Problem 39 - Precalculus with Calculus Previews

Question

Alternating Current The term \(\sin \omega t \sin (\omega t+\) \(\phi\) ) is encountered in the derivation of an expression for the power in an alternating-current circuit. Show that this term can be written as \(\frac{1}{2}[\cos \phi-\cos (2 \omega t+\phi)]\)

Step-by-Step Solution

Answer

The expression \( \sin \omega t \sin (\omega t + \phi) \) can be rewritten as \( \frac{1}{2}[\cos \phi - \cos (2\omega t + \phi)] \).

Step 1: Recognize the Trigonometric Identity

To rewrite the given expression \( \sin \omega t \sin (\omega t + \phi) \), we will use the trigonometric identity for the product of sines: \[ \sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)] \] which will help in simplifying the expression.

Step 2: Substitute Variables into the Identity

Let \( A = \omega t \) and \( B = \omega t + \phi \). Now, substitute \( A \) and \( B \) into the trigonometric identity:\[\sin \omega t \sin (\omega t + \phi) = \frac{1}{2} [\cos(\omega t - (\omega t + \phi)) - \cos(\omega t + (\omega t + \phi))]\]

Step 3: Simplify Each Cosine Argument

Let's simplify the arguments of each cosine term:- For \( \cos(\omega t - (\omega t + \phi)) \), simplify to \( \cos(-\phi) \).- For \( \cos(\omega t + (\omega t + \phi)) \), simplify to \( \cos(2\omega t + \phi) \).

Step 4: Apply Cosine of Negative Angle

Recall that \( \cos(-\phi) = \cos \phi \). Therefore, substitute back:\[\frac{1}{2} [\cos(-\phi) - \cos(2\omega t + \phi)] = \frac{1}{2} [\cos \phi - \cos(2\omega t + \phi)]\] This matches the desired expression.

Key Concepts

Alternating Current

Alternating Current (AC) refers to the flow of electric charge that periodically reverses direction. This is different from Direct Current (DC) where the flow is in a single direction. AC is the type of current typically used in households and businesses for electrical power. It is transported and distributed with ease because of the way power can be transmitted over long distances at high voltages, then easily reduced by transformers for safe domestic use.

• AC is often represented as a sine wave because its current and voltage vary over time in a smoothly oscillating manner.
• The frequency of AC systems varies by region; for example, in North America, it is 60 Hz, meaning it cycles 60 times per second, whereas, in Europe, it's typically 50 Hz.
• AC circuits involve changing the phase, amplitude, and frequency, unlike DC circuits which involve a constant flow.

An important aspect of AC is its ability to be transformed into different voltage levels with very little loss of power, making it economical for electric power distribution. By understanding AC, we also delve into aspects such as phase differences, which are crucial in power systems and drive the use of trigonometric identities in AC power equations.

Power in AC Circuits

Understanding power in AC circuits involves considering the relationship between voltage, current, and the phase difference between them. The power in an AC circuit is not constant because of the cyclically varying nature of AC voltage and current.

• The instantaneous power is given by the product of the instantaneous voltage and current.
• We define real power, reactive power, and apparent power to measure different energy components in AC circuits.

Real power (measured in watts) is the actual power consumed in the circuit, while reactive power (measured in VARs) is associated with the reactive components like inductors and capacitors, which do not consume energy but store it temporarily.
The term "apparent power," calculated as the product of the root-mean-square (RMS) voltage and current, combines both real power and reactive power components. This mixture requires using the trigonometric identities especially when the phase differences between voltage and current lead to sinusoidal variations that need simplification.

Trigonometric Simplification

Simplifying expressions using trigonometric identities is a critical skill in mathematics and physics, especially when dealing with AC circuits. The expression given in the exercise \( \sin \omega t \sin (\omega t + \phi) \) relies on a known trigonometric identity to be re-written more conveniently.

• Product-to-sum identities aid in transforming the product of two sine functions into a more manageable form.
• For the given exercise, the identity \[ \sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)] \] simplifies the expression effectively.

To apply this, you substitute the specific terms for A and B, where A = \( \omega t \) and B = \( \omega t + \phi \). This allows us to break the product into a difference of cosines.
The use of the property \( \cos(-\phi) = \cos \phi \) helps achieve the final simplified form: \[ \frac{1}{2} [\cos \phi - \cos(2\omega t + \phi)] \]. This simplification is crucial for calculating other elements in AC circuits such as power, where phase angles and amplitude modulation frequently arise.