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

Pythagorean Identities

INQUIRY ACTIVITY SUMMARY:

Fig 1 (http://www.regentsprep.org/Regents/math/algtrig/ATT9/pythagoreanid.htm)

1. Where does sin2x+cos2x=1 come from?

a) An "identity" by definition according to Mrs. Kirch, is a "proven fact or formula that is always true". The Pythagorean theorem is an identity because it will work with any triangle that is legitimately a right triangle. Because right triangles reoccur in the unit circle (in 30, 60, and 45 degree angle smaller triangles), the Pythagorean theorem can be found all throughout.

b) The Pythagorean theorem is side x squared plus side y squared equals the hypotenuse (c) squared. These terms only work when we have a right angle, and luckily, we do in all 3 types of our unit circle triangles (30, 60, and 90).

Fig 2 (http://image.tutorvista.com/cms/images/38/pythagorus-theorem.JPG)

c) We can make the Pythagorean theorem equal to 1 simply by dividing both dies by c^2. This can be then simplified by writing the squared outside parenthesis, signifying everything inside is squared. What's left inside the parenthesis should look very familiar :)

What's left after simplifying is (x/r)^2+(y/r)^2=1

d) The ratio for cosine on the unit circle is (x/r),with x being "adjacent" and r being "radius". In the unit circle, r is always 1 so cosine can also be represented as simply "x" (x/1).

e) The ratio for sine on the unit circle is (y/r), with y being "opposite" and r being "radius". In the unit circle, the radius is always equal to 1 so sine can also be represented as simply "y" (y/1). 

f) The inside of the parenthesis in part c can be represented by the ratios we found.  (x/r)^2+(y/r)^2=1 becomes (cosine)^2+(sine)^2=1. I can conclude that the Pythagorean theorem is always true because it is a THEOREM, therefore the relationship I just came up with after simplifying is also true.

Fig 3(http://www.regentsprep.org/Regents/math/algtrig/ATT9/pythagoreanid.htm)

g) sin2x+cos2x=1 is referred to as a Pythagorean identity because when a circle is formed out of the unit circle with a radius of 1, the legs of that triangle can have ratios that are over 1 (cosine and sine) and thus represented simply as x and y. The picture above depicts this sort of triangle in the unit circle Because it is a right triangle, x^2 (leg 1 and also cosinex) + y^2 (leg 2 and also sinex) = 1 (radius in unit circle). 

h)Our "magic pair" of a 30 degree (rad3/2, 1/2) triangle can prove the validity of the Pythagorean identity.

2. You can derive the two remaining Pythagorean Identities from sin2x+cos2x=1 easily with both secant and tangent (a) and cosecant and cotangent (b).

a)

 Fig 4 (http://www.regentsprep.org/Regents/math/algtrig/ATT9/pythagoreanid.htm)

 Using our fundamental Pythagorean identity of sin2x+cos2x=1 we can derive a second equation that contains tangent and secant. Simply divide the formula by cos^2x. According to our ratio identities, sinx/cosx=tanx. We also know that the inverse of cosine (1/cosx) is secantx. By plugging in these new values we get tan^2x+1=sec^2x.

b)

Fig 5 (http://www.regentsprep.org/Regents/math/algtrig/ATT9/pythagoreanid.htm)

Starting off we our fundamental pythegorean identity of sin2x+cos2x=1, we can divide by sinx^2x. We know cosx/sinx=cotangentx because it is a ratio identity. We also know the inverse of sine (1/sinx) is cosecant x. Therefore, if we plug in those new values we know 1+cot^2x=csc^2x.

INQUIRY ACTIVITY REFLECTION:

1. The connections that I see between Units N, O, P, and Q so far are... the same trigonometric ratios, the same validity tests that can be proven with our magic pair triangle coordinates, and the repeated use of the unit circle (with radius of 1 of course).

2. If I had to describe trigonometry in THREE words, they would be... ratios, identities, and derivations.

REFERENCES:

Fig 2: http://image.tutorvista.com/cms/images/38/pythagorus-theorem.JPG

Fig 1,3,4,5: http://www.regentsprep.org/Regents/math/algtrig/ATT9/pythagoreanid.htm

I/D# 2: Unit O Concept 7-8: How can we derive the patterns for our special right triangles?

Deriving the Special Square Triangles

Inquiry Activity Summary:

1. We can derive the 30-60-90 triangle from an equilateral triangle with a side length of one by using the pythagorean theorem. To create a 30-60-90 triangle, we need to form the 30 angle measure. To do this, the base can be divided equally in half. That would make the two portions 1/2 and 1/2 in side length. We already know the hypotenuse is 1 because it is an equilateral triangle. Therefore, we can find the height by using the pythagorean theorem. After doing so, the hypotenuse is found to be rad3/2. Because the base also has a denominator of 2, it would be smart to multiply all the side lengths by 2 to get fewer fractions. That would leave us with rad3 for the height (opposite to 60 degree angle measure), 1 for the base (opposite to the 30 degree angle measure) and 1/2 for the hypotenuse (opposite to the right angle). 

Multiplying by 2 does not alter their relationship because it keeps the proportions the same (it is done to all three sides). Our pattern found must include n because it would account for the change that would occur if our equilateral triangle did not have sides of 1. We keep the pattern the same because those are constants that represent the relationship between the sides.

2. We can derive the 45-45-90 triangle from a square with a side length of one by using the pythagorean theorem. To form a 45-45-90 triangle, we must cut the diagonal right corners of the square in half (in order to get the two 45 degree angle measures). That would leave us with an unknown value for the hypotenuse. We have the adjacent and opposite, we know they are 1 (it is a square). If we use the Pythagorean theorem, we find the hypotenuse is rad2. 

We must add the variable of n because the pattern constants represent the relationship between the sides when the sides are 1. To account for the change made when the sides aren't 1, we must add the n to the pattern values. The constants would remain consistent because it is being done equally.

Inquiry Activity Reflection:

1. Something I never noticed before about special right triangles is that their relationship between sides are always constant. I did not know that the "n" in their patterns stood for change in those constants and that it could equal any number as long as it stays consistent with the rest oft he triangle.

2. Being able to derive these patterns myself aids in my learning because it makes me identify and UNDERSTAND the relationship between the sides. Before, I only memorized those constants. Now I know where they come from and I know what the relationships between them mean. If I ever forget, I could derive the patterns themselves.

I/D# 1: Unit N Concept 7: How do special right triangles and the unit circle compare?

Deriving the Unit Circle

Inquiry Activity Summary:

1. 

Before Reduction (http://www.biology.arizona.edu/biomath/tutorials/trigonometric/graphics/trig_30_60_90.gif)

A 30° Triangle has three sides to it: adjacent (x), opposite (y), and hypotenuse (r). The opposite side is always x. The adjacent will be xradical3 and the hypotenuse will be 2x. To derive the unit circle from this triangle, the hypotenuse must equal 1. To get that to happen, all three sides have to be divided by 2x and simplified. After reduction the values become 1 for hypotenuse, radical3/2 for adjacent, and 1/2 for opposite. The resulting sides can be plotted and the coordinate near the 60° measure will be the coordinate used in the unit circle. The pair of coordinates can be used for the 30° angle measure and all its supplemental angles (150°, 210°, 330°). For the 30° degree triangle we know the coordinate is (radical3, 1/2). An example of all the steps being taken can be found in the bottom:

2. 

Triangle before reduction (http://www.biology.arizona.edu/biomath/tutorials/trigonometric/graphics/trig_45_45_90.gif)

A 45° Triangle is composed of two sides that are the same length (x and y) and a hypotenuse. The equal sides are labeled as x and the hypotenuse is xradical 2. To derive the unit circle values, we must get the hypotenuse to equal 1. By dividing every value by xradical2 we can accomplish this and simplify to get our distance values that can be plotted out. The final values after reducing are 1 for hypotenuse, and radical2/2 for both adjacent and opposite. The resulting coordinate near the 45° angle measure will be used for 45° in the circle and all its supplemental angles (135°, 225°, 315°). That value is (xradical2, xradical2). An example of this derivation can be seen in the following image:

3. 

Triangle before reduction (http://img.sparknotes.com/figures/B/b21b8be4e663f46072a3c3b19b657891/306090.gif)

A 60° Triangle has the same sides as a 30° triangle. The opposite side is always xradical3. The adjacent will be x and the hypotenuse will be 2x. Since the location of the degrees change, the same simplifications are taken but they end up in different values. The final measures after reducing are 1 for hypotenuse, 1/2 for adjacent, and radical3/2 for opposite. The coordinate pair near the 30 measure will be used in the unit circle, the pair is (1/2, radical3/2). This value is the same for 60° and all its supplementary angles (120°, 240°, 300°). An example can be found in this image:

4.

This activity helps me derive the Unit Circle because the triangles reflect in the circle throughout all four quadrants. If reflected upon the y-axis, the coordinates measure the same from the bottom up. If reflected from there down the x-axis, the coordinates measure the same from the top down. If shifted from the first quadrant down across the x-axis, the coordinates measure the same from the top down. The only things that change are the signs and those sign changes can are explained in section 5.

5.The triangle in this activity lies in quadrant I since both x y values in the (x,y) coordinates are positive. If you change quadrants, the coordinate values would change. If you drew the triangles in quadrant II, the x value would become negative and the y would remain positive. If you drew the triangles in quadrant III both the x and y values would become negative. If you drew the triangles in quadrant IV, the x value would remain positive and the y values would change to negative. To view an example of different quadrants and the change in coordinates that occur along with them view this image derived by myself from repeating triangles:

Inquiry Activity Reflection:

1. The coolest thing I learned from this activity was how easy it is to follow the triangles throughout the different quadrants.

2. This activity will help me in this unit because when I feel lost if I forget my coordinates or angle measures, I can always use the first five to figure out the rest of the unit circle or try to derive the triangles again if my mind if blank.

3. Something I never realized before about special right triangles and the unit circle is that the values in the unit circle correlate with the values found when the special right triangles' hypotenuse is reduced to one.

References:

30 degree triangle image: http://www.biology.arizona.edu/biomath/tutorials/trigonometric/graphics/trig_30_60_90.gif

60 degree triangle image:http://img.sparknotes.com/figures/B/b21b8be4e663f46072a3c3b19b657891/306090.gif

45 degree triangle image:http://www.biology.arizona.edu/biomath/tutorials/trigonometric/graphics/trig_45_45_90.gif