Problem 64
Question
Identify the period of each function. Then tell where two asymptotes occur for each function. $$ y=\tan \frac{\theta}{6} $$
Step-by-Step Solution
Verified Answer
The function \( y=\tan \frac{\theta}{6} \) has a period of \( 6\pi \) and its first two asymptotes occur at \( \theta = 3\pi \) and \( \theta = 15\pi \).
1Step 1: Find the period
The period of the standard tangent function is \( \pi \), but since the input is divided by 6 in this case, the period is stretched by a factor of 6. This can be obtained by taking the standard period, \( \pi \), and dividing it by the absolute value of the coefficient of \( \theta \), which here is \( \frac{1}{6} \). Therefore, the period of \( y=\tan \frac{\theta}{6} \) is \( \pi \) divided by \( \frac{1}{6} \), which equals \( \pi \times 6 = 6\pi \).
2Step 2: Find the asymptotes
Asymptotes for \( y=\tan(\theta) \) function occur at \( \frac{\pi}{2} + n\pi \) where n is an integer. So, as \( \theta \) in the formula is divided by 6, asymptotes for \( y=\tan \frac{\theta}{6} \) will occur when the part of the formula inside the parentheses, \( \frac{\theta}{6} \), equals \( \frac{\pi}{2} + n\pi \). Solving this for \( \theta \) gives \( \theta = 3\pi + 12n\pi \), where n is an integer. Therefore, the first two asymptotes occur when n = 0 and n = 1, which gives \( \theta = 3\pi \) and \( \theta = 15\pi \).
Key Concepts
Tangent FunctionPeriod of a FunctionVertical Asymptotes
Tangent Function
The tangent function, often written as \( y = \tan(\theta) \), is a fundamental trigonometric function. It relates to the sine and cosine functions through the equation \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \). This relationship shows that the tangent function is undefined whenever the cosine of \( \theta \) is zero, leading to vertical asymptotes. Because of this, the tangent function exhibits interesting behavior:
- The graph of the tangent function has a series of recurring patterns.
- It stretches indefinitely in the vertical direction (upward and downward), giving it an elongated look.
- Where the cosine is zero, the tangent goes through vertical asymptotes, creating breaks in its graph.
Period of a Function
A period of a function refers to the horizontal length of one complete cycle of the graph. For a standard tangent function \( y = \tan(\theta) \), this period is \( \pi \), meaning every \( \pi \) units horizontally, the graph repeats its pattern.
However, when dealing with a function like \( y = \tan\left(\frac{\theta}{6}\right) \), the period changes. You can determine the period of the altered tangent function by using the formula:
However, when dealing with a function like \( y = \tan\left(\frac{\theta}{6}\right) \), the period changes. You can determine the period of the altered tangent function by using the formula:
- \( \text{Period} = \frac{\pi}{|\text{coefficient of } \theta|} \)
Vertical Asymptotes
Vertical asymptotes are lines where a function moves towards positive or negative infinity. For tangent functions, these lines occur whenever the cosine part of the function equals zero. In the case with \( y = \tan\left(\frac{\theta}{6}\right) \), the asymptotes form based on the division of the angle \( \theta \) by 6.
- For a general tangent equation \( y = \tan(\theta) \), vertical asymptotes occur at \( \theta = \frac{\pi}{2} + n\pi \), where \( n \) is an integer.
- However, for \( y = \tan\left(\frac{\theta}{6}\right) \), this becomes the points \( \frac{\theta}{6} = \frac{\pi}{2} + n\pi \).
Other exercises in this chapter
Problem 64
Graph each function in the interval from 0 to 2\(\pi .\) Describe any phase shift and vertical shift in the graph. $$ y=\csc 2\left(\theta-\frac{\pi}{2}\right)
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Find the 27 th term of each sequence. $$ 59,48,37, \dots $$
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Sketch each angle in standard position. $$ -75^{\circ} $$
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An angle drawn in standard position has a terminal side that passes through the point \((\sqrt{2},-\sqrt{2}) .\) What is one possible measure of the angle? $$ \
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