1/28/09 - Proofs for Law of Sines, Law of Cosines, and more fun stuff...

Today was more of a theoretical class, we worked with 5 problems from the board, 3 proving the Law of Sines, the Law of Cosines, and the Pythargoren thereom, and 2 involving those theorems, some geometry and trig. We started, but didn't finish covering the last two problems, so you can try to figure them out on your own, and wait for the next post to check your work...

one fact we covered in class that was new to me: to inscribe a triangle in a circle, center the circle around the point of intersection of two of the triangle's perpendicular bisectors don't worry about the third, because all three intersect at the same point)

Law of Sines- 1/27/10

In class today, we took notes on the Law of Sines. This formula allows you to solve for the rest of a triangle if you already know an angle, angle, and a side (AAS), or if you know a side, side, and an angle (SSA). This formula allows you to solve for the unknowns of all triangles as long as you have enough given information. When you have an AAS triangle, there is one possible solution, but with SSA triangles, you can have 0, 1, or even 2 possible solutions, so it is important to check if triangle even exists.

The basic formula for the Law of Sines is:

 (a/sinA) = (b/sinB) = (c/sinC) 

Notes 1/26 - Law of cosines

1/26 - And we're back to school. . . Today we took notes on "the law of cosines." This formula allows you to solve for any triangle (find all values for angles and sides) if at least 3 of these values are known. The law of cosines works with any type of triangle - not just right triangles. One thing to remember in notation is that uppercase letters refer to angles and lowercase to sides. In solving for a triangle using the law of cosines it does not matter which value you look for first, as long as you can find the others taking that approach. Also, remember to check if the triangle is even possible in the first place. The following is the formula for law of cosines:

a^2 = b^2 + c^2 - 2bc*cosA

Today we studies the Double Angle ID's: sin(2A)=2sinAcosA and cos(2A)=cos^2A-sin^2A. Once you solve for both sin(2A) and cos(2A), you can solve for tan(2A)=sin(2A)/cos(2A). It's important to know that for that the 2 in 2 sinAcosA isn't the same as the 2 in its equivalent sin(2A). Its derived from the sum of the 2A represented in the equation 2(sinAcosA).

12/2/09 Sum/Difference Identities

In class on December 2nd we learned about the Sum and Difference identities. (Note: a large portion of these notes show how we derived the identities given angles and prior knowledge of the unit circle.)

The identities allow us to find the sine, cosine, and tangent of angles we previously could not (ie: 15 degrees), by adding or subtracting (ie: 60-45=15) using angles for which we can already find the trig functions.

Coefficients of Angles (Trigonometry)

We discussed in class that normally when using trig identities to the meaasure of an angle, you come out with two possible angles. However, a coefficient of an angle produces that double that number of solutions.
girl in the hall: "It took me three minutes to get ready today"
MV: "I can tell"

Notes: 11/30

Sorry it took me this long to post the notes. I had a few problems. Anyway, today in class we learned how to solve trig equations by using 3 main steps:
1.Get the 1st answer from the inverse trig function
2.Then the second answer from the reflection of the trig ID (Ex: Tan= 180 or 2pie+the angle)
3.Add or subtract 2 pie or 360 (depending on whether you are in degrees or radians) to both of your answers from 1 and 2 to get as many answers as needed.