6.07 Graphing reciprocal trigonometric functions

State the equations of the asymptotes on the interval \left[0, 2\pi\right].

We know that the asymptotes of y=\csc x are at multiples of x=\pi. If we can identify how {f\left(x\right)=\csc\left(3x\right)} has been transformed from y=\csc x, then we can determine how the asymptotes have been affected.

The given function is in the form f\left(x\right)=\csc\left(b\cdot x\right) where b=3 which means it has been horizontally compressed by a factor of \dfrac{1}{3}. This will affect the location of the asymptotes, causing them to occur at multiples of x=\dfrac{\pi}{3}.

On the interval \left[0,2\pi\right], the vertical asymptotes are located at x=0,\,\dfrac{\pi}{3},\,\dfrac{2\pi}{3},\,\pi,\,\dfrac{4\pi}{3},\,\dfrac{5\pi}{3},\,2\pi.

We can check this answer by seeing if the stated values make \csc\left(3x\right) undefined or if they make \sin\left(3x\right)=0.

• \sin\left(3\cdot 0\right)=\sin 0=0
• \sin\left(3\cdot \frac{\pi}{3}\right)=\sin \pi=0
• \sin\left(3\cdot \frac{2\pi}{3}\right)=\sin 2\pi=0
• \sin\left(3\cdot \pi\right)=\sin 3\pi=0
• \sin\left(3\cdot \frac{4\pi}{3}\right)=\sin 4\pi=0
• \sin\left(3\cdot \frac{5\pi}{3}\right)=\sin 5\pi=0
• \sin\left(3\cdot 2\pi\right)=\sin 6\pi=0

Sketch the graph of f \left(x\right) = \csc \left(3x\right).

In the previous part, we found that f\left(x\right)=\csc\left(3x\right) is the graph of y=\csc x that has been horizontally compressed by a factor of \dfrac{1}{3}, and its asymptotes occur at multiples of \dfrac{\pi}{3}.

Graphing the parent function, y=\csc x, we have:

Compressing this by a factor of \dfrac{1}{3} results in:

Another way we could have graphed the function is by first graphing y=\sin\left(3x\right), then drawing the vertical asymptotes through the x-intercepts, and drawing the U-shaped curves tangent to the maxima and minima of sine but in between the drawn asymptotes.

Explain whether the following key features are affected by the transformation:

• Period
• Midline
• Domain

As we found in part (a), the function has been compressed horizontally by a factor of \dfrac{1}{3} which affected the asymptotes of the parent function y=\csc x. From this, we also know the period and domain have been affected by the transformation.

Since the transformation only compressed the function horizontally, it had no effect on the midline. So, the midline remains at y=0.

Since the function was compressed by a factor of \dfrac{1}{3}, the period will be compressed by this factor. The period of the transformed function is \dfrac{2\pi}{3}.

Because the period and asymptotes of the function have been compressed, the domain has been affected too. The domain of the transformed function is x\neq \dfrac{n\pi}{3} where n is any integer.

State the values of \theta on the interval \left[0, 2\pi\right] for which f \left( \theta \right) is undefined.

We know that f \left( \theta \right) = \cot \theta = \dfrac{\cos \theta}{\sin \theta}, so any values where \sin \theta = 0 are where f \left( \theta \right) is undefined.

The values of \theta where f \left( \theta \right)=\cot\theta is undefined are \theta=0, \pi, and 2 \pi.

We can also use the graph of y=\tan\theta to determine where the undefined values are.

Since \cot \theta=\dfrac{1}{\tan \theta}, we know that f\left(\theta\right)=\cot\theta will be undefined where \tan\theta=0, and \tan\theta=0 at multiples of \theta=\pi.

Sketch the graph of f \left( \theta \right) = \cot \theta on the interval \left[ -2 \pi, 2 \pi \right].

We can complete the table of values by evaluating \cot\theta=\dfrac{\cos \theta}{\sin \theta}, then plot the points \left(\theta,\cot\theta\right) on a coordinate plane.

The asymptotes of the cotangent function occur at multiples of \pi, which is where the values of the sine function approach and become equal to 0.

To graph the function from -2\pi to 0, we can use the fact that angles that are coterminal on the unit circle will result in the same values of \cot \theta. For example, \theta= \dfrac{-\pi}{2} is coterminal to \alpha= \dfrac{3 \pi}{2}, and since \cot \left( \dfrac{3\pi}{2} \right)=0, \cot \left( \dfrac{-\pi}{2} \right) = 0 also. The following coterminal angles give the exact values:

\cot \left( \dfrac{-3 \pi}{2} \right) = \cot \left( \dfrac{\pi}{2} \right) = 0

\cot \left( \dfrac{-7 \pi}{4} \right) = \cot \left( \dfrac{\pi}{4} \right) = 1

\cot \left( \dfrac{-5 \pi}{4} \right) = \cot \left( \dfrac{3\pi}{4} \right) = -1

\cot \left( \dfrac{-3 \pi}{4} \right) = \cot \left( \dfrac{5\pi}{4} \right) = 1

\cot \left( \dfrac{- \pi}{4} \right) = \cot \left( \dfrac{7\pi}{4} \right) = -1

We can now graph the periodic function:

Compare the domain and range of f \left( \theta \right) = \cot \theta to g \left( \theta \right) = \tan \theta.

For the parent tangent function:

• Domain: x \neq \dfrac{\pi}{2} + n \pi, n is any integer

• Range: \left( -\infty, \infty \right)

For the cotangent function:

• Domain: x \neq n \pi, n is any integer

• Range: \left( -\infty, \infty \right)

The range for the tangent and cotangent functions is the same. However, the asymptotes of the functions are different, so the domain for the functions is different.

The excluded values of cotangent are at multiplies of \pi whereas the excluded values for the tangent function are at odd multiples of \dfrac{\pi}{2}.

Note that the domain for the functions relates to their asymptotes and the unit circle, and recall that since the tangent function is the ratio \dfrac{\sin \theta}{\cos \theta}, anywhere cosine is equal to 0 on the unit circle, tangent must be undefined.