#
The Inverse Cosine Function - Problem 3
*
*3,764 views

I want to graph a transformation of the inverse cosine function. I have a problem that asks me to graph y equals negative inverse cosine x plus pi over 2. I’m going to go with my usual method. I’m going to make a table of key points of inverse cosine and transform the points first.

Now I have my graph of cosine here in blue and I’ve jotted down the key points, -1 pi, zero pi over 2 and 1 zero and all I need to do is figure out what kind of transformation negative inverse cosine of x plus pi over 2 represents. Let me put that in here, negative inverse cosine of x plus pi over 2.

This minus sign just means that I’m going to be reflecting the graph across the x axis and we accomplish that by multiplying the y values here by -1. This plus pi over 2 means the graph’s going to get shifted up by pi over 2 and I do that by adding pi over 2 to the result. There’s really no horizontal transformation so I don’t have to do anything with the x coordinates so I can just write them down, -1, 0 and 1, right from here.

Let’s get ready to transform these points. We multiply by -1 and add pi over 2. We get negative pi plus pi over 2, negative pi over 2, times -1 plus pi over 2 is zero, times -1 plus pi over 2 is pi over 2. And so we have 3 points that we can put on our graph for the negative inverse cosine of x plus pi over 2. We’ve got -1, negative pi over 2, 0, 0 and 1, pi over 2. We want to draw this with the inverse cosine shape. Looks something like this.

This is the graph of y equals negative inverse cosine x plus pi over 2 but you may notice that this is also the graph of inverse sine. Y equals inverse sine of x. It turns out that this is an identity. Inverse sine of x equals negative inverse cosine of x plus pi over 2. This identity is actually related to the co-function identity.

Again whenever we graph transformations key points of the original parent graph, transform the points and then plot the points in your graph and draw a smooth curve.

## Comments (0)

Please Sign in or Sign up to add your comment.

## ·

Delete