Maybe I shouldn't have started this blog now, not with everything that's been going on.

Student Affairs Head Ronnie has a new problem for now: some of the students want their money refunded for now, and they promise to pay again in two weeks time. Not that the organizing committee has already spent the money; it's just that much additional work for the finance committee.

Yesterday in my TrigMat class we talked about the trigonometric functions of any angle. So far we have only been using acute angles, or those that are between zero and ninety degrees.

Now I clarified that all angles we would be discussing are greater than ninety degrees and in the standard position, so the first order of business is to find out what quadrant the angle's terminal side lies on.

From that information, the students will be able to determine whether the x and y values of the angle will be positive or negative. This led to teaching them about when the trigonometric functions are negative depending on its quadrant.

For comparison, I also gave them the equivalent or reference acute angle for each quadrant. In other words, I showed the angle less than ninety degrees that has the same numeric values of trigonometric functions as the one that is more than ninety degrees.

I even gave them the figure for determining graphically which are the reference angles: given four terminal sides, one in each quadrant, the angle measure they would take for each is the one whose rotation starts either upward or downward from the nearest part of the x axis: positive or negative.

After that I also talked about the functions for the quadrantal angles, which only had values of zero, one, negative one or undefined, since x and y are either equal to r, negative r or zero.

One typle of example I gave them were ones where in the exact (and fractional) value of one trigonometric function is given, and the quadrant the terminal side of the angle is located. From that information they are supposed to derive the three sides of the triangle and solve for the rest of the functions.

The second type of example I gave them was one wherein the value of a positive angle larger than ninety degrees is given. From that they are supposed to get the equivalent acute angle, which will turn out to be a special angle we have discussed previously. From there they should easily be able derive the trigonometric functions, and just give a negative sign to the ones appropriate to that quadrant.

Of course there were still some students for which this type of reverse analysis needed a little too much deduction on their part. When the tangent of one angle in the second quadrant was given as -7/5, placing the numerator in y and the denominator is x told them that the angle was in the fourth quadrant (where y is negative and x positive).

I had to remind them that -7/5 is the same as 7/(-5).