Let’s graph the transformation of the secant function. Y equals secant, theta minus pi over 2, and this is a pretty easy one. Let’s start with key points of secant. Now the key points are negative pi over 2, and pi over 2, these are places where secant is undefined because cosine is 0 there.
Then right in between 0 secant of 0 was 1, pi over 3, secant is 2 and negative pi over 3. Remember that secants and even functions, it will also be 2. So what happens when we transform to secant of theta minus pi over 2? Let’s make the substitution theta minus pi over 2 equals u.
U equals theta minus pi over 2 means, u plus pi over 2 equals theta. So to get the theta values I add pi over 2 to all these values. So let me add pi over 2 to negative pi over 2, I get 0. I add pi over 2 to negative pi over 3, I get pi over 6.
I pi over 2 to 0,pi over 2 added to pi over 3 and I get 5 pi over 6. And if I add pi over 2 to pi over 2 I get pi. Now what values do I put here? Well these are just secant u, same us this so I just copy these values down undefined and undefined, 1, 2 and 2.
I’m ready to graph let me start by plotting these vertical asymptotes. X equals 0 and x equals pi. Remember this will give me half period of the secant graph. x equals 0, and x equals pi.
So we start with the point pi over 2, 1 that’s right here. Pi over 6, 2 and 5, pi over 6 2 pi over 6 is a third the way from 0 to pi over 2 so pi over 62 is right here. 5 pi over 6 is 2/3 the way from pi over 2 to pi. So right here so pi over 6. 2 is here. So I graph and I get that familiar u shape of the secant function. Recall to get the second half period you take this half period flip it across the x axis hand shift it to the right half a period. In this case the period is 2 pi, so half a period would be pi. So I shift to the right pi. This point flipped across and shift it to the right, become -1. This point flipped across shifted to the right. I’m a third a way to the pi to the 3 and pi over 2 and I’m going to get -2.
Likewise, this point flipped across and shifted pi to the right gives me a point here. So I can plot that. And also you take this and flip it across the x axis and shift to the right pi, you get another asymptote at x equals 2 pi.
So you should remember that the secant function or the cosecant function, is going to have 3 asymptotes for every period. Left and the right and in the middle one. So this thing is also going to have an asymptote at negative pi.
Well we only have room for another half period so let me graph that. Remember, once you have a four period all you have to do to get more is to shift to the right or left of full period write 2 pi. So to get points over here, I have to shift these points 2 pie to the left.
So in this point for example at 3 pi over 2, -1 goes to negative pi over 2 -1. This point goes 2/3 the way between negative pi over 2 right here, and this point goes over here. So that I have it 1 and half periods of y equals secant theta minus pi over 2.
And you might recognize this function. This is the same as y equals cosecant theta. It’s really important that you know that the secant function and the secant function and the cosecant function have exactly the same shape. And to get cosecant, all you have to do is take the secant graph and shift it to the right pi over 2.
And that’s what this is secant shifted to the right pi over 2.