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Wednesday, March 22, 2006

Scribe 28.....

Today's class wasn't so bad. Thought it would have a lot of explaination on math, but it didn't!

Anyways, we started off the class with a pre-test. Quite a nice pre-test.


1) Simplify: sin(2x+ pi)

You have to use the "dance" or not! But you'll have a much more easier time remembering

Sin(2x+ pi) = (sin2x)(cos pi) + (cos2x)(sin pi)

Cos pi is -1. Sin pi is 0 which means (cos2x)(sin pi)=o. You're left with (sin2x)(-1) which equals to

(C) -sin2x



2)
The terminal arm of angle theta in standard position passes through point (m,n), where m>0, n>0. Determine the value of sin(pi + theta)

Since m and n are greater than 0, that must mean that the cooridinates are in Quadrant one. To figure out the third side, you use the pythagorean theorem.

You end up with: Square root of m^2 + n^2

sin(pi + theta)= (sin pi)(cos theta) + (cos pi)(sin theta)
= (0)(costheta) + (-1)(sintheta)
= -sintheta

Since n is sin, the result would be

-n/square root m^2 + n^2


3) In the triangle ABC, angle A and B are both acute. If cosA =1/2 and sin B=square root of three over two, then the value of cos(A-B) must be:

The easier way is to just use the pythagorean theorem to find the value. Either that or do it the long way. Take the exact values and plug it into the cos(a-b) formula. You guys know how to do this so I wont' show it.

(1/2)(1/2) + (root three/2)(root three/2)

1/4 + 3/4 = 1


Oh nevermind. It's better that you see the work. =)



4) Solve 2cos^2x + cosx - 1=0 algebraically x is greater than pi/2 and less than 3pi/2.

(2cosx-1)(cosx+1)=0

2cosx=1 cosx=-1
cosx=1/2

I didn't get to right down the right answers, so please tell me if these are wrong or not. These are my answers.

Answer: pi, 2pi/3, 4pi/3




5) Prove: sin2x/1 +cos2x = sec^2x-1/tanx

2sinxcosx/1 + cos^2x - sin^2x tan^2x/tanx
2sinxcosx/2cos^2x
(after all cancellations)

sinx/cosx

tanx tanx



That's what we did in the morning.




In the afternoon, Mr. K talked about this Delicious site. Well, it's not actually delicious. Well, if you think it is, what ever makes you happy works for me.

IMPORTANT: EVERYONE IN CLASS MUST SIGN UP FOR THIS BY TOMORROW

Create an account at this site- http://del.icio.us/
and install the button on your explorer. More preferably, you guys should use firefox because it's great and Mr. K said so. Listen to him. He's wise..
Also, Spring Break homework is to find a min. of one link for
-circular functions
-transformations
-identities

And post this onto your delicious account.


Then we had to solve these identities. We didn't have that much time to do them so I hope you guys understand this
.

sec^2x - secx=2

sec^2x- secx-2=0
(secx-2)(secx+1)=0
secx=2 secx=-1
cosx=1/2 cosx=-1

x= pi/3 +2kpi;kEI
x=5pi/3 + 2kpi;KEI



3sin^2theta-10sintheta-8=0

(3sintheta + 2)(sintheta -4)=0
3sintheta=-2 sintheta=4
sintheta=-2/3 undefined



cos^2x=sin^2x+sinx [0, 2pi]

1-sin^2 sin^x+sinx
0=2sin^2x+sinx-1


and from here, you have to use the quad. formula. Hopefully, Mr. K goes over this tomorrow too.

Anyways, that's it that's all.

Homework for you all: Page 278 from 27-49. All odd numbers.

THE SCRIBE FOR TOMORROW IS Jeffers---
MANNY!



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