Today, the practice problem couldn’t have been easier. Using the numbers 1,3,4,6, and 7 we had to come up with the largest even number that was still divisible by 3. Since any number using those values can be divided by 3, that didn’t really matter (since all the values add up to a number divisible by 3). Then, it was just a matter of order. 7 first but then there are only two even numbers left. We put the smaller one at the end so that the number would be even. So after 7 came six then in decreasing order, 6 then 3 then 1 and finally 4. Plain and simple.

First we went over finding an equation for a secant or cosecant graph. You first determine how far away the midline is from 0. Since theses types of lines work differently from sine and cosine, it is a matter of drawing in the companion function or just counting the distance between the highest point of the lower portion and the lowest part of the higher portion. The middle is the written down for the equation.

In this diagram, that vertical displacement is 0 because it goes 1 above and 1 below the x axis. Next, you need to find the vertical stretch. This is the distance from the vertical displacement which turns out to be 1. These numbers will be placed outside of the trig function in the equation.

Next is the issue for the horizontal changes. These will be placed inside the trig function in the equation. First we will look for the period. I will chose this to be a secant function to i will start at the 0 point because I see it is an exact point. I will follow the cosine revolution which starts high, hits the midline, goes low, hits the midline again, and goes high again. It starts at 0 and ends high at 2pi. 1 full revolution in radians is 2pi so dividing them would just get 1. There is no vertical displacement so it stays just x.

The final equation is y=sec(x) and not theta which refers to degrees.

If it were a cosecant graph, it could start at pi/2 (180°) so the equation would be y=csc(x-pi/2)

For more practice, check out this website and click on interactive tutorial. you can play around with the sliders and than write an equation for the function. Although it says secant, you could also make it a cosecant. This is a great way for at home practice. Although it isn’t in radians, why not practice transferring the degrees to radians?

http://www.analyzemath.com/Secant/Secant.html

The equation is simple:     x°/360=x radians/2?

A simple way is to remember is using the unit circle:

Next is what we did as the main learning portion of class. We used a cosine or sin equation to solve for radian values based on a y value. First we found equations using graphs but since we just reviewed that, I will skip it.

The first one we got was 5 = 9+ 7 cos 2pi/13 (x-4)

Using basic algebra, you subtract the 9 and divide by 7 and are left with -4/7 = cos 2pi/13 (x-4)

The way to get around the cosine is to do the inverse which switches the x and y values and since -4/7 is the y value, that is put inside of the function. So now we have cos-(-4/7) = 2pi/13 (x-4)

We have to round this answer so far because that cosine value isn’t a regular one like cos(1/2)

So then we divide by 2pi and multiply by 13  and add 4. This leaves us with x = 8.508…

This is not the only answer though. Since this function repeats, that same spot of 5 from the original equation will come up many times and so we must add or subtract this number by the period to get the other numbers. In the equation, it says that 2pi/13 of the revolutions will fit into one actual revolution which is just taking 2pi divided by the period. Since we divided the period by 2pi, we must divide the period by in, leaving us with 13. So 8.508…. + or – 13 are all solutions to this problem.

But there is another answer to this since 5 is the y value for more than 1 radian measure for each cycle. To solve for that, we must look at other places where the cosine measure is -4 and the radius measure is 7

Imagine the red line is 7 and the x distance or cosine distance to the green like is -4. We get the same sine value in the pink triangle below it. This is the other place on the graph where the line passes the 5 line. We can incorporate this into our equation with a negative of the angle we already have.

Simple, instead of using cos-(-4/7) which approximates to about 2.179 radians, it becomes -2.179 because we are using the negative form of the angle. So then inserting that back into the equation, it becomes:

-cos-(-4/7) = 2pi/13 (x-4) so then we solve once again for x and x = -.508 and then the same as the other answer, the additional answers are found in regular cycles of the period so all the answers are -.508  + or – 13.

But sine is not the same. The process is similar algebraically but when we get to solving for the other point on the graph that is not found from the original equation, it changes. On the prior image that shows the triangle that makes the negative angle, we will now be looking at the blue radius and red radius triangles. If the original sine is the angle of the blue triangle, we are looking for the angle to the red triangle. If x is the radian measure for the blue angle the equation is:

pi – x

So using the equation 0 = -1 + 2 sin 3 (x) and solving for the original angle, it is pi/6 so then:

pi – pi/6 = 5pi/6

So we replace that measure into the equation.

The first answer is pi/18 + or – 2pi/3

The second is 5pi/18 + or – 2pi/3

Here is a quick sample problem that takes is short problem and just talks through the answer using radians and pi.

This basically sums up what we did in class today.