Question

practice assignment 8.2 graphs of the other trigonometric functions
score: 3/9 answered: 5/9
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question 6
the graph above is a graph of what function?
$y = \tan(x)$
$y = \cot(x)$
$y = \cos(x)$
$y = \sec(x)$
$y = \sin(x)$
$y = \csc(x)$

Explanation:

Step1: Analyze the properties of each trigonometric function

• For \(y = \tan(x)\), its period is \(\pi\), and it has vertical asymptotes at \(x=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}\), and its graph is a set of "S - shaped" curves between vertical asymptotes.
• For \(y=\cot(x)=\frac{\cos(x)}{\sin(x)}\), its period is \(\pi\), and it has vertical asymptotes at \(x = k\pi,k\in\mathbb{Z}\), and its graph is a set of "reverse - S - shaped" curves between vertical asymptotes.
• For \(y=\cos(x)\), its period is \(2\pi\), and its range is \([- 1,1]\), and it is a smooth sinusoidal curve.
• For \(y=\sec(x)=\frac{1}{\cos(x)}\), its period is \(2\pi\), and it has vertical asymptotes at \(x=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}\). The graph of \(y = \sec(x)\) has U - shaped curves.
• For \(y=\sin(x)\), its period is \(2\pi\), and its range is \([-1,1]\), and it is a smooth sinusoidal curve.
• For \(y=\csc(x)=\frac{1}{\sin(x)}\), its period is \(2\pi\), and it has vertical asymptotes at \(x=k\pi,k\in\mathbb{Z}\).

Step2: Match the graph with the function

The given graph has vertical asymptotes at \(x=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}\) and has a pattern of U - shaped curves. The function \(y = \sec(x)=\frac{1}{\cos(x)}\) has vertical asymptotes at \(x=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}\) (since \(\cos(x) = 0\) at \(x=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}\)) and its graph is composed of U - shaped curves.

Answer:

\(y=\sec(x)\)