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

Inquiry Activity Summary:

The purpose of this activity was for us to understand where does the main Pythagorean Identity (sin²x + cos²x  = 1) come from in relation to the Unit Circle and the Pythagorean Theorem. By doing so, we would be able to understand how to derive the other two remaining Pythagorean Identities from sin²x + cos²x  = 1.

1. Where does sin²x + cos²x  = 1 come from? (Refer to Unit Circle and Pythagorean Theorem)

Before we do anything with this, we must understand what exactly is an identity. An identity is proven fact or formula that is always true. That is why the Pythagorean Theorem is also an identity because it is a proven formula that is always true.

Then, we remember that for the Unit Circle, the Pythagorean Theorem came in terms of x, y, and r instead of a, b, and c, making it x² + y² = r². Now we notice, that while the Pythagorean Identity is equal to 1, the Pythagorean Theorem that we now have is equal to r². So we'll divide both sides of the Pythagorean Theorem by r² to make one side equal to one. So it is now x²/r² + y²/r² = 1, which we can arrange into (x/r)² + (y/r)² = 1 because they still both still mean the same thing.

After that, remember how the ratio for cosine in the Unit Circle was x/r and the ratio for sine in the Unit Circle was y/r...See it yet. If cos is x/r and sin is y/r, then we can plug them into (x/r)² + (y/r)² = 1 to make it cos²θ + sin²θ = 1, which we were trying to find. That is why cos²θ + sin²θ = 1 is referred to as the Pythagorean Identity, because it is basically the Pythagorean Theorem moved around algebraically and manipulated to look like that, but it is still essentially the same thing. We can prove that it is an identity by substituting an ordered pair from the Unit Circle. Say we have a 45° angle, which has an ordered pair of (√2/2, √2/2). Sin45 would equal √2/2 and cos45 would equal √2/2. When we plug them into the identity, (√2/2)² + (√2/2)² = 1. (√2/2)² would equal 2/4 or 1/2, so 1/2 +1/2 = 1, which is true.

2. How do we derive the other two remaining Pythagorean Identities from sin²x + cos²x  = 1.

To derive the identity with secant and tangent by dividing  sin²x + cos²x  = 1 by cos²x on both sides. That becomes (y/r)²/(x/r)² + (x/r)²/(x/r)² = 1/(x/r)². When simplified, the r's will cancel out to make tangent (y/x), and the cosines will cancel to make 1. From the reciprocal identities, we know that 1/(x/r)² is equal to sec². That is how it ends up as tan²x + 1 = sec²x.

To derive the identity with cosecant and cotangent, Now instead of dividing  sin²x + cos²x  = 1 by cosine, we'll divide both sides by sin²x.The sines will cancel out to make 1. Cos/sin is the same as (x/r)²/(y/r)², so when it simplifies you end up with (x/y)² which is equal to cot²x because of the reciprocal identities. 1/sin²x is equal to cscx² due once again to the reciprocal identities. You end up with 1 + cot²x = csc²x.

Inquiry Activity Reflection:
1. “The connections that I see between Units N, O, P, and Q so far are…” that they all relate to the Unit Circle and trig functions in some way. They also have something to with triangles like the Pythagorean Theorem/Identity to solve for what we are looking for.
2. “If I had to describe trigonometry in THREE words, they would be…” triangles, ratios, and measures.

I/D#2: Unit O Concept 7-8: Deriving Patterns for Special Right Triangles

Inquiry Activity Summary:

The purpose of this activity was to help us understand why special right triangles have the patterns they do. For example, a 45-45-90 triangle has side lengths of n, n, and n√2 while a 30-60-90 triangle has side lengths of n, n√3, and 2n. Instead of memorizing two seemingly random patterns, we are now able to understand where these patterns come from so if we forget the patterns, we can easily figure it out ourselves, since we now know how to derive them.

1) 30°-60°-90° Triangle

We can construct a 30°-60°-90° triangle bu cutting an equilateral triangle of any side length (in this case to keep it simple, is 1) down its altitude to make 2 right triangles. As we all know equilateral triangles have equal sides and equal angles(which are 60°) and because we cut one angle and one side in half, there is now two 30° angles and the side length of the smaller triangles is 1/2. So far, we know two sides of the smaller triangle: hypotenuse of 1 and one side(opposite 30° angle) length of 1/2. We use Pythagorean theorem to solve for the last side to end up with √3/2. Now that we know all the sides and angles of the
30°-60°-90° triangle, we can start to see the relationship between the sides and how we got the pattern. The hypotenuse is 2 times the length of the shortest leg of 1/2 while the longer leg looks to √3 times 1/2. Therefore, if we multiply n to each of the sides and multiply by 2(to make the numbers easier), we get sides of n, n√3, and 2n. This applies to any number n might equal to. The pattern is still the same because n is just a constant.

2) 45°-45°-90° Triangle

We construct 45°-45°-90° triangle by cutting a square (side lengths 1 to keep it simple) down its diagonal, cutting the 90° angles in half and making two 45°-45°-90° triangles. Because the triangle still has 2 sides of the original square, we know their values are both 1. Then, we use the Pythagorean theorem to solve of the length of the hypotenuse, which in this case is just √2. Looking at our triangle with all of its sides and angles solve for, we can see the relationship of the sides already. The two legs are equal with a length of 1 while the hypotenuse is 1 times √2. Now if we multiply n to all 3 sides, we end up with sides of n, n, and n√2, making the pattern. N is again is just a constant that applies to any side length because the relationship between the sides is still the same. That way it is easier to remember the pattern rather than memorizing sets of numbers that apply this pattern.

Inquiry Activity Reflection:
1. "Something I never noticed before about special right triangles is..." that there patterns can be logically derived from a process involving Pythagorean theorem and cutting squares/triangles.
2. "Being able to derive these patterns myself aids in my learning because..." I can now easily and quickly calculate the side lengths of the SRTs. That means I don't have to memorize sets of numbers that could apply to these triangles or the pattern because if I forget the pattern, I can easily derive it again and use it to find the sides, instead of remembering a random pattern.