Show that the following function is a sinusoid by rewriting it in the form C(x) = A cos(𝜔x + φ) + B

 or S(x) = A sin(𝜔x + φ) + B.

f(x) = − 1/2sin(4x) − √3/2cos(4x)

To demonstrate that \( f(x) = -\frac{1}{2} \sin(4x) - \frac{\sqrt{3}}{2} \cos(4x) \) is a sinusoidal function, we can rewrite it in the form \( S(x) = A \sin(\omega x + \varphi) + B \), where \( A \), \( \omega \), \( \varphi \), and \( B \) are constants.

### Step 1: Identify the Parameters
The given function is:
\[
f(x) = -\frac{1}{2} \sin(4x) - \frac{\sqrt{3}}{2} \cos(4x)
\]
We can observe that this is a linear combination of \( \sin(4x) \) and \( \cos(4x) \) with coefficients \( -\frac{1}{2} \) and \( -\frac{\sqrt{3}}{2} \), respectively.

### Step 2: Rewrite the Function in Terms of a Single Sinusoidal Function
To rewrite the expression as a single sinusoidal function, we can use the identity:
\[
a \sin(4x) + b \cos(4x) = R \sin(4x + \varphi)
\]
where:
- \( R = \sqrt{a^2 + b^2} \) (the amplitude),
- \( \varphi = \arctan\left(\frac{b}{a}\right) \) (the phase shift).

In this case:
- \( a = -\frac{1}{2} \)
- \( b = -\frac{\sqrt{3}}{2} \)

#### Calculating \( R \):
\[
R = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2}
\]
\[
R = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1
\]

Thus, \( R = 1 \).

#### Calculating \( \varphi \):
\[
\varphi = \arctan\left(\frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}}\right) = \arctan\left(\frac{\sqrt{3}}{1}\right) = \arctan(\sqrt{3})
\]
Since \( \arctan(\sqrt{3}) = \frac{\pi}{3} \), we have \( \varphi = \frac{\pi}{3} \).

### Step 3: Substitute Back
Now we can rewrite \( f(x) \) as:
\[
f(x) = R \sin(4x + \varphi)
\]
Substituting \( R = 1 \) and \( \varphi = \frac{\pi}{3} \):
\[
f(x) = \sin\left(4x + \frac{\pi}{3}\right)
\]

### Final Answer
Thus, the function can be written as:
\[
f(x) = \sin\left(4x + \frac{\pi}{3}\right)
\]
which is in the form \( S(x) = A \sin(\omega x + \varphi) + B \) with:
- \( A = 1 \)
- \( \omega = 4 \)
- \( \varphi = \frac{\pi}{3} \)
- \( B = 0 \)

This confirms that \( f(x) \) is indeed a sinusoidal function.

---

Would you like more details on any part of the solution?

### Related Questions:
1. How do you determine the amplitude and phase shift of a sinusoidal function?
2. Can you explain why any linear combination of sine and cosine functions can be written as a single sinusoid?
3. How do we convert a cosine function to an equivalent sine function?
4. What is the effect of different values of \( A \) and \( \omega \) on the graph of a sinusoidal function?
5. How does the phase shift affect the graph of the sinusoidal function?

### Tip:
Remember, combining \( a \sin(x) + b \cos(x) \) into a single sinusoidal term often simplifies analysis, especially when graphing or solving equations.

Show that the following function is a sinusoid by rewriting it in the form C(x) = A cos(𝜔x + φ) + B or S(x) = A sin(𝜔x + φ) + B. f(x) = −1/2sin(4x) − √3/2cos(4x)

Explore the step-by-step process to rewrite the function f(x) = −1/2 sin(4x) − √3/2 cos(4x) in the form of a sinusoidal function S(x) = A sin(𝜔x + φ) + B. This guide covers key trigonometric concepts such as phase shift and amplitude, suitable for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation