Wednesday, March 19, 2014

I/D3: Unit Q - Pythagorean Identities


1. To understand where sin^2x+cos^2x=1 comes from, we must understand that what an identity is. An identity is a proven fact and formula that are always true. The Pythagorean Theorem is an identity because it meets these requirements. The Pythagorean Theorem is a^2+b^2=c^2. However, it can also be re-written as x^2+y^2=r^2. This is because when we look at a triangle in the unit circle, the horizontal leg is x, the vertical leg is y and the length of the hypotenuse is referred to as r. To further simplify the theorem, we want it to equal to one. To accomplish this, we will have to divide both sides of the equation by r^2, because r^2/r^2 is equal to 1. After completing this step, we are left with (x/r)^2+(y/r)^2=1. Isn't this beginning to look very similar? Because we know the unit circle, we know that cosine on it is x/r. We also know that the ratio for sine is y/r. Knowing this we can substitute Cos^2x and Sin^2x for (x/r)^2 and (y/r)^2 in the equation. We are now left with Cos^2x + Sin^2x=1. Now, we can understand that  Cos^2x + Sin^2x=1 is referred to as a Pythagorean Identity because it is both derived from the Pythagorean theorem and is also a formula that always proves true.

Let's choose 30* for on of the magic 3 ordered pairs from the unit circle to prove this identity true. We know  the ordered pair is (rad3/2, 1/2). We can plug rad3/2 in for cos^2x because cosine is x/r. We can also plug in 1/2 for sin^2x because sine is y/r. When we square both of these, we end up with 3/4+1/4 and this is equal to one, which simplifies to 1=1, proving this identity true.

2. From sin^2x+cos^2x=1, we can divide the entire equation by cos^2x. This leaves us with (sin^2x/cos^2x)+(cos^2x/cos^2x)=(1/cos^2x). Using our ratio identities, we know that (sin^2x/cos^2x) is equal to tan^2x, (cos^2x/cos^2x) is equal to 1, and (1/cos^2x) is equal to sec^2x. We can now rewrite our new equation as tan^2+1=sec^2x.

Additionally, we can divide the entire equation by sin^2x. This leaves us with (sin^2x/sin^2x)+(cos^2x/sin^2x)=(1/sin^2x). Using our knowledge or ratio identities, we know that (sin^2x/sin^2x) is equal to 1, (cos^2x/sin^2x) is equal to cot^2x, and (1/sin^2x) is equal to csc^2x.We can rewrite this new equation as 1+cot^2x=csc^2x.


1. “The connections that I see between Units N, O, P, and Q so far are…” that all triangles are interconnected with the unit circle's right triangles and that the ratios derived can be used to solve for non-right triangles as well. 2. “If I had to describe trigonometry in THREE words, they would be…” Relationships involving triangles!

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