Problem 61
Question
Identify the period of each function. Then tell where two asymptotes occur for each function. $$ y=\tan 6 \theta $$
Step-by-Step Solution
Verified Answer
The period of the function \(y = \tan 6\theta\) is \(\pi / 6\), and two asymptotes occur at \(\theta = \pi/12\) and \(\theta = 3\pi/12\).
1Step 1: Find the Period of the Function
The period of the function is determined by the coefficient of \(\theta\) in \(\tan 6\theta\). In the basic tangent function \(\tan \theta\), the period is \(\pi\). For \(\tan 6\theta\), the '6' tells us that the function completes six cycles in the space that the basic tangent function \(\tan \theta\) completes just one. Therefore, the period of the function \(\tan 6\theta\) is found by dividing the period of \(\tan \theta\) by the absolute value of the coefficient, 6. Hence, the period of \(\tan 6\theta\) is \(\pi / 6\).
2Step 2: Identify the Asymptotes of the Function
For the basic tangent function \(\tan \theta\), asymptotes occur at \(\theta = (2n + 1) \pi/2\), for any integer n. But for \(\tan 6\theta\), the asymptotes will occur six times as often. We take the equation for the basic tangent function asymptotes and replace \(\theta\) with \(6\theta\). Hence, the equation for the asymptotes of the function \(\tan 6\theta\) is given by \(6\theta = (2n + 1) \pi/2\). To find \(\theta\), we need to divide both sides of this equation by 6.
3Step 3: Solving for Theta
Dividing both sides of the equation for the asymptotes \(6\theta = (2n + 1) \pi/2\) by 6 gives \(\theta = (2n + 1) \pi/12\). Hence, for \(\tan 6\theta\), two asymptotes occur at \(\theta = \pi/12\) and \(\theta = 3\pi/12\), assuming n to be 0 and 1.
Key Concepts
Period of Tangent FunctionAsymptotesTangent Function Transformations
Period of Tangent Function
The period of a trigonometric function is how often it repeats its pattern. For the basic tangent function, \( an \theta\), this period is \(\pi\). This means that every \(\pi\) radians, the tangent function starts to repeat its cycle.
However, in the function \(\tan 6\theta\), the '6' represents a transformation. It affects the period by speeding up the frequency of cycles. The function \( an 6\theta\) completes 6 cycles in the same interval where \(\tan \theta\) completes one. To find the new period, we divide the original period \((\pi)\) by 6. Therefore, the period of \(\tan 6\theta\) is \(\pi / 6\).
A shorter period means the function completes its up-and-down cycle faster. This is important in graphing and understanding the behavior of the tangent function.
However, in the function \(\tan 6\theta\), the '6' represents a transformation. It affects the period by speeding up the frequency of cycles. The function \( an 6\theta\) completes 6 cycles in the same interval where \(\tan \theta\) completes one. To find the new period, we divide the original period \((\pi)\) by 6. Therefore, the period of \(\tan 6\theta\) is \(\pi / 6\).
A shorter period means the function completes its up-and-down cycle faster. This is important in graphing and understanding the behavior of the tangent function.
Asymptotes
Asymptotes are lines that a graph approaches but never actually touches. For the basic \(\tan \theta\) function, these occur at \((2n + 1)\frac{\pi}{2}\), where \(n\) is any integer. These are points where the tangent function is undefined and the graph shoots off to infinity.
When we consider \(\tan 6\theta\), the frequency of asymptotes increases because the period is shorter. The location of asymptotes now depends on the equation \(6\theta = (2n + 1) \frac{\pi}{2}\). By solving for \(\theta\), we find where these vertical asymptotes will appear.
To find them, divide both sides by 6, resulting in \(\theta = (2n + 1) \frac{\pi}{12}\). This means every \(\pi/12\), and its multiples, you'll encounter an asymptote. \(n = 0\) gives the first asymptote at \(\pi/12\), and \(n = 1\) gives a second one at \(3\pi/12\).
When we consider \(\tan 6\theta\), the frequency of asymptotes increases because the period is shorter. The location of asymptotes now depends on the equation \(6\theta = (2n + 1) \frac{\pi}{2}\). By solving for \(\theta\), we find where these vertical asymptotes will appear.
To find them, divide both sides by 6, resulting in \(\theta = (2n + 1) \frac{\pi}{12}\). This means every \(\pi/12\), and its multiples, you'll encounter an asymptote. \(n = 0\) gives the first asymptote at \(\pi/12\), and \(n = 1\) gives a second one at \(3\pi/12\).
Tangent Function Transformations
Transformations of trigonometric functions, like tangent, involve changes that alter their shape or position. There are several types of transformations.
- Horizontal Stretch/Compression: This is related to the constant multiplying \(\theta\). In \(\tan 6\theta\), the '6' compresses the function horizontally. The period reduces. That's why we see cycle repetitions every \(\pi/6\).
- Vertical Shifts and Stretches: Sometimes modifications outside the trigonometric expression alter the graph vertically. However, in \(\tan 6\theta\), there are no vertical transformations as there are no additional terms.
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