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Thursday, March 11, 2010

Scribe Post: March 11th 2010

DO NOT READ UNLESS YOU HAVE TAKEN THE TEST!!!

...SERIOUSLY. =)


Today in class we started going over section 2.1.

This section of the book includes:



The definition of a Unit Circle, that r (the distance to the origin) = l

Picture of a Unit Circle:
http://wpcontent.answers.com/wikipedia/commons/thumb/c/c9/Unit_circle_angles.svg/720px-Unit_circle_angles.svg.png

Using our knowledge of the Unit Circle, we learned more about the Trigonometric Functions.
In the first part of this section, the book showed you how to solve a function without a calculator, just like what we had to do on the first ten questions on the test yesterday.

This box (as found in the book) was extremely helpful in figuring out how to do this:
'If alpha is an angle in standard position whose terminal side intersects the unit circle at point (x,y), then
sin alpha = y cos alpha = x tan alpha = y/x csc alpha = 1/y sec alpha = 1/x cot alpha = x/y provided that no denominator is zero.'

So all you have to do is take the points given to you on the unit circle in coordination with the degree or radian given, and use the x &/or y according to the function.

Example: sin 45 degrees

The answer to this is the square root of two divided by two because that is the y point that the unit circle gives you.

if the problem was csc 45 degrees, than the answer would be 1/ square root of two divided by two.



We also started going over graphing trigonometric functions. We learned that when graphing the functions in an xy-coordinate system (normal graph) you use x as the independent variable and y as the dependent variable. So, for example, if sin (x)= y, and you used 30 degrees as your x, than sin (30 degrees)= 1/2. x, 30 degrees, is the independent variable (or input) and y, 1/2, is the dependent variable (or output). Another example JoJo used in class was if you used day light or moon light and time Proxy-Connection: keep-alive
Cache-Control: max-age=0

day. In this the time of day would be your imput, and your output would be the amount of sunlight or moonlight there is left. This only works with things that move within a rotation, like the unit circle.

Hope this helps you guys understand!

Tomorrow's scribe is Rebekah. =)

1 comment:

  1. Hey, as you probably know the unit circle is a little skewed... AS in I can't see all of it.

    ReplyDelete

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