5.4. More Trigonometric Graphs

Graph of $y = sec x = \frac{1}{cos x}$

$y = sec x$ is undefined whenever $cos x$ becomes zero. Therefore, the domain of the secant function is all real numbers except $x = \frac{(2 n + 1) π}{2}$ , where $n$ is an integer. The secant function has vertical asymptotes whenever the function is undefined. The equations of the vertical asymptotes are

$x = \frac{(2 n + 1) π}{2}$ , where $n$ is an integer. The period of the secant function is $2 π$ .

The range of the secant function is $(- \infty, - 1] \cup [1, \infty) .$

The secant function is an even function because $sec (- x) = \frac{1}{cos (- x)} = \frac{1}{cos x} = sec x .$

Example

Find the period of $y = 4 sec (2 π x)$ and graph it.

Solution

Period is 1( $= \frac{2 π}{2 π})$ . Since $y = 4 sec (2 π x) = \frac{4}{cos (2 π x)}$ , we draw the graph of $y = cos (2 π x)$ first. Then we draw the vertical asymptotes of $y = 4 sec (2 π x),$ whenever $y = cos (2 π x)$ is zero. Obviously, the lowest part of the U shape part of the graph of $y = 4 sec (2 π x)$ is $4$ and the highest part of the n shape part of the graph is $- 4$ . We complete the graph by alternating the U shape and n shape graphs.

Graph of $y = csc x = \frac{1}{sin x}$

$y = csc x$ is undefined whenever $sin x$ becomes zero. Therefore, the domain of the cosecant function is all real numbers except $x = n π$ , where $n$ is an integer. The cosecant function has vertical asymptotes whenever the function is undefined. The equations of the vertical asymptotes are

$x = n π$ , where $n$ is an integer. The period of the cosecant function is $2 π$ .

The range of the cosecant function is $(- \infty, - 1] \cup [1, \infty) .$

The cosecant function is an odd function because $csc (- x) = \frac{1}{sin (- x)} = \frac{1}{- s i n x} = - c s c x .$

Example

Find the period of $y = 3 csc (\frac{π}{2} x)$ and graph it.

Solution

The period is $4 (= \frac{2 π}{\frac{π}{2}} = \frac{2 π}{1} ∙ \frac{2}{π})$ . Since $y = 3 csc (\frac{π}{2} x) = \frac{3}{sin (\frac{π}{2} x)}$ , we draw the graph of $y = sin (\frac{π}{2} x)$ first. Then we draw the vertical asymptotes of $y = 3 csc (\frac{π}{2} x)$ whenever $y = sin (\frac{π}{2} x)$ is zero. Obviously, the lowest part of the U shape part of the graph of $y = 3 csc (\frac{π}{2} x)$ is $3$ and the highest part of the n shape part of the graph is $- 3$ . We complete the graph by alternating the U shape and n shape graphs.