Find Amplitude, Period, and Phase Shift y=10sin(theta/6+30)-5
The problem provided relates to trigonometry and involves analyzing a transformed sine function to determine its amplitude, period, and phase shift. The particular function given is "y = 10sin(θ/6 + 30) - 5." The task is to:
Find the Amplitude: This part of the question requires identifying the coefficient that determines how far the sine wave oscillates above and below its central axis.
Find the Period: This task involves calculating the time it takes for one complete cycle of the sine wave to occur, which is directly affected by the coefficient in front of the variable θ.
Find the Phase Shift: The phase shift is the horizontal displacement of the sine wave from its standard position, determined by the constant added or subtracted within the sine function's argument.
The problem does not request solving for any specific values of θ, only the general features of the sinusoidal function as modified by the numerical coefficients and constants provided.
$y = 10 sin \left(\right. \frac{\theta}{6} + 30 \left.\right) - 5$
Answer
Solution:
Step 1:
Identify the parameters $a$, $b$, $c$, and $d$ from the standard trigonometric form $a \sin(b\theta + c) + d$ to determine amplitude, period, phase shift, and vertical shift.
- Amplitude coefficient $a = 10$
- Frequency coefficient $b = \frac{1}{6}$
- Phase shift constant $c = 30$ (note the sign change due to the form used)
- Vertical shift $d = -5$
Step 2:
Calculate the amplitude as the absolute value of $a$.
- Amplitude: $|a| = |10| = 10$
Step 3:
Compute the period using the formula $T = \frac{2\pi}{|b|}$.
Step 3.1:
Determine the period for the sine function $10 \sin\left(\frac{\theta}{6} + 30\right)$.
Step 3.1.1:
Use the period formula $T = \frac{2\pi}{|b|}$.
Step 3.1.2:
Substitute $b$ with $\frac{1}{6}$ to find the period.
- $T = \frac{2\pi}{|\frac{1}{6}|}$
Step 3.1.3:
Since $\frac{1}{6}$ is positive, we can disregard the absolute value.
- $T = \frac{2\pi}{\frac{1}{6}}$
Step 3.1.4:
Calculate the period by multiplying $2\pi$ by the reciprocal of $\frac{1}{6}$.
- $T = 2\pi \cdot 6$
Step 3.1.5:
Simplify the multiplication to get the period.
- Period: $T = 12\pi$
Step 3.2:
The constant term $-5$ does not affect the period of the sine function.
Step 4:
Determine the phase shift using the formula $\text{Phase Shift} = \frac{c}{b}$.
Step 4.1:
Calculate the phase shift with $\text{Phase Shift} = \frac{c}{b}$.
Step 4.2:
Insert the values for $c$ and $b$ into the phase shift equation.
- Phase Shift: $\frac{30}{\frac{1}{6}}$
Step 4.3:
Multiply the numerator by the reciprocal of the denominator to find the phase shift.
- Phase Shift: $30 \cdot 6$
Step 4.4:
Complete the multiplication to obtain the phase shift.
- Phase Shift: $180$ (This is a shift to the right, opposite to the original solution's direction)
Step 5:
Summarize the properties of the trigonometric function.
- Amplitude: $10$
- Period: $12\pi$
- Phase Shift: $180$ (to the right)
- Vertical Shift: $-5$
Knowledge Notes:
The amplitude of a trigonometric function is the coefficient in front of the sine or cosine function, which determines the height of the wave's peaks and troughs.
The period of a sine or cosine function is the length of one complete cycle of the wave, which can be calculated using the formula $T = \frac{2\pi}{|b|}$, where $b$ is the frequency coefficient.
The phase shift of a trigonometric function is the horizontal shift along the x-axis and is calculated by the formula $\text{Phase Shift} = \frac{c}{b}$, where $c$ is the phase shift constant in the function's equation.
The vertical shift is the constant $d$ added or subtracted from the trigonometric function, which moves the graph up or down along the y-axis.
In the context of the given problem, the trigonometric function is a sine function with modifications to its amplitude, period, phase shift, and vertical shift due to the coefficients and constants present in the equation $y = 10\sin\left(\frac{\theta}{6} + 30\right) - 5$.
