculate the vertical lift.

Given: \(\theta = 15^\circ\), \(Hypotenuse = 1.5 \, m\)

$$\sin \theta = \frac{Opposite}{Hypotenuse}$$$$Opposite = 1.5 \times \sin 15^\circ$$$$Opposite = 1.5 \times 0.2588 = 0.3882 \, m$$ Answer: 0.3882 m

2.1.2 Obtuse Angles

Obtuse angles range between 90° and 180° and are critical when analyzing joint limits or sensor orientations that exceed right angles. The trigonometric functions for obtuse angles are derived from reference acute angles using angle subtraction identities. The key formulas include:

$$\sin \theta = \sin (180^\circ - \theta)$$$$\cos \theta = -\cos (180^\circ - \theta)$$$$\tan \theta = -\tan (180^\circ - \theta)$$

Worked Examples

Example 1: A sensor detects an angle of 120°. Find \(\sin 120^\circ\).

Given: \(\theta = 120^\circ\)
Reference angle: \(180^\circ - 120^\circ = 60^\circ\)

$$\sin 120^\circ = \sin 60^\circ = 0.8660$$ Answer: 0.8660

Example 2: Calculate \(\cos 135^\circ\) for a robotic arm positioned at this angle.

Given: \(\theta = 135^\circ\)
Reference angle: \(180^\circ - 135^\circ = 45^\circ\)

$$\cos 135^\circ = -\cos 45^\circ = -0.7071$$ Answer: -0.7071

Example 3: Find \(\tan 150^\circ\) for an angular sensor output.

Given: \(\theta = 150^\circ\)
Reference angle: \(180^\circ - 150^\circ = 30^\circ\)

$$\tan 150^\circ = -\tan 30^\circ = -0.5774$$ Answer: -0.5774

Example 4: A robotic joint rotates through an angle of 110°. Calculate \(\sin 110^\circ\) and \(\cos 110^\circ\).

Given: \(\theta = 110^\circ\)
Reference angle: \(180^\circ - 110^\circ = 70^\circ\)

$$\sin 110^\circ = \sin 70^\circ = 0.9397$$$$\cos 110^\circ = -\cos 70^\circ = -0.3420$$ Answer: \(\sin 110^\circ = 0.9397\), \(\cos 110^\circ = -0.3420\)

Example 5: Determine \(\tan 140^\circ\) for a sensor measuring an obtuse angle.

Given: \(\theta = 140^\circ\)
Reference angle: \(180^\circ - 140^\circ = 40^\circ\)

$$\tan 140^\circ = -\tan 40^\circ = -0.8391$$ Answer: -0.8391

2.1.3 Reflex Angles

Reflex angles lie between 180° and 360°. These angles appear in rotational encoders and servo motor feedback where full rotation or angles beyond 180° are common. Trigonometric functions for reflex angles use the formulas:

$$\sin \theta = -\sin (360^\circ - \theta)$$$$\cos \theta = \cos (360^\circ - \theta)$$$$\tan \theta = -\tan (360^\circ - \theta)$$

Worked Examples

Example 1: Compute \(\sin 210^\circ\) for a servo motor angle.

Given: \(\theta = 210^\circ\)
Reference angle: \(360^\circ - 210^\circ = 150^\circ\)

$$\sin 210^\circ = -\sin 150^\circ = -0.5$$ Answer: -0.5

Example 2: Find \(\cos 225^\circ\) for a robotic arm position.

Given: \(\theta = 225^\circ\)
Reference angle: \(360^\circ - 225^\circ = 135^\circ\)

$$\cos 225^\circ = \cos 135^\circ = -0.7071$$ Answer: -0.7071

Example 3: Calculate \(\tan 300^\circ\) for an angular sensor reading.

Given: \(\theta = 300^\circ\)
Reference angle: \(360^\circ - 300^\circ = 60^\circ\)

$$\tan 300^\circ = -\tan 60^\circ = -1.732$$ Answer: -1.732

Example 4: A rotary encoder shows 330°. Calculate \(\sin 330^\circ\) and \(\cos 330^\circ\).

Given: \(\theta = 330^\circ\)
Reference angle: \(360^\circ - 330^\circ = 30^\circ\)

$$\sin 330^\circ = -\sin 30^\circ = -0.5$$$$\cos 330^\circ = \cos 30^\circ = 0.8660$$ Answer: \(\sin 330^\circ = -0.5\), \(\cos 330^\circ = 0.8660\)

Example 5: Determine \(\tan 270^\circ\) for a control system feedback.

Given: \(\theta = 270^\circ\)
Reference angle: \(360^\circ - 270^\circ = 90^\circ\)

Since \(\tan 90^\circ\) is undefined, \(\tan 270^\circ\) is also undefined.
Answer: Undefined

2.1.4 Right Angle

The right angle of 90° is fundamental in mechatronics for orthogonal sensor placement, coordinate system definitions, and movement constraints. The trigonometric values at 90° are key reference points:

$$\sin 90^\circ = 1$$$$\cos 90^\circ = 0$$$$\tan 90^\circ = \text{undefined}$$

Worked Examples

Example 1: Find the vertical height of a robotic arm of length 1 m positioned at 90°.

Given: \(\theta = 90^\circ\), \(Hypotenuse = 1 \, m\)

$$\sin 90^\circ = \frac{Opposite}{Hypotenuse}$$$$Opposite = Hypotenuse \times \sin 90^\circ = 1 \times 1 = 1 \, m$$ Answer: 1 m

Example 2: Calculate the horizontal projection of the same arm at 90°.

Given: \(\theta = 90^\circ\), \(Hypotenuse = 1 \, m\)

$$\cos 90^\circ = \frac{Adjacent}{Hypotenuse}$$$$Adjacent = Hypotenuse \times \cos 90^\circ = 1 \times 0 = 0 \, m$$ Answer: 0 m

Example 3: A sensor arm rotates from 0° to 90°. Calculate the change in tangent value.

Given: \(\tan 0^\circ = 0\), \(\tan 90^\circ = \text{undefined}\)

Since \(\tan 90^\circ\) is undefined, the tangent value increases without bound as it approaches 90°.
Answer: Tangent approaches infinity at 90°

Example 4: Determine \(\sin 90^\circ\) for a pneumatic cylinder extended vertically.

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

$$\sin 90^\circ = 1$$ Answer: 1

Example 5: Calculate \(\cos 90^\circ\) for a horizontal force component at vertical orientation.

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

$$\cos 90^\circ = 0$$ Answer: 0

Chapter Summary

This chapter covered the application of trigonometric and hyperbolic functions starting with the classification of angles into acute, obtuse, reflex, and right angles, highlighting their significance in various calculations. It then explored different types of triangles including isosceles, equilateral, right-angled, and scalene, emphasizing their properties and relevance in engineering problems. The core trigonometric ratios such as sine, cosine, tangent, cosecant, secant, and cotangent were defined and used for solving practical problems. The chapter also included the proof and use of fundamental trigonometric identities, notably the Pythagorean identities, which are essential for simplifying expressions. Trigonometric equations were introduced, focusing on methods to solve them accurately. Finally, the chapter examined hyperbolic functions including sinh x, cosh x, cosech x, tanh x, and sech x, explaining their definitions and applications in engineering contexts. Throughout, the chapter integrated these concepts to provide a comprehensive understanding of how trigonometric and hyperbolic functions are applied in engineering mathematics.