### Video Transcript

Find the minimum and maximum values of the function π of π equals 13 over six sin nine π by drawing the function.

There are two main elements to this question. Firstly, we need to be able to recall what the graph of a trigonometric function β in particular the sine function β looks like. Secondly, we need to know how to apply a number of transformations to this function. Letβs start by considering the graph of the function π of π equals sin π.

sin π is a periodic function. This means it repeats after a certain amount of time. In this case, it repeats every 360 degrees. We might now wish to draw the unit circle to help us identify the key coordinates for sin π. So at this stage, it is really useful to know the shape and the π¦-intercept by heart. Sin of zero is zero. This graph passes through the origin.

We can now add the classic wave graph that we associate with both the sine and cosine functions, insuring the first full period finishes at the point 360, zero. The maximum and minimum values of the sine graph fall at π of π equals one and minus one. We can work backwards from the end of the period to see that the graph must have a root at the halfway point, where π equals 180 degrees.

By further considering the symmetry of the sine curve, we can see that the maximum occurs at the point where π equals 90 degrees and the minimum occurs where π equals 270 degrees. Now that we have the graph of π of π equals sin π, you might wish to pause the video and consider what transformations might be required to answer this question. If you are unsure, you can have a search on our website for transformations of functions before continuing with this video.

For any function π of π₯, we can complete the following transformations: π times π of π₯ is a stretch in the π¦-direction by a scale factor of π; π of π times π₯ is a stretch in the extraction by a scale factor of one over π. Comparing these to the function given, we can see that the sine function has been stretched in the π¦-direction by a scale factor of 13 over six and the extraction by a scale factor of one over nine.

Letβs draw this on the graph. We can see that the maximum point now occurs when π₯ is a ninth of 90 degrees β thatβs 10 degrees. It has been stretched vertically by a scale factor of 13 over six taken up to π of π₯ equals 13 over six. Similarly, the minimum point now occurs when π₯ is a ninth of 270 degrees β thatβs 30 degrees. The vertical stretch takes it down to π of π₯ equals minus 13 over six.

Our minimum value is therefore minus 13 over six and our maximum is 13 over six.