Trigonometry - Graphing Tangent and Cotangent

Which of the following best describes where the asymptotes are located on a tangent graph?

Angle measures where the cosine is 0, such as

Angle measures where the sine is 0, such as .

Angle measures where the sine and cosine are equal, such as .

Angle measures where the tangent cannot be calculated.

Explanation

In trigonometry,

$tangent = \frac{opposite}{adjacent}$ .

It may also be thought of as $\frac{sine}{cosine}$ .

This is because

$sine = \frac{opposite}{hypotenuse}$ and $cosine = \frac{adjacent}{hypotenuse}$ , so

$\frac{sine}{cosine}=\frac{opposite}{hypotenuse}\div \frac{adjacent}{hypotenuse}$

$\frac{sine}{cosine}=\frac{opposite}{hypotenuse}\cdot \frac{hypotenuse}{adjacent}=\frac{opposite}{adjacent}}=tangent$ .

This means that whenever cosine is 0, tangent is undefined, because it would be evaluated by dividing by 0.

Which of the following is the graph of $tan(3(x+\frac{\pi}{3}) +1$ ?

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Explanation

We will begin by considering the general graph of and apply transformations step by step to produce a graph of $tan(3(x+\frac{\pi}{3}) +1$ . The graph of is

Screen shot 2020 08 27 at 11.37.25 am

The general equal of a tangent transformation equation is . A is the amplitude of the graph of tangent. Here, so we do not need to apply a transformation here. Next, we will consider the period. The period of the tangent function is equal to $\frac{\pi}{B}$ . So the period of our graph would be

Period = $\frac{\pi}{B}$

Period = $\frac{\pi}{3}$

So the period is shortened from $\pi$ to $\frac{\pi}{3}$ .

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Now, we will consider $C = -\frac{\pi}{3}$ . is the phase shift of our graph. So we will shift our graph $\frac{\pi}{3}$ units to the left. This does not change our graph visually due to the period now being $\frac{\pi}{3}$ . Lastly, we will consider . is the vertical shift of our graph, and so we must shift our graph 1 unit up.

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And we are left with the graph of $tan(3(x+\frac{\pi}{3}) +1$ .

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Which of the following is the graph of ?

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Explanation

To derive the graph of , recall that $tan(x) = \frac{sin(x)}{cos(x)}$ . The graph of is

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and the graph of is

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Vertical asymptotes will occur in the graph of whenever . This is because the denominator of the tangent function will be equal to zero whenever the cosine function is equal to zero and then the entire function will be undefined at those points. Wherever cosine crosses the x-axis a vertical asymptote will occur. If we overlay the sine and cosine graphs we see the following:

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So our tangent graph will follow the same form as the sine and cosine graphs when they are increasing, but will have vertical asymptotes wherever cosine crosses the x-axis.

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And we are left with our graph of

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Which of the following is the graph of ?

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Explanation

To derive the graph of recall that $cot(x) = \frac{1}{tan(x)}$ . So the tangent and cotangent graphs are reciprocals of one another. We will consider the tangent graph since it is one we are more familiar with:

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Now we will simply invert the tangent graph to get the cotangent graph

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And we are left with our cotangent graph

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Which of the following is the graph of $4cot(x + \frac{2\pi}{5})$ .

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Explanation

First, we will consider the graph of and apply transformations step-by-step. The graph of is

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The general form of a cotangent transformation function is . For our function $4cot(x + \frac{2\pi}{5})$ , and so we need to increase the amplitude 4 units.

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here so we do not need to make any changes to the period of this graph. $C = -\frac{2\pi}{5}$ , giving us a negative phase shift of $\frac{2\pi}{5}$ units.

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This leaves us with our graph of $4cot(x + \frac{2\pi}{5})$ .

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True or False: The period of tangent and cotangent function is $\pi$ .

True

False

Explanation

This is because $tan(x) = \frac{sin(x)}{cos(x)}$ and $cos(n\pi) = 0$ causing the tangent function to be undefined at these points and forming a vertical asymptote. This is also true for the cotangent function because $cot(x) = \frac{1}{tan(x)}$ so wherever is zero or undefined, cotangent will be as well.

Which of the following is not a solution to the following equation?

$\frac{\pi }{4}$

$\frac{3\pi }{4}$

$\frac{7\pi }{4}$

$\frac{-5\pi }{4}$

...

Explanation

We can factor the original expression as follows:

$tan^2 x + 3tan\ x + 2 = (tan\ x + 1)(tan\ x + 2) = 0{}$

So from this equation we conclude either that:

$tan\ x + 1 = 0; \ \ tan\ x = -1; \ \ x = \frac{3\pi }{4} + n\pi$

$tan\ x + 2 = 0; \ \ tan\ x = -2; \ \ x \approx 2.0344 + n\pi$

So any number that is not some integer multiple of $\pi$ away from these two solutions is not a solution to the original equation.

The only such choice is $\frac{\pi}{4}{}$ , which is $\frac{3\pi }{4} + n\pi, \ n = - \frac{1}{2}{}$ ; n is not an integer, therefore it is not a solution.

The following is a graph of which function?

Screen shot 2015 07 20 at 12.13.22 pm

$2 \tan (x+\pi) + 1$

$cot (x + \pi)$

$cot (x + \pi) + 1$

$sin(x+\pi) - 3$

$tan(x+\pi)$

Explanation

The graph looks to have infinite range, but multiple vertical asymptotes. That means we can limit our choices to tangent and cotangent graphs.

Furthermore, we observe that the graph starts at the bottom and increases from left to right, consistent with tangent graphs. So we narrow our focus to the choices involving tangents.

To decide between the remaining two graphs, observe that y-intercept (where x=0) of our graph is (0,1).

Now $tan(x + \pi){}$ evaluated at is $tan (\pi) = 0$ , which means that we need a vertical shift of unit.