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This chapter focuses on the application of trigonometric and hyperbolic functions in Automotive Mechatronics, an essential skill for professionals working with vehicle sensors, actuators, and control systems. Understanding half-angle formulas enables precise calculation of angles and signal processing, which are critical in diagnostics, suspension geometry, and robotic arm positioning. Mastery of these formulas improves the ability to analyze and design systems that rely on angular measurements and transformations in automotive mechatronics.
The half-angle formulas allow calculation of trigonometric functions of half an angle from the functions of the whole angle. These formulas are especially useful in Automotive Mechatronics when dealing with angular resolution in sensors or when resolving complex angles into simpler components. Using half-angle identities simplifies calculations required in steering geometry, camshaft timing, and signal modulation in vehicle control units.
The half-angle formula for sine expresses \(\sin \frac{\theta}{2}\) in terms of \(\cos \theta\). This is useful when the cosine of an angle is known, and the sine of half that angle is needed, such as in calculating intermediate shaft angles in gear trains or resolving forces in suspension linkages.
The half-angle formula for sine is:
$$\sin \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{2}}$$
The sign depends on the quadrant in which \(\frac{\theta}{2}\) lies.
Example 1: A steering sensor detects an angle \(\theta = 60^\circ\). Calculate \(\sin 30^\circ\) using the half-angle formula.
Given: \(\theta = 60^\circ\)
$$\sin \frac{60^\circ}{2} = \sin 30^\circ = \pm \sqrt{\frac{1 - \cos 60^\circ}{2}}$$
Calculate \(\cos 60^\circ\):
$$\cos 60^\circ = 0.5$$
Substitute:
$$\sin 30^\circ = \pm \sqrt{\frac{1 - 0.5}{2}}$$
$$= \pm \sqrt{\frac{0.5}{2}} = \pm \sqrt{0.25}$$
$$= \pm 0.5$$
Since \(30^\circ\) is in the first quadrant, sine is positive.
Answer: 0.5
Example 2: In a camshaft angle measurement, \(\theta = 120^\circ\). Calculate \(\sin 60^\circ\) using the half-angle formula.
Given: \(\theta = 120^\circ\)
$$\sin \frac{120^\circ}{2} = \sin 60^\circ = \pm \sqrt{\frac{1 - \cos 120^\circ}{2}}$$
Calculate \(\cos 120^\circ\):
$$\cos 120^\circ = -0.5$$
Substitute:
$$\sin 60^\circ = \pm \sqrt{\frac{1 - (-0.5)}{2}}$$
$$= \pm \sqrt{\frac{1 + 0.5}{2}} = \pm \sqrt{\frac{1.5}{2}}$$
$$= \pm \sqrt{0.75} = \pm 0.866$$
Since \(60^\circ\) is in the first quadrant, sine is positive.
Answer: 0.866
Example 3: A robotic arm joint rotates through \(\theta = 225^\circ\). Calculate \(\sin 112.5^\circ\) using the half-angle formula.
Given: \(\theta = 225^\circ\)
$$\sin \frac{225^\circ}{2} = \sin 112.5^\circ = \pm \sqrt{\frac{1 - \cos 225^\circ}{2}}$$
Calculate \(\cos 225^\circ\):
$$\cos 225^\circ = -\frac{\sqrt{2}}{2} \approx -0.707$$
Substitute:
$$\sin 112.5^\circ = \pm \sqrt{\frac{1 - (-0.707)}{2}}$$
$$= \pm \sqrt{\frac{1 + 0.707}{2}} = \pm \sqrt{\frac{1.707}{2}}$$
$$= \pm \sqrt{0.8535} = \pm 0.924$$
Since \(112.5^\circ\) lies in the second quadrant where sine is positive:
Answer: 0.924
Example 4: In a vehicle suspension angle calculation, \(\theta = 300^\circ\). Find \(\sin 150^\circ\) using the half-angle formula.
Given: \(\theta = 300^\circ\)
$$\sin \frac{300^\circ}{2} = \sin 150^\circ = \pm \sqrt{\frac{1 - \cos 300^\circ}{2}}$$
Calculate \(\cos 300^\circ\):
$$\cos 300^\circ = 0.5$$
Substitute:
$$\sin 150^\circ = \pm \sqrt{\frac{1 - 0.5}{2}}$$
$$= \pm \sqrt{\frac{0.5}{2}} = \pm \sqrt{0.25}$$
$$= \pm 0.5$$
Since \(150^\circ\) lies in the second quadrant, sine is positive.
Answer: 0.5
Example 5: Calculate \(\sin 7.5^\circ\) given \(\theta = 15^\circ\) for a precision sensor calibration.
Given: \(\theta = 15^\circ\)
$$\sin \frac{15^\circ}{2} = \sin 7.5^\circ = \pm \sqrt{\frac{1 - \cos 15^\circ}{2}}$$
Calculate \(\cos 15^\circ\):
Using cosine approximation or calculator:
$$\cos 15^\circ \approx 0.9659$$
Substitute:
$$\sin 7.5^\circ = \pm \sqrt{\frac{1 - 0.9659}{2}}$$
$$= \pm \sqrt{\frac{0.0341}{2}} = \pm \sqrt{0.01705}$$
$$= \pm 0.1305$$
Since \(7.5^\circ\) is in the first quadrant, sine is positive.
Answer: 0.1305
The half-angle formula for cosine is useful to find the cosine of half an angle when the cosine of the whole angle is known. This is often applied in calculating angular displacements in sensors and actuators that operate within restricted angle ranges, such as throttle position sensors or steering angle sensors.
The half-angle formula for cosine is:
$$\cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}}$$
The sign depends on the quadrant of \(\frac{\theta}{2}\).
Example 1: Calculate \(\cos 30^\circ\) given \(\theta = 60^\circ\).
Given: \(\theta = 60^\circ\)
$$\cos \frac{60^\circ}{2} = \cos 30^\circ = \pm \sqrt{\frac{1 + \cos 60^\circ}{2}}$$
Calculate \(\cos 60^\circ\):
$$\cos 60^\circ = 0.5$$
Substitute:
$$\cos 30^\circ = \pm \sqrt{\frac{1 + 0.5}{2}} = \pm \sqrt{\frac{1.5}{2}}$$
$$= \pm \sqrt{0.75} = \pm 0.866$$
Since \(30^\circ\) lies in the first quadrant, cosine is positive.
Answer: 0.866
Example 2: Find \(\cos 75^\circ\) given \(\theta = 150^\circ\).
Given: \(\theta = 150^\circ\)
$$\cos \frac{150^\circ}{2} = \cos 75^\circ = \pm \sqrt{\frac{1 + \cos 150^\circ}{2}}$$
Calculate \(\cos 150^\circ\):
$$\cos 150^\circ = -0.866$$
Substitute:
$$\cos 75^\circ = \pm \sqrt{\frac{1 - 0.866}{2}} = \pm \sqrt{\frac{0.134}{2}}$$
$$= \pm \sqrt{0.067} = \pm 0.259$$
Since \(75^\circ\) is in the first quadrant, cosine is positive.
Answer: 0.259
Example 3: A camshaft sensor reads \(\theta = 225^\circ\). Calculate \(\cos 112.5^\circ\).
Given: \(\theta = 225^\circ\)
$$\cos \frac{225^\circ}{2} = \cos 112.5^\circ = \pm \sqrt{\frac{1 + \cos 225^\circ}{2}}$$
Calculate \(\cos 225^\circ\):
$$\cos 225^\circ = -0.707$$
Substitute:
$$\cos 112.5^\circ = \pm \sqrt{\frac{1 - 0.707}{2}} = \pm \sqrt{\frac{0.293}{2}}$$
$$= \pm \sqrt{0.1465} = \pm 0.382$$
Since \(112.5^\circ\) is in the second quadrant, cosine is negative.
Answer: -0.382
Example 4: Calculate \(\cos 7.5^\circ\) given \(\theta = 15^\circ\) for an angular sensor calibration.
Given: \(\theta = 15^\circ\)
$$\cos \frac{15^\circ}{2} = \cos 7.5^\circ = \pm \sqrt{\frac{1 + \cos 15^\circ}{2}}$$
Calculate \(\cos 15^\circ\):
$$\cos 15^\circ = 0.9659$$
Substitute:
$$\cos 7.5^\circ = \pm \sqrt{\frac{1 + 0.9659}{2}} = \pm \sqrt{\frac{1.9659}{2}}$$
$$= \pm \sqrt{0.98295} = \pm 0.991$$
Since \(7.5^\circ\) is in the first quadrant, cosine is positive.
Answer: 0.991
Example 5: Find \(\cos 150^\circ\) given \(\theta = 300^\circ\) for a suspension angle calculation.
Given: \(\theta = 300^\circ\)
$$\cos \frac{300^\circ}{2} = \cos 150^\circ = \pm \sqrt{\frac{1 + \cos 300^\circ}{2}}$$
Calculate \(\cos 300^\circ\):
$$\cos 300^\circ = 0.5$$
Substitute:
$$\cos 150^\circ = \pm \sqrt{\frac{1 + 0.5}{2}} = \pm \sqrt{\frac{1.5}{2}}$$
$$= \pm \sqrt{0.75} = \pm 0.866$$
Since \(150^\circ\) is in the second quadrant, cosine is negative.
Answer: -0.866
The half-angle formula for tangent is particularly useful in Automotive Mechatronics for calculating angular velocities and sensor signal phases when the angle is halved. It provides a direct way to compute \(\tan \frac{\theta}{2}\) from either sine and cosine of \(\theta\) or from cosine alone.
The half-angle formulas for tangent are:
$$\tan \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} = \frac{\sin \theta}{1 + \cos \theta} = \frac{1 - \cos \theta}{\sin \theta}$$
The sign depends on the quadrant of \(\frac{\theta}{2}\).
Example 1: Calculate \(\tan 30^\circ\) given \(\theta = 60^\circ\) using the formula \(\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}\).
Given: \(\theta = 60^\circ\)
Calculate \(\sin 60^\circ\) and \(\cos 60^\circ\):
$$\sin 60^\circ = 0.866$$
$$\cos 60^\circ = 0.5$$
Substitute:
$$\tan 30^\circ = \frac{0.866}{1 + 0.5} = \frac{0.866}{1.5} = 0.577$$
Answer: 0.577
Example 2: Find \(\tan 75^\circ\) given \(\theta = 150^\circ\) using the formula \(\tan \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}}\).
Given: \(\theta = 150^\circ\)
Calculate \(\cos 150^\circ\):
$$\cos 150^\circ = -0.866$$
Substitute:
$$\tan 75^\circ = \sqrt{\frac{1 - (-0.866)}{1 + (-0.866)}} = \sqrt{\frac{1 + 0.866}{1 - 0.866}} = \sqrt{\frac{1.866}{0.134}}$$
Calculate numerator and denominator:
$$\frac{1.866}{0.134} = 13.93$$
$$\sqrt{13.93} = 3.735$$
Since \(75^\circ\) is in the first quadrant, tangent is positive.
Answer: 3.735
Example 3: A crankshaft sensor measures \(\theta = 225^\circ\). Calculate \(\tan 112.5^\circ\) using \(\tan \frac{\theta}{2} = \frac{1 - \cos \theta}{\sin \theta}\).
Given: \(\theta = 225^\circ\)
Calculate \(\cos 225^\circ\) and \(\sin 225^\circ\):
$$\cos 225^\circ = -0.707$$
$$\sin 225^\circ = -0.707$$
Substitute:
$$\tan 112.5^\circ = \frac{1 - (-0.707)}{-0.707} = \frac{1 + 0.707}{-0.707} = \frac{1.707}{-0.707} = -2.414$$
Since \(112.5^\circ\) is in the second quadrant, tangent is negative.
Answer: -2.414
Example 4: Calculate \(\tan 7.5^\circ\) given \(\theta = 15^\circ\) using \(\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}\).
Given: \(\theta = 15^\circ\)
Calculate \(\sin 15^\circ\) and \(\cos 15^\circ\):
$$\sin 15^\circ = 0.2588$$
$$\cos 15^\circ = 0.9659$$
Substitute:
$$\tan 7.5^\circ = \frac{0.2588}{1 + 0.9659} = \frac{0.2588}{1.9659} = 0.1317$$
Answer: 0.1317
Example 5: For a suspension angle \(\theta = 300^\circ\), find \(\tan 150^\circ\) using \(\tan \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}}\).
Given: \(\theta = 300^\circ\)
Calculate \(\cos 300^\circ\):
$$\cos 300^\circ = 0.5$$
Substitute:
$$\tan 150^\circ = \sqrt{\frac{1 - 0.5}{1 + 0.5}} = \sqrt{\frac{0.5}{1.5}} = \sqrt{0.333} = 0.577$$
Since \(150^\circ\) is in the second quadrant, tangent is negative.
Answer: -0.577
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Create a free accountThis chapter began by exploring the half-angle formula and factor formula, providing essential tools for simplifying and transforming trigonometric expressions. It then reviewed the fundamental trigonometric functions and their application in parametric equations, which describe curves using parameters. The chapter addressed relative and absolute measures assessments, followed by detailed calculations of these measures in various contexts. The focus shifted to hyperbolic functions with a clear definition and exercises to reinforce understanding, before examining their properties and methods for evaluation using hyperbolic identities. Osborne's Rule was introduced as a technique to relate trigonometric and hyperbolic functions. The equation involving the sum of hyperbolic sine terms, Ashx + Bshx = C, was analyzed for its solution approach. Concepts of one-to-one relationships in functions were discussed to establish the basis for inverse functions, which led to studying inverse functions specifically for trigonometric functions along with their graphical representations. The chapter concluded by defining inverse hyperbolic functions and their significance in solving equations involving hyperbolic terms.
Calculate \(\sin 15^\circ\) using the half-angle formula given \(\sin 30^\circ = 0.5\). (2 marks)
Simplify \(\sin 75^\circ + \sin 15^\circ\) using the factor formula. (2 marks)
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