Kami Export - Hussien Aldubaishi - Graphing Sine and Cosine ... - Free Printable

Problem Analysis:

The task involves identifying the correct mathematical expressions for three given waveforms. Each waveform is described by a trigonometric function, and we need to match the waveforms with their corresponding equations.

#### Key Observations:
1. Waveform Characteristics:
- The waveforms are sinusoidal or cosine-based.
- The amplitude, frequency, and phase shift of each waveform must be determined from the graphs.

2. Equation Forms:
- The general form of a sinusoidal function is:
\[
y(t) = A \sin(\omega t + \phi) \quad \text{or} \quad y(t) = A \cos(\omega t + \phi)
\]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency (\( \omega = 2\pi f \)),
- \( \phi \) is the phase shift.

3. Given Equations:
- Equation (I): \( y_1(t) = 4 \sin(2t + \frac{\pi}{6}) \)
- Equation (II): \( y_2(t) = 4 \cos(2t + \frac{\pi}{6}) \)
- Equation (III): \( y_3(t) = 4 \cos(2t - \frac{\pi}{3}) \)

#### Step-by-Step Solution:

---

Waveform 1:

- Amplitude: The peak value of the waveform is 4, so \( A = 4 \).
- Frequency: The period of the waveform appears to be \( \pi \). Since the period \( T \) is related to the angular frequency \( \omega \) by \( \omega = \frac{2\pi}{T} \), we have:
\[
\omega = \frac{2\pi}{\pi} = 2
\]
- Phase Shift: At \( t = 0 \), the waveform starts at a positive value, which suggests it is a cosine function (since \( \cos(0) = 1 \)). The phase shift can be determined by comparing the starting point to the standard cosine function. The waveform starts slightly before its maximum, indicating a phase shift of \( -\frac{\pi}{3} \).

Thus, the equation for Waveform 1 is:
\[
y_3(t) = 4 \cos(2t - \frac{\pi}{3})
\]

---

Waveform 2:

- Amplitude: The peak value of the waveform is 4, so \( A = 4 \).
- Frequency: The period of the waveform is \( \pi \), so \( \omega = 2 \).
- Phase Shift: At \( t = 0 \), the waveform starts at zero and is increasing, which suggests it is a sine function. The phase shift can be determined by comparing the starting point to the standard sine function. The waveform starts at zero and increases, indicating a phase shift of \( \frac{\pi}{6} \).

Thus, the equation for Waveform 2 is:
\[
y_1(t) = 4 \sin(2t + \frac{\pi}{6})
\]

---

Waveform 3:

- Amplitude: The peak value of the waveform is 4, so \( A = 4 \).
- Frequency: The period of the waveform is \( \pi \), so \( \omega = 2 \).
- Phase Shift: At \( t = 0 \), the waveform starts at a positive value, which suggests it is a cosine function. The phase shift can be determined by comparing the starting point to the standard cosine function. The waveform starts slightly after its maximum, indicating a phase shift of \( \frac{\pi}{6} \).

Thus, the equation for Waveform 3 is:
\[
y_2(t) = 4 \cos(2t + \frac{\pi}{6})
\]

---

Final Answer:

Matching the waveforms to their equations:
- Waveform 1 corresponds to Equation (III): \( y_3(t) = 4 \cos(2t - \frac{\pi}{3}) \)
- Waveform 2 corresponds to Equation (I): \( y_1(t) = 4 \sin(2t + \frac{\pi}{6}) \)
- Waveform 3 corresponds to Equation (II): \( y_2(t) = 4 \cos(2t + \frac{\pi}{6}) \)

\[
\boxed{\text{Waveform 1: III, Waveform 2: I, Waveform 3: II}}
\]