Section 5.6 #78

Look at the note on page 452 for a clue as to how to graph this function. You can use piecewise notation to restrict the domain. But first we analyze secant x on the restricted domain of [0, p /2), (p /2, p ] and note that this function is undefined on at p /2 as the cosine is zero there. Also, the first portion of the domain has a range of [1, ¥ ), while the second portion has a range of (-¥ , -1]. The range of the secant function will be the domain of the arcsecant function.

So the domain of the arcsecant function is (-¥ , -1], [1, ¥ ). And the range of the arcsecant function is [0, p /2), (p /2, p ], with a horizontal asymptote at p /2. Plotting this on your calculator as cos^-1 (1/x) produces:

Do you notice a relationship between arccos(x) and arcsec(x)? What are the domains of each? Recall that with this restriction the domain of the secant function, [0, p /2), (p /2, p ], we can use our calculator to find arcsec(x):

arcsec(x) = arccos(1/x).

Essential Trogonometry, Math Links

by Jack Tompkins