Math Problem Statement

Find the amplitude, period, and phase shift of the function, and graph one complete period. a) y = 3 cos ( x + π/ 4)  b) y = sin 1/ 2 ( x − π/ 4)  c) y = 2 sin (2x − π)

Solution

Let's break down each trigonometric function to find the amplitude, period, and phase shift, then summarize for each case.


a) y=3cos(x+π4)y = 3 \cos \left( x + \frac{\pi}{4} \right)

Step 1: Amplitude
The amplitude is the coefficient of the cosine function.
Amplitude=3\text{Amplitude} = 3

Step 2: Period
The standard period of cos(x)\cos(x) is 2π2\pi. Since there's no coefficient of xx, the period remains the same:
Period=2π\text{Period} = 2\pi

Step 3: Phase Shift
To find the phase shift, we look at the expression inside the cosine. The general form is cos(bx+c)\cos(bx + c). The phase shift is calculated as:
Phase Shift=cb\text{Phase Shift} = -\frac{c}{b} Here, c=π4c = \frac{\pi}{4} and b=1b = 1, so:
Phase Shift=π41=π4\text{Phase Shift} = -\frac{\frac{\pi}{4}}{1} = -\frac{\pi}{4}

So, this function has a phase shift of π4-\frac{\pi}{4}, meaning it's shifted to the left by π4\frac{\pi}{4}.


b) y=sin(12(xπ4))y = \sin \left( \frac{1}{2} \left( x - \frac{\pi}{4} \right) \right)

Step 1: Amplitude
The amplitude is 1 since the coefficient of the sine function is 1.

Step 2: Period
The period of the sine function is 2π2\pi, but here the function has a coefficient of 12\frac{1}{2} inside the parentheses. The period is calculated as:
Period=2πb\text{Period} = \frac{2\pi}{b} where b=12b = \frac{1}{2}, so:
Period=2π12=4π\text{Period} = \frac{2\pi}{\frac{1}{2}} = 4\pi

Step 3: Phase Shift
The phase shift is given by cb\frac{c}{b}, where c=π4c = \frac{\pi}{4} and b=12b = \frac{1}{2}:
Phase Shift=π412=π2\text{Phase Shift} = \frac{\frac{\pi}{4}}{\frac{1}{2}} = \frac{\pi}{2}

So, the phase shift is π2\frac{\pi}{2} to the right.


c) y=2sin(2xπ)y = 2 \sin(2x - \pi)

Step 1: Amplitude
The amplitude is 2 since the coefficient of the sine function is 2.
Amplitude=2\text{Amplitude} = 2

Step 2: Period
The period is calculated as:
Period=2πb\text{Period} = \frac{2\pi}{b} Here, b=2b = 2, so:
Period=2π2=π\text{Period} = \frac{2\pi}{2} = \pi

Step 3: Phase Shift
The phase shift is given by cb\frac{c}{b}, where c=πc = \pi and b=2b = 2:
Phase Shift=π2\text{Phase Shift} = \frac{\pi}{2}

So, the phase shift is π2\frac{\pi}{2} to the right.


Summary:

  1. a) y=3cos(x+π4)y = 3 \cos \left( x + \frac{\pi}{4} \right)

    • Amplitude: 33
    • Period: 2π2\pi
    • Phase Shift: π4-\frac{\pi}{4} (left)
  2. b) y=sin(12(xπ4))y = \sin \left( \frac{1}{2} \left( x - \frac{\pi}{4} \right) \right)

    • Amplitude: 11
    • Period: 4π4\pi
    • Phase Shift: π2\frac{\pi}{2} (right)
  3. c) y=2sin(2xπ)y = 2 \sin(2x - \pi)

    • Amplitude: 22
    • Period: π\pi
    • Phase Shift: π2\frac{\pi}{2} (right)

Would you like me to provide graphs for each function or additional details on these results?

Here are some follow-up questions to deepen your understanding:

  1. How does changing the amplitude affect the shape of the graph?
  2. What happens to the period if the coefficient of xx inside the sine or cosine function increases?
  3. How does the phase shift impact the starting point of the graph?
  4. Can you explain how the period differs between sine and cosine functions?
  5. How would you graph a cosine function with a negative amplitude?

Tip: The period of a trigonometric function depends on the frequency, which is the coefficient of xx inside the function.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Amplitude
Period
Phase Shift

Formulas

Amplitude = |A| (from y = A sin(Bx + C) or y = A cos(Bx + C))
Period = 2π/B (from y = A sin(Bx + C) or y = A cos(Bx + C))
Phase Shift = -C/B (from y = A sin(Bx + C) or y = A cos(Bx + C))

Theorems

Basic Properties of Sine and Cosine Functions

Suitable Grade Level

Grades 10-12