I/D #3: Unit Q Concept 1: Using Fundamental Identities to Simplify or Verify Expressions

INQUIRY SUMMARY ACTIVITY

1. Where does sin^2x + cos^2x=1 come from?

To begin this is a Pythagorean identity. An identity is a proven fact or formula that is always true. The first thing we did was write out the Pythagorean theorem as show below. Then we replaced a^2+b^2=C^2 with x^2+y^2=r^2 ( as shown above) We are able to replace it with these variables because it consists of the same characteristic as the unit circle, were using all the variables from the first quadrant. At this moment our formula  is  x^2 + y^2 = r^2 and we want to get to sin2x+cos2x=1. To do that, we divide everything by r^2. If we divide everything by r^2, we get this: x^2/r^2 + y^2/r^2 = 1 (r^2/r^2).  So r^2/r^2 becomes 1 and we can put (x/r) ^2+ (y/r)^2=1. Then we should notice that these variables are the ratios of  cosine and sine. Now that we know this we can replace then and get a final answer of cos^2x+sin^2x=1.


Does it really work?

 

           In order to verify that it works we can plug in the 30,45 and 60 degree angles. ( All values from the unit circle) In this case I choose 60*. The 1/2^2 becomes 1/4 and the rad3/2 becomes 3/4 because the squared cancels the 3 and you just multiply 2 by 2. This gives us an answer of 1.

 

2. The other Pythagorean Identities.

 

 

 

 

 The picture above shows the steps of how to get tan^2x + 1 = sec^2x. First, you will have to divide both sides by cos^2x. For this one, we want tangent and secant. We can look at that as y/r times r/x. So as a result we have y/x which is tangent. Cosine divided by cosine is simply 1. And 1 divided by cos2x  which is sec2x because sec x = 1/cos x, we just powered up because everything is being multiplied in this reciprocal identity. (look at the picture above)

 

The other Pythagorean identity is 1+cot^2x=csc^2x. In this identity we have cot and csc.  We divide everything by sin2x . We divide cos^2x by sin^2x and we get cot^2x. We divide 1 by sin^2x and we end up with csc2x. This is how we get the answer.

 

INQUIRY ACTIVITY REFLECTION 

1.THE CONNECTIONS I SEE WITH UNIT N, O, P AND Q SO FAR ARE the trig functions and the ratios. Also the magic three from the unit circle to derive the things above.

2. IF I HAD TO DESCRIBE TRIGONOMETRY IN THREE WORDS, THEY WOULD BE sin, cos, and tan.

 





I/D #2: Unit O - How can we derive the patterns for our special rights triangle?

INQUIRY ACTIVITY SUMMARY:

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30-60-90 Triangle:        To begin we have to know that it is an equilateral triangle. This means that all sides and angles are equal. With that in mind we realize that the triangle has three angles, and all the angles in a triangle add up to 180 so what we do is divide 180 by 3 and get an answer of 60.
Then we cut straight down on the triangle and that gives us an angle of 30* on the top because since we divide it symmetrically the angle dives into two. If we were to add 30+30 we would get an answer of 60.  Then we created a 90* angle on the bottom left because the line created a perpendicular angle with the bottom of the triangle. ( as shown in the first picture). Since its an equilateral angle we know that all sides are equal and in this case it has a measurement of 1 on all three sides. The bottom side is 1/2 because its half of one. To find the other side (height) we use the Pythagorean Theorem ( a^2+b^2=c^2). We plug in 1/2 for a, were looking for b, and we plug in for 1 for c. This gives us an answer of rad3/2 for b. However to make the sides easier do deal with we give all sides  a variable, we will use n, and were going to give that n a value of 2n. We give it a value of two because we don't want to deal with fractions. (see picture 2) Once we have done that we get a hypotenuse of 2n a horizontal side of n, and a vertical side of n rad3. Remember n just represents any number and it keeps the relationship consistent.

    2. 45-45-90 triangle

             For the 45-45-90 triangle we begin with a square. A square has four equal sides and four 90* angles.  (see the first picture)First we drew a diagonal which  cut the 90* angle into two  pieces and that left us with  a 45 degree angle. Since all sides are equal the horizontal and the vertical sides are 1. Then we use the Pythagorean theorem to find c ( diagonal). We use 1 for a and b and that gives us an answer of rad2 (picture number 2)  Then we add n because n represents any number and it remains the relationship consistent.

INQUIRY ACTIVITY REFLECTION
       1. Something I never noticed before about right triangles is how we were able to find the ratios from the equilateral triangle and a square.
        2. Being able to derive these triangles myself aids in my learning because  I can understand how a special right triangle works and use this for concepts 7 and 8.

I/D# 1: Unit N Concept 7: The Unit Circle

• INQUIRY ACTIVITY SUMMARY

                 1.        Describe the 30* triangle:
                                  

              The first thing I did was label the triangle according to the Special Right Triangle. The hypotenuse was labeled as 2x, the horizontal side x rad3, and the vertical x.  Next we had to simplify the hypotenuse to make it one. In order to do that we had to divide all three sides by 2x. As seen in the picture above we can see that when we divided x by 2x the x's canceled and it became 1/2. For the horizontal side the x's also canceled and it simplified to rad 3/2.  Afterwards, I labeled the hypotenuse r, the horizontal value x, and the vertical value y. Then I drew a coordinate plane. We began by making the labeled angle side the origin (0,0). Then near the 90 angle I labeled it as ( rad3/2,0) because we moved rad 3/2 on the x axis and we didn't move any units up. The top vertices is labeled as ( rad3 /2,1/2) because we moved right rad3/2 and up 1/2.

                2. Describe the 45* Triangle:
                          

 

 

               The first thing I did was label the triangle using the rules of Special Right Triangle. For this triangle the hypotenuse is x rad 2 and the horizontal and vertical side are both x. Then we had to make the hypotenuse 1. In order to do this we had to divided xrad2 by xrad2(hypotenuse) to obtain an answer of one. Then I have to divide the two other sides by xrad2 and since both are x, when we divide by xrad2, we get rad2/2. However, sine there is a square root in the bottom we had to simplify. The next step was to label the hypotenuse r, the horizontal value x and the vertical value y. Afterwards we also had to draw a coordinate plane. We also labeled the origin at the given labeled angle (0,0). As showed in the second triangle, we labeled the 90* vertex (rad2/2,0) because we moved rad2/2 along the x- axis and 0 unit along the y axis. Finally for the top vertex the point would be (rad2/2,rad2/2 because we moved along the x and y axis rad2/2. 

               3.  Describe the 60* triangle:

 

                                          

 

                  This is very similar to the 30* triangle but for the 60* triangle the sides are switched. The hypotenuse still being 2x, the vertical side x radical 3, and the horizontal side as x. We first divide everything by 2x and we get a vertical side of radical 3/2, a horizontal side of 1/2, and hypotenuse of one (just like the 30* angle and after simplifying).Next we place the triangle on the first quadrant giving us ordered pairs of (1/2, radical 3\2), (1/2,0), and (0,0). 

 

            4. This activity helps me derive the unit circle because it gives me the degrees, points, and ordered pairs that are located at quadrant one of the unit circle. These three triangles are important because there reflected on the unit circle. The vertices on the top are the points used all around the circle but they have different x and y values. ( explained below)

 

 

 

 

 

             5.  The 30*,45*,and 60* triangles that we did in the activity lie on the first quadrant. Since the first quadrant is reflected on the second quadrant, the second is reflected on the third quadrant,  and the third on the fourth quadrant. The values of the vertices change. In the first quadrant both values are positive, in the second quadrant the x value is negative. In the third quadrant the values of both the x and y are negative. For the fourth quadrant the y value is negative.

 

INQUIRY ACTIVITY REFLECTION

The coolest thing I learned from this activity was how the triangles even when shifted to different quadrants are the same as the first quadrant.

 

This activity will help me in this unit because with just knowing the ordered pairs from these three triangles, I can easily narrow down the answers for degrees that have the share reference angles.

Something I never realized before about special right triangles and the unit circle is that I can see the unit circle as triangle to find the vertices.