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Sinusoidal Fucntion

Introduction

A sinusoidal function is a mathematical function that combines sine and cosine terms, typically written as \( f(x) = a\sin(x) + b\cos(x) \). It's characterized by a smooth, repetitive oscillation that can be graphically represented as a wave-like pattern. The constants \( a \) and \( b \) determine the shape and orientation of the wave. This type of function is widely used in mathematics to represent periodic phenomena.

Theorem: The function \( f(x) = a\sin(x) + b\cos(x) \) can be rewritten as \( f(x) = \sqrt{a^2 + b^2}\sin(x + \phi) \), where \( \phi \) is a phase shift such that \( \cos(\phi) = \frac{a}{\sqrt{a^2 + b^2}} \) and \( \sin(\phi) = \frac{b}{\sqrt{a^2 + b^2}} \).

Proof:

  1. Factor Out the Common Term: We express \( f(x) \) by factoring out the common term \( \sqrt{a^2 + b^2} \):

    \[ f(x) = \sqrt{a^2 + b^2}\left(\frac{a}{\sqrt{a^2 + b^2}}\sin(x) + \frac{b}{\sqrt{a^2 + b^2}}\cos(x)\right) \]

  2. Definition of \( \phi \): We define \( \phi \) such that \( \cos(\phi) = \frac{a}{\sqrt{a^2 + b^2}} \) and \( \sin(\phi) = \frac{b}{\sqrt{a^2 + b^2}} \). It follows from the Pythagorean identity that \( \cos^2(\phi) + \sin^2(\phi) = 1 \), and thus:

    \[ \left(\frac{a}{\sqrt{a^2 + b^2}}\right)^2 + \left(\frac{b}{\sqrt{a^2 + b^2}}\right)^2 = \frac{a^2}{a^2 + b^2} + \frac{b^2}{a^2 + b^2} = 1 \]

  3. Applying the Angle Sum Identity: Using the angle sum identity for sine, \( \sin(x + \phi) \), we can rewrite \( f(x) \) as:

    \[ f(x) = \sqrt{a^2 + b^2}\left(\frac{a}{\sqrt{a^2 + b^2}}\sin(x) + \frac{b}{\sqrt{a^2 + b^2}}\cos(x)\right) \]
    \[ = \sqrt{a^2 + b^2}\sin(x + \phi) \]

The angle \( \phi \) is determined such that its sine and cosine are proportional to \( a \) and \( b \), and their squares add up to 1, as per the Pythagorean identity. This completes the proof that \( a\sin(x) + b\cos(x) \) can be expressed in the form \( \sqrt{a^2 + b^2}\sin(x + \phi) \).

Express in terms of sine

To express \( 3\sin(x) + 5\cos(x) \) in the form \( \sqrt{a^2 + b^2}\sin(x + \phi) \) with the appropriate \( \phi \), we follow these steps:

  1. Calculate the magnitude \( \sqrt{a^2 + b^2} \):

    \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34} ]

  2. Determine \( \cos(\phi) \) and \( \sin(\phi) \) using \( a \) and \( b \):

    \[ \cos(\phi) = \frac{3}{\sqrt{34}} \]
    \[ \sin(\phi) = \frac{5}{\sqrt{34}} \]

  3. Now we rewrite \( 3\sin(x) + 5\cos(x) \) as:

    \[ 3\sin(x) + 5\cos(x) = \sqrt{34}\left(\frac{3}{\sqrt{34}}\sin(x) + \frac{5}{\sqrt{34}}\cos(x)\right) \]
    \[ = \sqrt{34}\sin(x + \phi) \]

So, \( 3\sin(x) + 5\cos(x) \) is equivalent to \( \sqrt{34}\sin(x + \phi) \) where \( \cos(\phi) = \frac{3}{\sqrt{34}} \) and \( \sin(\phi) = \frac{5}{\sqrt{34}} \).

Express in terms of cosine

The function \( 3\sin(x) + 5\cos(x) \) can also be expressed in terms of cosine with a phase shift. Following a similar process as with sine, we would express it in the form \( \sqrt{a^2 + b^2}\cos(x + \psi) \), where \( \psi \) is a phase angle such that \( \sin(\psi) = \frac{a}{\sqrt{a^2 + b^2}} \) and \( \cos(\psi) = \frac{b}{\sqrt{a^2 + b^2}} \).

Using the given function:

  1. Calculate the magnitude \( \sqrt{a^2 + b^2} \):

    \[ \sqrt{3^2 + 5^2} = \sqrt{34} \]

  2. Determine \( \cos(\psi) \) and \( \sin(\psi) \) using \( a \) and \( b \):

    \[ \sin(\psi) = \frac{3}{\sqrt{34}} \]
    \[ \cos(\psi) = \frac{5}{\sqrt{34}} \]

  3. Now we rewrite \( 3\sin(x) + 5\cos(x) \) as:

    \[ 3\sin(x) + 5\cos(x) = \sqrt{34}\left(\frac{3}{\sqrt{34}}\sin(x) + \frac{5}{\sqrt{34}}\cos(x)\right) \]
    \[ = \sqrt{34}\cos(x + \psi) \]

So, \( 3\sin(x) + 5\cos(x) \) is equivalent to \( \sqrt{34}\cos(x + \psi) \) where \( \sin(\psi) = \frac{3}{\sqrt{34}} \) and \( \cos(\psi) = \frac{5}{\sqrt{34}} \). The value of \( \psi \) ensures that when \( x = -\psi \), the cosine function starts at its maximum value, similar to how \( \phi \) was determined for the sine function. The exact value of \( \psi \) would be the phase shift required to align the cosine function with the given linear combination of sine and cosine.

Example

Express the function \( f(x) = 4\sin(x) - 5\cos(x) \) in the form \( \sqrt{a^2 + b^2}\sin(x - \phi) \).

Solution:

  1. Determine the Magnitude: The magnitude is \( \sqrt{4^2 + 5^2} = \sqrt{41} \).

  2. Rewrite the Function:

    \[ f(x) = \sqrt{41}\left(\frac{4}{\sqrt{41}}\sin(x) - \frac{5}{\sqrt{41}}\cos(x)\right) \]

  3. Identify Cosine and Sine Components of \( \phi \):

    Here, \( \cos(\phi) = \frac{4}{\sqrt{41}} \) and \( \sin(\phi) = \frac{5}{\sqrt{41}} \).

  4. Express in the Sinusoidal Form: We use the angle sum identity for sine:

    \[ \sin(x - \phi) = \sin(x)\cos(\phi) - \cos(x)\sin(\phi) \]

  5. Substitute the Values of \( \cos(\phi) \) and \( \sin(\phi) \):

    \[ \sin(x - \phi) = \sin(x)\frac{4}{\sqrt{41}} - \cos(x)\frac{5}{\sqrt{41}} \]

  6. Combine into the Single Function:

    \[ f(x) = \sqrt{41}\sin(x - \phi) \]

In this solution, \( f(x) = 4\sin(x) - 5\cos(x) \) is expressed as \( \sqrt{41}\sin(x - \phi) \) where \( \cos(\phi) = \frac{4}{\sqrt{41}} \) and \( \sin(\phi) = \frac{5}{\sqrt{41}} \). This aligns the original function with a single sinusoidal function by adjusting the phase angle \( \phi \) accordingly.

Sinusoidal function and waves

The transformation of \( a\sin(x) + b\cos(x) \) to \( \sqrt{a^2 + b^2}\sin(x + \phi) \) shows that the sinusoidal fucntion is basically a sine function. The maximum and minimum value of sine is +1 and -1 respectively. It is easy to see that this function has the maximum and minimum value of \( \sqrt{a^2 + b^2}\). We define the following terms in context of usefulness of sinusoidal function in Physics, where they are used to represent a wave. If you are not acquainted with waves yet, you will soon get to learn.

  1. Amplitude (\( \sqrt{a^2 + b^2} \)): This represents the peak value (maximum or minimum) of the wave. The amplitude is determined by the coefficients \( a \) and \( b \), and it reflects the combined effect of the sine and cosine components on the wave's overall intensity or strength.

  2. Phase Shift (\( \phi \)): This represents the horizontal shift of the wave. It's the angle by which the standard sine wave is shifted to align with the given combination of sine and cosine waves. The phase shift is determined by the ratio of \( a \) and \( b \), and it essentially tells us how the wave is displaced along the horizontal axis from its standard position.

In physics, this transformation is extremely useful in representing waves because many physical phenomena involve wave-like behavior that can be modeled as a combination of sine and cosine functions. By converting to a single sinusoidal function, it simplifies the analysis and understanding of the wave's behavior. This is particularly helpful in areas like signal processing, acoustics, and electromagnetism, where waves are a fundamental concept, and their amplitude and phase shift carry significant information about the physical system.

Read here.

Maximum and Minimum values

We always have a choice of converting a sinusoidal function \( f(x) = a\sin(x) + b\cos(x) \) to \( f(x) = \sqrt{a^2 + b^2}\sin(x + \phi) \) or \( f(x) = \sqrt{a^2 + b^2}\cos(x + \phi) \) i.e. either in sine form or cosine. For the maximum and minimum analysis it is good to write in cosine form. Let us understand this by taking the following examples:

Examples

Find the maximum and minimum value of \(7\sin(x) - \cos(x) \). Also find the values of \(x\) for which the maximum and minimum occur.

Solution:

  1. Factor Out \( \sqrt{50} \):

    \[ f(x) = \sqrt{50}\left(\frac{7}{\sqrt{50}}\cos(x) - \frac{1}{\sqrt{50}}\sin(x)\right) \]

  2. Identify \( \cos(\phi) \) and \( \sin(\phi) \):

    \[ \cos(\phi) = \frac{7}{\sqrt{50}}, \quad \sin(\phi) = \frac{1}{\sqrt{50}} \]

  3. Express as Single Cosine Function:

    \[ f(x) = \sqrt{50}\cos(x + \phi) \]

  4. Maximum and Minimum Values: The cosine function has a maximum value of 1 and a minimum of -1. Therefore, the maximum value of \( f(x) \) is \( \sqrt{50} \), and the minimum value is \( -\sqrt{50} \).

  5. Determine \( x \) for Maximum and Minimum:

    • The maximum value of a cosine function occurs at \( x = -\phi \). This is because \( \cos(x + \phi) \) reaches its maximum value when \( x + \phi = 2k\pi \), where \( k \) is an integer. for \(k=0\) we get one of the values of \( x \) as \( x = -\phi \).
    • The minimum value occurs at \( x = \pi - \phi \). This is because the minimum value of \( \cos(x + \phi) \) is reached when \( x + \phi = (2k + 1)\pi \), for \(k=0\) one such \( x \) being \( x = \pi - \phi \).

Find the maximum and minimum value of \(12\sin(x) + 4\cos(x) \). Also find the values of \(x\) for which the maximum and minimum occur.

Solution:

  1. Convert to Cosine Form: Express the function by factoring out \( \sqrt{12^2 + 4^2} \):

    \[ f(x) = \sqrt{160}\left(\frac{12}{\sqrt{160}}\sin(x) + \frac{4}{\sqrt{160}}\cos(x)\right) \]

    Simplify the square root:

    \[ f(x) = 4\sqrt{10}\left(\frac{12}{4\sqrt{10}}\sin(x) + \frac{4}{4\sqrt{10}}\cos(x)\right) \]
    \[ f(x) = 4\sqrt{10}\left(\frac{3}{\sqrt{10}}\sin(x) + \frac{1}{\sqrt{10}}\cos(x)\right) \]

  2. Determine \( \sin(\phi) \) and \( \cos(\phi) \):

    \[ \sin(\phi) = \frac{3}{\sqrt{10}}, \quad \cos(\phi) = \frac{1}{\sqrt{10}} \]

  3. Express as a Single Cosine Function:

    \[ f(x) = 4\sqrt{10}\cos(x - \phi) \]

  4. Maximum and Minimum Values: The maximum value is the amplitude \( 4\sqrt{10} \), and the minimum value is \( -4\sqrt{10} \).

  5. Values of \( x \) for Maximum and Minimum: The maximum occurs at \( x = \phi \) and the minimum occurs at \( x = \pi + \phi \).