Find Trigonometric Functions Given Their Graphs (Phase Shift)

1. Amplitude (\(a\)):
The maximum y-value is 2. Therefore, \( |a| = 2 \). Let's use \( a = 2 \).

2. Period (\(b\)):
Looking at the graph's scale, 4 small horizontal divisions equal \( \pi \), so 1 small division = \( \dfrac{\pi}{4} \).
A full wave cycle spans exactly 16 small divisions.
Period \( P = 16 \times \dfrac{\pi}{4} = 4\pi \).
Using \( P = \dfrac{2\pi}{b} \), we solve: \( 4\pi = \dfrac{2\pi}{b} \implies b = \dfrac{1}{2} \).

3. Phase Shift (\(c\)):
A standard sine wave crosses the origin \( (0,0) \) heading up. Our graph crosses the x-axis heading up at 1 small division to the left of the y-axis.
Phase shift = \( -\dfrac{\pi}{4} \) (left shift).
The formula for phase shift is \( -\dfrac{c}{b} \). Substitute the knowns:
\( -\dfrac{\pi}{4} = -\dfrac{c}{1/2} \implies c = \dfrac{1}{2} \cdot \dfrac{\pi}{4} = \dfrac{\pi}{8} \).

Conclusion:
\[ \mathbf{y = 2 \sin \left( \dfrac{x}{2} + \dfrac{\pi}{8} \right)} \]

1. Amplitude (\(a\)):
The maximum value is halfway between 1 and 2. Therefore, \( |a| = 1.5 \). Let \( a = 1.5 \).

2. Period (\(b\)):
One full cycle (e.g., from peak to peak) spans exactly 4 units on the x-axis.
Period \( P = 4 \).
\( 4 = \dfrac{2\pi}{b} \implies 4b = 2\pi \implies b = \dfrac{\pi}{2} \).

3. Phase Shift (\(c\)):
The graph crosses the x-axis heading upwards at \( x = 1 \). This is a shift of 1 unit to the right.
Phase shift = \( 1 \).
\( 1 = -\dfrac{c}{b} \implies c = -b \).
Substitute \( b \): \( c = -\dfrac{\pi}{2} \).

Conclusion:
\[ \mathbf{y = 1.5 \sin\left( \dfrac{\pi x}{2} - \dfrac{\pi}{2} \right)} \]

1. Amplitude (\(a\)):
The maximum y-value is 10. \( |a| = 10 \). Let \( a = 10 \).

2. Period (\(b\)):
5 small divisions equal \( \pi \), so 1 small division = \( \dfrac{\pi}{5} \).
One full period spans 8 small divisions.
Period \( P = \dfrac{8\pi}{5} \).
\( \dfrac{8\pi}{5} = \dfrac{2\pi}{b} \implies 8\pi b = 10\pi \implies b = \dfrac{10}{8} = \dfrac{5}{4} \).

3. Phase Shift (\(c\)):
The graph crosses the center axis heading up at 2 divisions to the right of the y-axis.
Phase shift = \( 2 \times \dfrac{\pi}{5} = \dfrac{2\pi}{5} \) (right shift).
\( \dfrac{2\pi}{5} = -\dfrac{c}{b} \implies c = -b\left(\dfrac{2\pi}{5}\right) \).
Substitute \( b \): \( c = -\left(\dfrac{5}{4}\right)\left(\dfrac{2\pi}{5}\right) = -\dfrac{2\pi}{4} = -\dfrac{\pi}{2} \).

Conclusion:
\[ \mathbf{y = 10 \sin\left(\dfrac{5x}{4} - \dfrac{\pi}{2}\right)} \]

1. Amplitude (\(a\)):
The maximum value is 2. \( |a| = 2 \). Let \( a = 2 \).

2. Period (\(b\)):
12 small divisions equal \( \pi \), so 1 small division = \( \dfrac{\pi}{12} \).
One full period spans 8 divisions.
Period \( P = 8 \times \dfrac{\pi}{12} = \dfrac{2\pi}{3} \).
\( \dfrac{2\pi}{3} = \dfrac{2\pi}{b} \implies \mathbf{b = 3} \).

3. Phase Shift (\(c\)):
The graph starts its upward crossing 1 division to the right.
Phase shift = \( \dfrac{\pi}{12} \).
\( \dfrac{\pi}{12} = -\dfrac{c}{b} \implies c = -b\left(\dfrac{\pi}{12}\right) = -3\left(\dfrac{\pi}{12}\right) = -\dfrac{\pi}{4} \).

Conclusion:
\[ \mathbf{y = 2 \sin \left( 3x - \dfrac{\pi}{4} \right)} \]

1. Amplitude (\(a\)):
The maximum y-value is 4. \( |a| = 4 \). Let \( a = 4 \).

2. Period (\(b\)):
4 small divisions equal \( \dfrac{\pi}{2} \), so 1 small division = \( \dfrac{\pi}{8} \).
A full cycle spans 12 small divisions.
Period \( P = 12 \times \dfrac{\pi}{8} = \dfrac{3\pi}{2} \).
\( \dfrac{3\pi}{2} = \dfrac{2\pi}{b} \implies 3\pi b = 4\pi \implies b = \dfrac{4}{3} \).

3. Phase Shift (\(c\)):
A standard cosine function starts at a maximum when \( x = 0 \). Our graph's maximum is shifted 3 small divisions to the right.

Phase shift = \( 3 \times \dfrac{\pi}{8} = \dfrac{3\pi}{8} \).
\( \dfrac{3\pi}{8} = -\dfrac{c}{b} \implies c = -b\left(\dfrac{3\pi}{8}\right) \).
Substitute \( b \): \( c = -\left(\dfrac{4}{3}\right)\left(\dfrac{3\pi}{8}\right) = -\dfrac{12\pi}{24} = -\dfrac{\pi}{2} \).

Conclusion:
\[ \mathbf{y = 4 \cos\left(\dfrac{4x}{3} - \dfrac{\pi}{2}\right)} \]

1. Amplitude (\(a\)):
The maximum value is 3. \( |a| = 3 \). Let \( a = 3 \).

2. Period (\(b\)):
The length of one full cycle on the x-axis is exactly 1 unit.
Period \( P = 1 \).
\( 1 = \dfrac{2\pi}{b} \implies b = 2\pi \).

3. Phase Shift (\(c\)):
The maximum is shifted to the right by 0.5 units.
Phase shift = \( \dfrac{1}{2} \).
\( \dfrac{1}{2} = -\dfrac{c}{b} \implies c = -\dfrac{b}{2} \).
Substitute \( b \): \( c = -\dfrac{2\pi}{2} = -\pi \).

Conclusion:
\[ \mathbf{y = 3 \cos(2\pi x - \pi)} \]

1. Amplitude (\(a\)):
The maximum y-value is 40. \( |a| = 40 \). Let \( a = 40 \).

2. Period (\(b\)):
4 small divisions equal \( \dfrac{\pi}{2} \), so 1 small division = \( \dfrac{\pi}{8} \).
One full period spans 8 divisions.
Period \( P = 8 \times \dfrac{\pi}{8} = \pi \).
\( \pi = \dfrac{2\pi}{b} \implies \pi b = 2\pi \implies \mathbf{b = 2} \).

3. Phase Shift (\(c\)):
The maximum is shifted 3 divisions to the right.
Phase shift = \( 3 \times \dfrac{\pi}{8} = \dfrac{3\pi}{8} \).
\( \dfrac{3\pi}{8} = -\dfrac{c}{b} \implies c = -b\left(\dfrac{3\pi}{8}\right) \).
Substitute \( b = 2 \): \( c = -2\left(\dfrac{3\pi}{8}\right) = -\dfrac{6\pi}{8} = -\dfrac{3\pi}{4} \).

Conclusion:
\[ \mathbf{y = 40 \cos\left(2x - \dfrac{3\pi}{4}\right)} \]

a) Finding the Period:
For a standard sine wave, the horizontal distance between an x-intercept (where it crosses the axis heading up) and the very next maximum is exactly one quarter of a full period.
Quarter Period = \( \dfrac{11\pi}{5} - \dfrac{6\pi}{5} = \dfrac{5\pi}{5} = \pi \).
Therefore, the full period \( P = 4 \times \pi = \mathbf{4\pi} \).

b) Finding \(b\), \(c\), and the Equation:
Use the period formula to find \(b\):
\( P = \dfrac{2\pi}{|b|} \implies 4\pi = \dfrac{2\pi}{b} \implies 4\pi b = 2\pi \implies \mathbf{b = \dfrac{1}{2}} \).

Find the phase shift \(c\):
The upward-crossing x-intercept represents the "start" of the sine cycle. This occurs at \( x = \dfrac{6\pi}{5} \), which is a shift to the right.
Phase shift = \( \dfrac{6\pi}{5} \).
\( \dfrac{6\pi}{5} = -\dfrac{c}{b} \implies c = -b\left(\dfrac{6\pi}{5}\right) \).
Substitute \(b\): \( c = -\left(\dfrac{1}{2}\right)\left(\dfrac{6\pi}{5}\right) = \mathbf{-\dfrac{3\pi}{5}} \).

Conclusion:
\[ \mathbf{y = \sin\left( \dfrac{1}{2}x - \dfrac{3\pi}{5} \right)} \]

1. Amplitude (\(a\)):
The amplitude is \( \dfrac{5 - (-5)}{2} = 5 \). Because the problem demands \( a < 0 \), we must use a reflected cosine wave. Therefore, \( a = -5 \).

2. Period (\(b\)):
The horizontal distance between a minimum and the next maximum is exactly half a period.
Half Period = \( \dfrac{5\pi}{6} - \dfrac{\pi}{6} = \dfrac{4\pi}{6} = \dfrac{2\pi}{3} \).
Full Period \( P = 2 \times \dfrac{2\pi}{3} = \dfrac{4\pi}{3} \).
\( \dfrac{4\pi}{3} = \dfrac{2\pi}{b} \implies 4\pi b = 6\pi \implies b = \dfrac{6}{4} = \dfrac{3}{2} \).

3. Phase Shift (\(c\)):
A reflected cosine wave \( (- \cos) \) naturally starts at its minimum. Since our graph's minimum occurs at \( x = \dfrac{\pi}{6} \), the graph is shifted to the right by \( \dfrac{\pi}{6} \).
Phase shift = \( \dfrac{\pi}{6} \).
\( \dfrac{\pi}{6} = -\dfrac{c}{b} \implies c = -b\left(\dfrac{\pi}{6}\right) \).
Substitute \( b = \dfrac{3}{2} \): \( c = -\left(\dfrac{3}{2}\right)\left(\dfrac{\pi}{6}\right) = -\dfrac{3\pi}{12} = -\dfrac{\pi}{4} \).

Conclusion:
\[ \mathbf{y = -5\cos\left(\dfrac{3}{2}x - \dfrac{\pi}{4}\right)} \]

1. Understand the Co-function Identity:
The sine function is simply the cosine function shifted horizontally. We know the identity:
\( \sin(\theta) = \cos\left(\theta - \dfrac{\pi}{2}\right) \).

2. Substitute the Argument:
Treat our entire inside expression \( \left(2x - \dfrac{\pi}{3}\right) \) as \( \theta \).
\( y = 3\cos\left( \left(2x - \dfrac{\pi}{3}\right) - \dfrac{\pi}{2} \right) \).

3. Simplify the Phase Shift:
Find a common denominator for the constants (which is 6).
\( -\dfrac{\pi}{3} - \dfrac{\pi}{2} = -\dfrac{2\pi}{6} - \dfrac{3\pi}{6} = -\dfrac{5\pi}{6} \).

Conclusion:
The equivalent cosine function is:
\[ \mathbf{y = 3\cos\left(2x - \dfrac{5\pi}{6}\right)} \]

1. Relate Maximum to the Cosine Function:
The function \( y = 8\cos(\theta) \) reaches its maximum when the inside argument \( \theta \) is equal to \( 0, 2\pi, 4\pi, -2\pi \), etc.
Therefore, at the maximum point, we can set the inside argument equal to \( 2k\pi \):
\[ 4x + c = 2k\pi \]

2. Substitute the Known x-value:
We know a maximum occurs at \( x = \dfrac{\pi}{12} \):
\[ 4\left(\dfrac{\pi}{12}\right) + c = 2k\pi \] \[ \dfrac{\pi}{3} + c = 2k\pi \] \[ c = 2k\pi - \dfrac{\pi}{3} \]

3. Find Positive Values for \( c \):
We can test integer values for \( k \) to find positive results for \( c \).
If \( k = 0 \): \( c = - \dfrac{\pi}{3} \) (Negative, discard).
If \( k = 1 \): \( c = 2\pi - \dfrac{\pi}{3} = \dfrac{6\pi}{3} - \dfrac{\pi}{3} = \mathbf{\dfrac{5\pi}{3}} \).
If \( k = 2 \): \( c = 4\pi - \dfrac{\pi}{3} = \dfrac{12\pi}{3} - \dfrac{\pi}{3} = \mathbf{\dfrac{11\pi}{3}} \).

Conclusion:
Two possible positive values for \( c \) are \( \dfrac{5\pi}{3} \) and \( \dfrac{11\pi}{3} \). Both create a valid equation that describes the exact same graph.