I/D #3: Unit Q: Concept 1: Pythagorean Identities

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There are the reciprocal identities that I will constantly be referring to.

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These are the ratio identities that I will constantly be referring to.


1. Identities are proven facts or formulas that are always true. The Pythagorean Identity sin²x+cos²x=1 comes from using the Pythagorean theorem.  The Pythagorean theorem, instead of using a²+b²=c², we will replace the letters with x²+y²=r², which is the same exact thing when placed on a coordinate plane. Then we want the right side to equal 1 so we divide both sides by r² and we get(x/r)²+(y/r)²=1. When looking at what we now have, we can make to changes which we learned previously. First, we can change (x/r)² to cos² (cosine) because the equation for cosine on the unit circle is x/r. Also, we can change (y/r)² to sin² (sine) since that is the equation for sine on the unit circle. Now, to make sure that this is really identity, we must apply the unit circle to this to get the Let's plug in an angle of 45* to this identity. First, we find cos of 45 and sin of 45 which turn out to be √2/2. So we plug it into the Pythagorean Identity that we found. So it is going to be (√2/2)²+(√2/2)²=1. When we distribute the power and add both of them together, our answer becomes 4/4 which equal 1. Remember to put x or theta because that means that it can be any angle. Now we have one of our Pythagorean Identity which is cos²x+sin²x=q.

©Kelsea

©Kelsea

2. There are two remaining Pythagorean Identities, 1+tan²x=sec²x and 1+cot²x=csc²x. To obtain the first equation I mentioned, first divide both sides by cos²x. By doing so we get cos²x/cos²x+sin²x/cos²x=1/cos²x. Now when we look at our ratio identities and our reciprocal identities, we can see that some of the fractions we have created are equal to some sort of single identity: sin²x/cos²x=tan²x and 1/cos²x=sec²x. So we can replace the fractions with these single identities to get the equation 1=tan²x=sec²x.  We now have one last Pythagorean Identity to find. To do so, we divide by sin²x. Now our identity will be cos²x/sin²x+sin²x/sin²x=1/sin²x. Again, the fractions correspond with some of the ratio and reciprocal identities. cos²x/sin²x=cot²x and 1/sin²x=csc²x. Now replacing the fraction identity with a single identity, we get the last Pythagorean Identity which is cot²x+1=csc²x.

©Kelsea

References:

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I/D2: Unit O - How we derive the patterns for our special right triangle?

Inquiry Activity Summary

[Property of Kelsea]

1. A 30-60-90 triangle is derived from an equilateral which is cut in half. An equilateral is a triangle where all sides are equal as well as all angles. All the angles in the equilateral triangle are 60* and all the sides are 1. When we cut the triangle in half, we get different angles and sides. half of the triangle is now made up of angles which are 90*, 60* and 30*. The 60* is the one that was not affected by the cut. The 30* is the one at the top where the angle was cut in half, which makes sense since half of 60* is 30*. The 90* comes from the bottom where the side is perfectly perpendicular to the bottom side. The sides are now different as well. The hypotenuse, the slanted on, is not affected by the cut so is remains 1. The bottom one, which is cut in half becomes 1/2. However, there is no given value for the side that has been cut. In order to find that side, we use the Pythagorean Theorem [a^2 + b^2 = c^2]. The hypotenuse is c, the bottom side is a, and w solve for b. Once we plug it all in and solve for b, the answer should be b=√3/2. Now the sides should all be as follows: a=1/2, b=√3/2, and c=1. Now, since nobody is very fond of fractions, we can multiply all the sides by 2 to eliminate the fractions. Now the sides should be a=1, b=√3, and c=2. Now we are almost done! Now we must add an n to each side number because although the angles will remain, the sides are subject to change. That means n can be any number and the angles will remain the same. So finally, one last time, lets list what the angles are: a= n, b=n√3 [n should not be behind it], and c=2n. Our 30-60-90 triangle is now derived and conquered!

[Property of Kelsea]

[Property of Kelsea]

[Property of Kelsea]

[Property of Kelsea]

2. A 45-45-90 triangle is derived when a square is cut diagonally. A square is a shape where all four sides are equal [in this case we will set them equal to 1] and all sides are 90*. When cut diagonally, the two angles change, and one side changes. Since it is cut diagonally, the two angles affected will become 45* which makes sense because half of 90 is 45. The other angle remains 90* since it was not affected by the cut and it is perfectly perpendicular to the bottom side. Two of the sides were not affected, the ones that are not slanted, so they remain 1. However, one was, the hypotenuse. In order to find the hypotenuse, we must once again use the Pythagorean Theorem [a^2 + b^2 = c^2] to find the hypotenuse. Hypotenuse will be c, the bottom side will be a, and the other side will be b. Now we plug our numbers into the equation and solve for c. The answer should b √2. Now all our angles should be as follows: a=1, b=1, c=√2. Yay! Wait ... we are forgetting something that we did in the 30-60-90 triangle. What is it? Oh Right! Our lovely buddy n. Add n to all the sides because, as stated before, the angles can remain the same but the sides are subject to change! So now, our sides should be a=n, b=n [It is 1n but since any number multiplied by 1 is that number, we do not need the 1], and c=n√2 [remember, no n behind the radical!]. Congratulations! The angles are now derived and we now know how to derive the 45-45-90 triangle! Good Job!

[Property of Kelsea]

[Property of Kelsea]

[Property of Kelsea]

Inquiry Activity Reflection

1. "Something I never noticed before about special right triangles is ..."  that the sides are found using the Pythagorean Theorem equation. [Hypotenuse for 45-45-90 and the b for the 30-60-90]
2. "Being able to derive these patterns myself aids in my learning because..." if I ever forget what the side is, I can derive it using the what I learned from this activity, which is to cut a square diagonally for a 45-45-90 or an equilateral triangle for a 30-60-90 and then use the Pythagorean Theorem to find the angle that is missing.

I/D #1: Unit N Concept 7: How Do SRT And UC Relate?

Inquiry Activity Summary

[Kelsea Del Campo; worksheet]

1. The 30* triangle is one of the special right triangles that is taught to us in Geometry. This means that each side has a special equation that goes with it. The hypotenuse is 2x, the horizontal angle is x, and the vertical angle is x radical 3. Now, we have to identify which is the x and y. The horizontal is x and the vertical is y. The hypotenuse is r, which refers to being a reference angle, the angle from the terminal side to the closest x axis. Now if we set the reference angle equal to one, we have to do that to all of the other sides by dividing by 2x. This means that we have to divide by 2x on the horizontal and vertical side. This means our hypotenuse will equal 1, our vertical angle equal to 1/2, and our horizontal angle equal to radical 3 over 2. Next I drew a coordinate plane where the triangle was in the first quadrant. Then I labeled the vertices into ordered pairs. The edge is (0,0) since it is at the middle of the graph. The one to its right is ( radical3/2, 0) because the x is radical3/2.  The one above it is (radical3/2, 1/2). The last ordered pair is a point on the unit circle and its angle is 30*. This can help us identify the ordered pair on the unit circle with a reference angle of 30*, depending on which quadrant one or both will be negative. 

[Kelsea Del Campo: worksheet]

2.  The 45* angle is also a special right triangle. The hypotenuse is x radical2, the vertical side is x, and the vertical side is also x. Now we also have to identify x and y. X will be the vertical side and y will be the horizontal side. The hypotenuse is, again, r. We have to set the hypotenuse equal to 1 again so we divide all three sides by x radical2. After doing so, our r should equal 1, our y should equal radical2/2, and our x should equal radical2/2. Next draw a coordinate plane with the triangle in the first quadrant. Label the verices as ordered pairs. The edge is (0,0), the one beside it is (radical2/2, 0), and the one above it is (radical2/2, radical2/2). The last ordered pair is a point on the unit circle with an angle of 45* and can help us find the ordered pairs for an angle with a reference angle of 45*, depending on which quadrant one or both will be negative. 

[Kelsea Del Campo: worksheet]

3. The 60* angle is very similar to the 30* except that everything is switched. This means that the vertical side is x radical3 but remains y. The horizontal is now x but remains the x. The hypotenuse remains the same. Set the hypotenuse equal to 1 by dividing by 2x and divide the rest by 2x as well. Now the hypotenuse should equal 1, the y should equal radical3/2, and the x should equal 1/2.Next, draw a coordinate plane where the triangle is on the first quadrant. Then identify the vertices as ordered pairs. The edge should be (0,0), the one beside it should be (1/2,0) and the one above it should be (1/2, radical3/2). As before, the last ordered pair is a point on the unit circle and is the same ordered pair for that with a reference angle of 60*, depending on which quadrant one or both will be negative.

4. This activity helps us derive from the unit circle because when we look at the unit circle and look at all reference angles of 30*, it has the same ordered pair, depending on the quadrant a negative will be present. Any angle with a reference angle with 45* has the same ordered pair, depending on the angle there will be a negative on the x or y, or both. Lastly, for any angle with a reference angle of 60*, it will have the same ordered pair, and depending on the quadrant, the x or y or both will be negative.

[Kelsea Del Campo: phone]

5. The triangle drawn in this activity all lie in the first quadrant. If we draw the triangle in the second quadrant, it will be a mirrored image of the first quadrant, the third quadrant will be a mirrored image of the second quadrant, and the fourth quadrant will be a mirrored image of the third quadrant. The closest angle to the x-axis is going to be an angle with a reference angle of 30*, the middle will that with a reference angle of 45*, and the one closest to the y-axis will e the one with the reference angle of 60*. The x's in the second quadrant will be negative, x and y will be negative in the third quadrant, and the y in the fourth quadrant will be negative. The 45* angle is shown in the second quadrant and is like a mirror image of the one in the first quadrant. It has the same ordered pair but its x is negative. The 30* angle is in the third quadrant and is upside down and the x and y of the ordered pair is negative. Lastly, the 60* angle us also upside down but only the y is negative.

[Kelsea Del Campo: phone]

[Kelsea Del Campo: phone]

Inquiry Activity Reflection

1. "The coolest thing I learned from this activity..." was how the special right triangle corresponds are are very similar in each quadrant. I didn't learn in Algebra II where these numbers came from so that was also interesting.

2. "This activity will help me in this unit because..." we are currently memorizing the unit circle to help us the sin, cos, and tan so it will help in finding the answer. 

3. "Something I never realized before about special right triangles and the unit circle is ..." that we get the ordered pairs from the special unit circle when we set the hypotenuse equal to 1 and that the triangle can be found in the unit circle itself.