Brightstorm is like having a personal tutor for every subject
See what all the buzz is aboutCheck it out
The Inverse Sine Function - Problem 3 3,879 views
Let’s graph the transformation of the inverse sine function. The problem says graph y equals negative inverse sine of x plus pi over 2. Now let’s recall what the graph of inverse sine looks like.
This graph in blue is the graph of inverse sine and whenever I transform graphs I like to use key points and the key points I’m going to use are these three points, it's -1, negative pi over 2, 0, 0 and 1, pi over 2 and I’ve got those plotted out on this table. In order to graph my new function I need to think about what the transformations are going to do to these points.
When I look at this, I think reflection across the x axis and then shift down pi over 2. That means that only the y coordinates here are going to be affected. The reflection will change the sines like multiplying by -1 and then shifting up pi over 2 is like adding pi over 2. The x values will remain unchanged. -1, 0 and 1. And here I multiply by -1, I get pi over 2 plus pi over 2 gives me pi, times -1 plus pi over 2 is pi over 2. Times -1 plus pi over 2 is zero. So I get these three points which I can plot. -1, let me change colors here, negative one pi is up here. Zero, pi over 2 right here and 1 zero is right here. I want to try to mimic the shape of the inverse sine graph, so something like this.
This is the graph of y equals negative inverse sine of x plus pi over 2 and what’s really interesting about this graph is that it’s exactly the graph of y equals inverse cosine of x. that tells us that there’s an identity between these two and that is the inverse cosine of x equals negative inverse sine of x plus pi over 2. Which is kind of interesting. It means that if you want to evaluate and inverse cosine and your inverse cosine button’s broke on your calculator let’s say you calculate inverse sine times -1 plus pi over 2. That’s a transformation of the inverse sine graph.