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Trigonometric and hyperbolic functions are fundamental in chemical engineering calculations involving periodic phenomena, oscillations, wave propagation, and thermal processes. In Kenya's chemical plants and research laboratories, these functions help model reactor behavior, fluid flow, and heat exchange. This chapter develops the understanding of trigonometric identities and equations, followed by the introduction and application of hyperbolic functions, which arise naturally in chemical engineering problems such as diffusion and reaction kinetics.
Trigonometric functions relate angles to ratios of sides in right-angled triangles and extend to periodic functions describing oscillations and waves. Chemical engineers use these functions to analyze alternating currents, vibrations in equipment, and phase angles in control systems.
The primary trigonometric functions are sine, cosine, and tangent, defined for an angle \(\theta\) as:
$$\sin \theta = \frac{\text{Opposite side}}{\text{Hypotenuse}}$$
$$\cos \theta = \frac{\text{Adjacent side}}{\text{Hypotenuse}}$$
$$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\text{Opposite side}}{\text{Adjacent side}}$$
Example 1: A chemical pipeline has a valve handle that rotates through an angle of \(30^\circ\). Calculate \(\sin 30^\circ\), \(\cos 30^\circ\), and \(\tan 30^\circ\).
Given: \(\theta = 30^\circ\)
$$\sin 30^\circ = 0.5$$
$$\cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866$$
$$\tan 30^\circ = \frac{0.5}{0.866} \approx 0.577$$
Answer: \(\sin 30^\circ = 0.5\), \(\cos 30^\circ = 0.866\), \(\tan 30^\circ = 0.577\)
Example 2: In a distillation column, the vapor-liquid equilibrium angle is \(45^\circ\). Find the sine, cosine, and tangent values.
Given: \(\theta = 45^\circ\)
$$\sin 45^\circ = \frac{\sqrt{2}}{2} \approx 0.707$$
$$\cos 45^\circ = \frac{\sqrt{2}}{2} \approx 0.707$$
$$\tan 45^\circ = \frac{0.707}{0.707} = 1$$
Answer: \(\sin 45^\circ = 0.707\), \(\cos 45^\circ = 0.707\), \(\tan 45^\circ = 1\)
Example 3: A reactor stirrer blade forms an angle of \(60^\circ\) with the horizontal. Calculate the sine, cosine, and tangent.
Given: \(\theta = 60^\circ\)
$$\sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866$$
$$\cos 60^\circ = 0.5$$
$$\tan 60^\circ = \frac{0.866}{0.5} = 1.732$$
Answer: \(\sin 60^\circ = 0.866\), \(\cos 60^\circ = 0.5\), \(\tan 60^\circ = 1.732\)
Trigonometric identities are equations involving trig functions true for all values in their domains. These identities simplify calculations in chemical engineering models.
Key identities include:
$$\sin^2 \theta + \cos^2 \theta = 1$$
$$\tan \theta = \frac{\sin \theta}{\cos \theta}$$
$$\csc \theta = \frac{1}{\sin \theta}, \quad \sec \theta = \frac{1}{\cos \theta}, \quad \cot \theta = \frac{1}{\tan \theta}$$
Example 1: Verify the Pythagorean identity for \(\theta = 45^\circ\).
Given: \(\theta = 45^\circ\)
Calculate \(\sin^2 45^\circ + \cos^2 45^\circ\):
$$\sin 45^\circ = 0.707, \quad \cos 45^\circ = 0.707$$
$$\sin^2 45^\circ + \cos^2 45^\circ = (0.707)^2 + (0.707)^2 = 0.5 + 0.5 = 1$$
Answer: Identity holds true as sum equals 1.
Example 2: Calculate \(\tan 60^\circ\) using \(\sin 60^\circ\) and \(\cos 60^\circ\).
Given: \(\sin 60^\circ = 0.866\), \(\cos 60^\circ = 0.5\)
$$\tan 60^\circ = \frac{0.866}{0.5} = 1.732$$
Answer: \(\tan 60^\circ = 1.732\)
Example 3: Find \(\csc 30^\circ\) and \(\sec 30^\circ\).
Given: \(\sin 30^\circ = 0.5\), \(\cos 30^\circ = 0.866\)
$$\csc 30^\circ = \frac{1}{0.5} = 2$$
$$\sec 30^\circ = \frac{1}{0.866} = 1.1547$$
Answer: \(\csc 30^\circ = 2\), \(\sec 30^\circ = 1.1547\)
Trigonometric equations involve solving for angles where the trig function equals a known value. These arise in chemical engineering when determining phase angles or optimizing equipment orientation.
Example 1: Solve for \(\theta\) if \(\sin \theta = 0.5\) and \(0^\circ \leq \theta \leq 360^\circ\).
Using sine inverse:
$$\theta = \sin^{-1} 0.5 = 30^\circ$$
Since sine is positive in QI and QII:
$$\theta = 30^\circ \quad \text{or} \quad 180^\circ - 30^\circ = 150^\circ$$
Answer: \(\theta = 30^\circ, 150^\circ\)
Example 2: Solve for \(\theta\) given \(\cos \theta = -0.5\) for \(0^\circ \leq \theta \leq 360^\circ\).
$$\theta = \cos^{-1} (-0.5) = 120^\circ$$
Cosine is negative in QII and QIII:
$$\theta = 120^\circ, 240^\circ$$
Answer: \(\theta = 120^\circ, 240^\circ\)
Example 3: Find \(\theta\) if \(\tan \theta = 1\) for \(0^\circ \leq \theta \leq 360^\circ\).
$$\theta = \tan^{-1} 1 = 45^\circ$$
Tangent positive in QI and QIII:
$$\theta = 45^\circ, 225^\circ$$
Answer: \(\theta = 45^\circ, 225^\circ\)
Half-angle formulas express trig functions of half an angle in terms of the full angle, useful in integration and solving equations in chemical process controls.
Formulas:
$$\sin \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{2}}$$
$$\cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}}$$
$$\tan \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} = \frac{\sin \theta}{1 + \cos \theta} = \frac{1 - \cos \theta}{\sin \theta}$$
Example 1: Calculate \(\sin 15^\circ\) using \(\sin \frac{\theta}{2}\) formula, given \(\theta = 30^\circ\).
Given: \(\theta = 30^\circ\), \(\cos 30^\circ = 0.866\)
$$\sin 15^\circ = \sqrt{\frac{1 - 0.866}{2}} = \sqrt{\frac{0.134}{2}} = \sqrt{0.067} = 0.2588$$
Answer: \(\sin 15^\circ = 0.2588\)
Example 2: Calculate \(\cos 75^\circ\) using half-angle formula with \(\theta = 150^\circ\).
Given: \(\theta = 150^\circ\), \(\cos 150^\circ = -0.866\)
$$\cos 75^\circ = \sqrt{\frac{1 + (-0.866)}{2}} = \sqrt{\frac{0.134}{2}} = \sqrt{0.067} = 0.2588$$
Answer: \(\cos 75^\circ = 0.2588\)
Example 3: Find \(\tan 22.5^\circ\) using \(\tan \frac{\theta}{2}\) formula with \(\theta = 45^\circ\).
Given: \(\theta = 45^\circ\), \(\cos 45^\circ = 0.707\), \(\sin 45^\circ = 0.707\)
Using formula:
$$\tan 22.5^\circ = \frac{\sin 45^\circ}{1 + \cos 45^\circ} = \frac{0.707}{1 + 0.707} = \frac{0.707}{1.707} = 0.4142$$
Answer: \(\tan 22.5^\circ = 0.4142\)
Factor formulas express sums or differences of sine or cosine functions as products, useful in signal processing and wave interference analysis in chemical instrumentation.
Formulas:
$$\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$$
$$\sin A - \sin B = 2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}$$
$$\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$$
$$\cos A - \cos B = -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}$$
Example 1: Simplify \(\sin 50^\circ + \sin 70^\circ\).
Given: \(A = 50^\circ\), \(B = 70^\circ\)
$$\sin 50^\circ + \sin 70^\circ = 2 \sin \frac{50^\circ + 70^\circ}{2} \cos \frac{50^\circ - 70^\circ}{2}$$
$$= 2 \sin 60^\circ \cos (-10^\circ) = 2 \times 0.866 \times 0.9848 = 1.705$$
Answer: \(\sin 50^\circ + \sin 70^\circ = 1.705\)
Example 2: Simplify \(\cos 40^\circ - \cos 80^\circ\).
Given: \(A = 40^\circ\), \(B = 80^\circ\)
$$\cos 40^\circ - \cos 80^\circ = -2 \sin \frac{40^\circ + 80^\circ}{2} \sin \frac{40^\circ - 80^\circ}{2}$$
$$= -2 \sin 60^\circ \sin (-20^\circ) = -2 \times 0.866 \times (-0.342) = 0.592$$
Answer: \(\cos 40^\circ - \cos 80^\circ = 0.592\)
Example 3: Simplify \(\sin 120^\circ - \sin 60^\circ\).
Given: \(A = 120^\circ\), \(B = 60^\circ\)
$$\sin 120^\circ - \sin 60^\circ = 2 \cos \frac{120^\circ + 60^\circ}{2} \sin \frac{120^\circ - 60^\circ}{2}$$
$$= 2 \cos 90^\circ \sin 30^\circ = 2 \times 0 \times 0.5 = 0$$
Answer: \(\sin 120^\circ - \sin 60^\circ = 0\)
Compound angle formulas relate trig functions of sums or differences of angles to products of trig functions of individual angles. They are critical in phase shift analysis in chemical reactors.
Formulas:
$$\sin (A \pm B) = \sin A \cos B \pm \cos A \sin B$$
$$\cos (A \pm B) = \cos A \cos B \mp \sin A \sin B$$
$$\tan (A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$$
Example 1: Calculate \(\sin 75^\circ\) using \(\sin (45^\circ + 30^\circ)\).
Given: \(A = 45^\circ\), \(B = 30^\circ\)
$$\sin 75^\circ = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ$$
$$= 0.707 \times 0.866 + 0.707 \times 0.5 = 0.612 + 0.354 = 0.966$$
Answer: \(\sin 75^\circ = 0.966\)
Example 2: Calculate \(\cos 105^\circ\) using \(\cos (60^\circ + 45^\circ)\).
Given: \(A = 60^\circ\), \(B = 45^\circ\)
$$\cos 105^\circ = \cos 60^\circ \cos 45^\circ - \sin 60^\circ \sin 45^\circ$$
$$= 0.5 \times 0.707 - 0.866 \times 0.707 = 0.354 - 0.612 = -0.258$$
Answer: \(\cos 105^\circ = -0.258\)
Example 3: Calculate \(\tan 15^\circ\) using \(\tan (45^\circ - 30^\circ)\).
Given: \(A = 45^\circ\), \(B = 30^\circ\)
$$\tan 15^\circ = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} = \frac{1 - 0.577}{1 + 1 \times 0.577} = \frac{0.423}{1.577} = 0.268$$
Answer: \(\tan 15^\circ = 0.268\)
Double angle formulas express trig functions of twice an angle in terms of the functions of the original angle. They are applied in oscillation frequency calculations in chemical sensors.
Formulas:
$$\sin 2\theta = 2 \sin \theta \cos \theta$$
$$\cos 2\theta = \cos^2 \theta - \sin^2 \theta = 2 \cos^2 \theta - 1 = 1 - 2 \sin^2 \theta$$
$$\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}$$
Example 1: Calculate \(\sin 60^\circ\) using \(\sin 2 \times 30^\circ\).
Given: \(\theta = 30^\circ\), \(\sin 30^\circ = 0.5\), \(\cos 30^\circ = 0.866\)
$$\sin 60^\circ = 2 \times 0.5 \times 0.866 = 0.866$$
Answer: \(\sin 60^\circ = 0.866\)
Example 2: Calculate \(\cos 90^\circ\) using \(\cos 2 \times 45^\circ\).
Given: \(\theta = 45^\circ\), \(\cos 45^\circ = 0.707\), \(\sin 45^\circ = 0.707\)
Using \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\):
$$\cos 90^\circ = (0.707)^2 - (0.707)^2 = 0.5 - 0.5 = 0$$
Answer: \(\cos 90^\circ = 0\)
Example 3: Calculate \(\tan 30^\circ\) using \(\tan 2 \times 15^\circ\).
Given: \(\theta = 15^\circ\), \(\tan 15^\circ = 0.268\)
$$\tan 30^\circ = \frac{2 \times 0.268}{1 - (0.268)^2} = \frac{0.536}{1 - 0.072} = \frac{0.536}{0.928} = 0.577$$
Answer: \(\tan 30^\circ = 0.577\)
Parametric equations express coordinates \(x\) and \(y\) as functions of a parameter \(t\), often an angle, useful in describing trajectories and oscillations in chemical reactor design.
General form:
$$x = f(t), \quad y = g(t)$$
Example with trigonometric functions:
$$x = a \cos t, \quad y = b \sin t$$
which describes an ellipse.
Example 1: Determine the coordinates of a point on the ellipse \(x = 3 \cos t\), \(y = 2 \sin t\) at \(t = \frac{\pi}{3}\).
Given: \(a=3\), \(b=2\), \(t = \frac{\pi}{3} = 60^\circ\)
Calculate:
$$x = 3 \cos 60^\circ = 3 \times 0.5 = 1.5$$
$$y = 2 \sin 60^\circ = 2 \times 0.866 = 1.732$$
Answer: Coordinates: \((1.5, 1.732)\)
Example 2: Find the point on the curve \(x = 5 \cos t\), \(y = 5 \sin t\) at \(t = 90^\circ\).
Given: \(a=5\), \(b=5\), \(t = 90^\circ\)
Calculate:
$$x = 5 \cos 90^\circ = 5 \times 0 = 0$$
$$y = 5 \sin 90^\circ = 5 \times 1 = 5$$
Answer: Coordinates: \((0, 5)\)
Example 3: For \(x = 4 \cos t\), \(y = 3 \sin t\), find the coordinates at \(t = 150^\circ\).
Given: \(a=4\), \(b=3\), \(t = 150^\circ\)
Calculate:
$$x = 4 \cos 150^\circ = 4 \times (-0.866) = -3.464$$
$$y = 3 \sin 150^\circ = 3 \times 0.5 = 1.5$$
Answer: Coordinates: \((-3.464, 1.5)\)
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Create a free accountThis chapter explored the fundamental concepts and applications of trigonometric functions, including the use of key trigonometric identities to simplify expressions and solve equations. Important formulas such as the half-angle, factor, compound angle, and double angle formulas were examined to enhance problem-solving skills. Parametric equations were introduced as a means to represent curves and motion in engineering contexts. The chapter then transitioned to hyperbolic functions, defining them and discussing their distinctive properties alongside methods for evaluating these functions. Hyperbolic identities and Osborne's Rule were presented to relate hyperbolic and trigonometric functions effectively. The concept of one-to-one relationships in functions was explained to establish the foundation for solving equations like \( a \sinh x + b \cosh x = C \). Finally, inverse functions for both one-to-one and trigonometric functions were covered to enable the determination of angles and hyperbolic values from given function outputs.
Calculate the value of \(\sin 45^\circ\) using the half-angle formula. (2 marks)
A chemical pipe is inclined at an angle \(\theta\) to the horizontal such that \(\tan \theta = \frac{3}{4}\). Calculate \(\sin \theta\) and \(\cos \theta\). (3 marks)
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