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Chapter 5: Problem 9

Without graphing, determine the amplitude and period of each function. State the period in degrees and in radians. a) \(y=2 \sin x\) b) \(y=-4 \cos 2 x\) c) \(y=\frac{5}{3} \sin \left(-\frac{2}{3} x\right)\) d) \(y=3 \cos \frac{1}{2} x\)

Short Answer

Expert verified
a) Amplitude: 2, Period: 360掳 or 2\pi radians. b) Amplitude: 4, Period: 180掳 or \pi radians. c) Amplitude: \(\frac{5}{3}\), Period: 540掳 or 3\pi radians. d) Amplitude: 3, Period: 720掳 or 4\pi radians.

Step by step solution

01

Identify amplitude and period for y=2 \sin x

For the function \(y = A \sin(Bx + C) + D\), the amplitude is \(|A|\) and the period is \(\frac{2\pi}{B}\). For \(y = 2 \sin x\), A = 2 and B = 1. Therefore, the amplitude is \(|2| = 2\) and the period is \(\frac{2\pi}{1} = 2\pi\) radians or \(360\) degrees.
02

Identify amplitude and period for y=-4 \cos 2x

For the function \(y = A \cos(Bx + C) + D\), the amplitude is \(|A|\) and the period is \(\frac{2\pi}{B}\). For \(y = -4 \cos 2x\), A = -4 and B = 2. Therefore, the amplitude is \(|-4| = 4\) and the period is \(\frac{2\pi}{2} = \pi\) radians or \(180\) degrees.
03

Identify amplitude and period for y=\frac{5}{3} \sin \left( -\frac{2}{3} x \right)

For the function \(y = A \sin(Bx + C) + D\), the amplitude is \(|A|\) and the period is \(\frac{2\pi}{|B|}\). For \(y = \frac{5}{3} \sin \left( -\frac{2}{3} x \right)\), A = \(\frac{5}{3}\) and B = \(-\frac{2}{3}\). Therefore, the amplitude is \(\left| \frac{5}{3} \right| = \frac{5}{3}\) and the period is \(\frac{2\pi}{\left| -\frac{2}{3} \right|} = \frac{2\pi}{\frac{2}{3}} = 3\pi\) radians or \(540\) degrees.
04

Identify amplitude and period for y=3 \cos \frac{1}{2} x

For the function \(y = A \cos(Bx + C) + D\), the amplitude is \(|A|\) and the period is \(\frac{2\pi}{B}\). For \(y = 3 \cos \frac{1}{2} x\), A = 3 and B = \(\frac{1}{2}\). Therefore, the amplitude is \(|3| = 3\) and the period is \(\frac{2\pi}{\frac{1}{2}} = 4\pi\) radians or \(720\) degrees.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Amplitude
The amplitude of a trigonometric function measures how far the function's values oscillate above and below its central axis. For the general functions of the form \(y = A \, \sin(Bx + C) + D\) or \(y = A \, \cos(Bx + C) + D\), the amplitude is given by \(|A|\). This value determines the height of the peaks and the depth of the troughs of the sine or cosine wave.
For example:
  • In the function \(y = 2 \, \sin x\), the amplitude is \(|2| = 2\).
  • In the function \(y = -4 \, \cos 2x\), the amplitude is \(|-4| = 4\).
  • In the function \(y = \frac{5}{3} \, \sin \left( -\frac{2}{3} x \right)\), the amplitude is \(\left| \frac{5}{3} \right| = \frac{5}{3}\).
  • In the function \(y = 3 \, \cos \frac{1}{2} x\), the amplitude is \(|3| = 3\).
The larger the amplitude, the taller the graph of the function will be.
Keep this in mind as you analyze or graph trigonometric functions.
Period of a function
The period of a trigonometric function is the distance required for the function to complete one full cycle of its pattern. For the general functions of the form \(y = A \, \sin(Bx + C) + D\) or \(y = A \, \cos(Bx + C) + D\), the period is calculated as \(\frac{2\pi}{|B|}\).
This important characteristic tells you how long it takes for the function to repeat its cycle.
Here are some examples:
  • In \(y = 2 \, \sin x\), the period is \(\frac{2\pi}{1} = 2\pi\) radians or 360 degrees.
  • In \(y = -4 \, \cos 2x\), the period is \(\frac{2\pi}{2} = \pi\) radians or 180 degrees.
  • In \(y = \frac{5}{3} \, \sin \left( -\frac{2}{3} x \right)\), the period is \( \frac{2\pi}{\left| -\frac{2}{3} \right|} = \frac{2\pi}{\frac{2}{3}} = 3\pi\) radians or 540 degrees.
  • In \(y = 3 \, \cos \frac{1}{2} x\), the period is \(\frac{2\pi}{\frac{1}{2}} = 4\pi\) radians or 720 degrees.
Understanding the period helps in graphing and analyzing repeating behaviors in real-world scenarios.
Always note the effect of coefficient \(B\) on the period of the function.
Radians and Degrees
Radians and degrees are two units used to measure angles. When working with trigonometric functions, it's essential to be comfortable converting between these units.
A full circle is \(360\) degrees, which is equivalent to \(2\pi\) radians.
Here are a few important conversions to remember:
  • \(180\) degrees is equal to \(\pi\) radians.
  • To convert from degrees to radians, use the formula: \( \text{radians} = \text{degrees} \times \frac{\pi}{180} \).
  • To convert from radians to degrees, use the formula: \( \text{degrees} = \text{radians} \times \frac{180}{\pi} \).
Let鈥檚 see how these conversions work in practice:
  • If the period of a trigonometric function is \(2\pi\) radians, you can convert it to degrees: \(2\pi \times \frac{180}{\pi} = 360\) degrees.
  • If the period is \(\pi\) radians, it converts to \(\pi \times \frac{180}{\pi} = 180\) degrees.
  • If you have \(4\pi\) radians, this translates to \(4\pi \times \frac{180}{\pi} = 720\) degrees.
Familiarity with both units and the ease of conversion can help you better understand and communicate results related to trigonometric functions.

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Most popular questions from this chapter

Golden, British Columbia, is one of the many locations for heliskiing in Western Canada. When skiing the open powder, the skier leaves behind a trail, with two turns creating one cycle of the sinusoidal curve. On one section of the slope, a skier makes a total of 10 turns over a \(20-s\) interval. a) If the distance for a turn, to the left or to the right, from the midline is \(1.2 \mathrm{m}\) determine the function that models the path of the skier. b) How would the function change if the skier made only eight turns in the same 20-s interval?

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Sketch the graph of each function over the interval \(\left[-360^{\circ}, 360^{\circ}\right] .\) For each function, clearly label the maximum and minimum values, the \(x\) -intercepts, the \(y\) -intercept, the period, and the range. a) \(y=2 \cos x\) b) \(y=-3 \sin x\) c) \(y=\frac{1}{2} \sin x\) d) \(y=-\frac{3}{4} \cos x\)
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