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

INQUIRY ACTIVITY SUMMARY:
     Pythagorean identities come from both the Unit Circle and the Pythagorean Theorem. First to clarify, an identity is "an equation that is true no matter what values are chosen". The Pythagorean Theorem is seen as an identity because no matter what two values you have and when you solve it it will give you the third value. If you check your work by using the found value and a value given, the answer will be that same given value you did not use. To get the Pythagorean identity, sin^2(theta) + cos^2(theta) = 1, I will work it out using the Unit Circle and the Pythagorean Theorem and I will show and explain it in the following pictures:

First I'll show it by using the Unit Circle using the first quadrant since sine, cosine, tangent, co-secant, secant, and cotangent are all positives. If we plot a triangle in the quadrant we can already see the connection between the two. Th height will be "y" since it follows the y-axis, "x" will be base since it's on the x-axis, and the hypotenuse will be "r" because the hypotenuse is the radius of the circle. When it gets plugged in to the Pythagorean Theorem "a" will be "y", "b" will be "x" and "r" will be "c".

When I use "x,y and r" in the Pythagorean Theorem it will make sense if it's written out like y^2 + x^2 = 6^2 because when we try to make it equal one then it will make sense. If I take the sine of angle A it will be y/r, the cosine of it will be x/r, and "r" will stay as "r". Now I can show how (y/r)^2 + (x/r)^2 = r^2 can equal one but there is a faster and easier way to get one. If I write out "opp" as opposite, "adj" for adjacent, and "hyp" for hypotenuse and then in Pythagorean form, it's opp^2 + adj^2 = hyp^2. If I divide both side by hyp^2 then the hyp^2 will be 1 and it will be opp^2/ hyp^2 + adj^2/ hyp^2 = 1. If I pug it in to the other equation I said, opp^2/ hyp^2 is the cosine of angle A and adj^2/ hyp^2 is the sine of angle A. So cos^2(theta) + sin^2(theta) = 1 is one of three Pythagorean identities.

To find the second one, which is the identity with secant and tangent, I will use cos^2(theta) + sin^2(theta) = 1 and divide both sides by cos^2. The cosines will divide to be 1, sin^2/cos^2 is the Ratio identity of tangent(theta), and 1/cos^2 is the Reciprocal identity of sec(theta). So we end up with 1 + tan^2(theta) = sec^2(theta).

    

To find the third Pythagorean identity that has cotangent and co-secant, it will be just like the second one except that I will divide it by sine^2. If I divide both side by sin^2 to both sides then cos^2/ sin^2 will be the Ratio identity of cotangent, sin^2/ sin^2 will divide to 1, and 1/ sin^2 is the Reciprocal identity of csc(theta).

INQUIRY ACTIVITY REFLECTION: 
1) The connections that I see between Unit N, O, P, and Q so far are the angles that can be found in the quadrants of the Unit Circle and how the triangles that are made in the quadrants can be used with the Law of Sine And the Law of Cosine to find any missing angle or side length.
2) If I had to describe trigonometry in THREE words, they will be hard, understandable, and progressive.