April 14 Scribe Post

Class begin with more work on the trigonometric identities packet

sin2x = 2sinxcosx

cos2x = (cos^2)x - (sin^2)x
this can also be written as:
1-(sin^2)x - (sin^2)x
thus,
cos2x = 1- 2(sin^2)x
written solely in terms of cosine,
cos2x = 2(cos^2)x - 1

To derive the identity for tan2x:
tan(x+x)
= (tanx + tanx)/(1-tanxtanx)
tan2x = (2tanx)/(1-(tan^2)x)

Next, we derived identities for sin(x/2) and cos(x/2)
cosx = 1- 2(sin^2)(x/2)
2(sin^2)(x/2) = 1 - cosx
sin(x/2) = +/- sqrt((1-cosa)/(2))
sin(x/2) = +/- sqrt((1-cosa)/(2))

cosx = 2(cos^2)(x/2) - 1
2(cos^2)(x/2) = 1 + cosx
cos(x/2) = +/- sqrt((1+cosx)/(2))
cos(x/2) = +/- sqrt((1+cosx)/(2))

Finally, we derived an identity for tan(x/2)
(+/- sqrt((1-cosx)/(2)))/(+/- sqrt((1+cosx)/(2)))
= +/- sqrt((1 - cosx)/(1 + cosx)) * (sqrt(1-cosx))/(sqrt(1-cosx))
= +/- sqrt(((1 - cosx)^2)/(1-(cos^2)x))
= +/- sqrt((1-cosx)^2/(sin^2)x)
= (1-cosx)/(sinx)
tan(x/2) = (1-cosx)/(sinx)
This concluded the work we did on the trigonometric identities packet [sorry if it's hard to visualize, but I couldn't get the online equation editor's html code to show up on the blog for some reason].

HW Questions We Reviewed:

13.
cos2x - cosx = 0
cos2x = cosx
(cos^2)x - (sin^2)x = cosx
cos^2(x) - (1 - (cos^2)x) - cosx = 0
2(cos^2)x - cosx - 1 = 0
(2cosx + 1)(cosx - 1) = 0
2cosx + 1 = 0
x = 2π/3, 4π/3
or
cosx - 1 = 0
x = 0

49.
sinu = 5/13, π/2 <> cos u = -12/13
sin(u/2) = sqrt((1-cosu)/2) = sqrt((1+12/13)/2) = (5sqrt(26))/26
cos(u/2) = sqrt((1+cosu)/2) = sqrt((1- 12/13)/2) = (sqrt(26)/26
tan(u/2) = (sinu)/(1+cosu) = (5/13)/(1-12/13) = 5

19.
6sinxcosx = 3(2sinxcosx) = 3sin2x

25.
tanu = 3/4, o <> sinu = 3/5 and cosu = 4/5
sin2u = 2sinucosu = 2(3/5)(4/5) = 24/25
cos2u = (cos^2)u - (sin^2)u = 16/25 - 9/25 = 7/25
tan2u = (2tanu)/(1-(tan^2)u) = (2(3/4))/(1-(9/16)) = (3/2)(16/7) = 24/7

11.
4sinxcosx = 1
2sin2x = 1
sin2x = 1/2
2x = π/6 + 2πk
x = π/12 + πk
x = π/12, 13π/12
or
2x = 5π/6 + 2πk
x = 5π/12 + πk
x = 5π/12, 17π/12

23.
sinu = -4/5, π <> cosu = -3/5
sin2u = 2sinucosu = 2(-4/5)(-3/5) = 24/25
cos2u = (cos^2)u - (sin^2)u = 9/25 - 16/25 = -7/25
tan2u = (2tanu)/(1 - (tan^2)u) = (2(4/3))/(1 - (16/9)) = (8/3)(-9/7) = -24/7

We finished off class with the Chapter 5 Quiz #2

Scribe for next class will be Nate

Feb. 3 Scribe Post

We started off the class with a review on angular and linear speed.
The example of a bicycle wheel was used.
A 29 inch wheel at 2 revolutions a second.
Linear Speed?
Take into account that distance = speed * time

182.2 (inches/second) = 10.35 mph

note: Stoichiometry can take care of all unit conversion before you really start the problem:


Angular Speed?


Additionally, it was proven that: radius * angular speed = linear speed, which will be useful for problems in the future.

We finished up class with the first Unit 4 quiz in Ms. Ferlauto’s room.

Scribe Post 10/27/09

At the start of class we had a quiz on the 2.1-2.2 homework. This took us up to 10:05, and from there we began going over some of the more recent homework problems, including pg. 159 (13, 23, 49, 25, 59, and 37). Problem 37 segued nicely into a discussion of the remainder theorem, which as we learned can be used to quickly find specific values of a function such as f(5). Next, we did a review of complex numbers, proceeding with a Venn-diagram of the various types of numbers and the relationships between them. This lead discussion towards terms such as rational, irrational, pure imaginary, etc.; we also reviewed operations involving complex numbers, including a brief discourse on complex conjugates. This took us right up to 10:50, our homework being p. 167/17, 19, 21, 29, 33, 37-51 odd, 65, 71. I haven't gotten a chance to talk to anyone about being the next scribe yet, so I'll do that at the beginning of next class.