Links for Friday, May. 7

http://www.ies.co.jp/math/java/vector/vsum/vsum.html

http://jwelker.lps.org/lessons/geogebra/vector_projection.html

Today we started with a homework quiz.
Then we proved why

We used Pythagorean theorem for the right triangle:

and we got that:
Then we used Pythagorean theorem for the left triangle:

also we used to get
Then we substituted everything from the left triangle in the equation we got for the right triangle. So the final equation is:
It works with SSS and SAS.

The next part of the unit were vectors.
In math the vector is a segment with direction, also called directed line segment:
The first point of the segment is called initial point (A), and the last point is called terminal point (B). a, b are components of the vector -
The length is sometimes referred to as magnitude,
Using Pythagorean theorem we could write it as:
The angle of the vector, θ is called direction angle.

examples:
1)Find the component form for a vector with initial point (-2,4) and terminal point (3,1).
The answer is <5,>

2)Find the component form for and θ=45º
, so
, so
The answer is

The rest of the class we went over hw.

Today in class we did a lot of things. We started with a warm up that asked us to find the triangle with these given sides and angle:
A=26º
a=5
b=8
This gave us an ASS (angle, side, side) case, which we know presents an ambiguous case. An ambiguous case provides you with information that you could use to create two different triangles. During this warm up we were asked to solve each angle and side of this triangle along with its area. Students could have solved for two different triangles with these given sides and angle. After the warm up we explored and proved the ambiguous case by constructing a visual of these possible triangles on Geogebra.

As you can see in the visuals there are two different triangles with values:
A=26º
a=5
b=8
This proves the ambiguous case.
We calculated the other possible sides and angles using the theorem that:




By Monday we are to prove that (when given SSS or SAS):



(and also find the largest angle when given SSS)

And for Bonus points we have the option to prove Heron's Area Formula (when given SSS) of:
, where s is the half perimeter.

Homework for 4/29

Some of us didn't understand problems 41 and 43 from the previous hw. Could you please post them here? Thanks.
Petra

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

Weds. March 10, Last Day of Unit 5

Today, we began class by finding the equations for two mystery graphs. The first one had the equation which caused the graph to look like this:

The second graph's equation was which caused the second graph to look like this:

After discovering the equations for the two mystery graphs; we turned our attention to number 21 and 22 of Francois and his Pedometer. For number one we noted that =. After evaluating number twenty two we discovered that his set for all distances was 2.42 + 2k where k is an integer but also at 3.86 + 2k where k is also an integer.

Following the warm-up, our class reviewed Quiz 4 of Unit 5. The two numbers that we paid special attention to were number 1 and 4. For number 1, we noted that if you take and you added you would get , but if you added another you would end up with . This means, that all you have to do next is find the sin() which is . For number 4, one thing to keep in mind is that in functions, there is an opposite order of operations. This means that the easiest way to solve the problem is by finding your a, b, and c value then plugging them into a equation which looks like this: , the next step is to distribute your which makes a=2, b= and c=-.

Once we finished going over the Unit 5 Quiz 4 and any questions students had trouble on, we took the review quiz. The quiz will only count towards your quarter grade if you want it to. The quiz questions were the exact same as the ones on the previous quizzes. Next class, there will be the unit test covering chapter four. Mr. O'Brien suggested to students that they should challenge themselves while studying for the test by working on more complex and conceptual questions than the basic review.

The next class Scribe will be Tyler.

Scribe post

I'm sorry this is late but I've been having troubles with this posting thing.
To start the class we did a warm up where we had to evaluate

cos^-1(√3/3)
a) you would be looking for an angle and since you are looking for an angle where the inverse cos is a -√(3)/2 then it would be 150˚it is also possible to get an answer of 5π/6 algebraically you could do x=cos
cosx=-√3/2
sin(cos^-1 √5/5)
b) in this one the answer will be a ratio ø=cos^-1 sqrt(5)/5
cosø=sqrt(5)/5 sinø=(√20)/5

we also had to evaluate the graphs of
y=arscin(x-1)
a)y=arcsin(x-1) this is just a sin graph only with the restricted domain to make it a function and the negative 1 makes it move 1 unit to the right and the minimum and maximum is π/2
y=tan(3x-6/π)
b) you can set this up as an inequality because the period of tan is π so -π/2<3x-π/6<π/2>2π/9 y="sec^-1" secy="x" cosy="1/x" y="arccos^-1(1/x"

Thursday, March 4 Class

We went over quiz, noting several things...

First, when Mr. O’brien says to graph both and angle and its reference angle in standard position, the initial sides of both angles must be on the x-axis. Second, much as sin(90º-θ)= cosθ, sec(90º-θ)= cscθ

After going over the quiz, we went on to discuss the inverse trig functions.

y=sin^-1x

y=cos^-1x

y=tan^-1x

Notation problems: sin-1x does not equal (sinx)-1

Because of this potential confusion, there are other notations to represent the inverse functions:

arcsin (x)
arccos (x)
arctan (x)

We saw that on grapher, arcsin(x) isn’t a function. On our graphing calculators we saw the inverse trig function is a portion of the entire function. We consider the inverse trig functions to be a collection of all the points possible while still being a function.
The same applies to Cosine, which doesn’t pass the horizontal line test any more than the Sine function does. A single Tangent wave passes the horizontal line test, and that is all that’s graphed of the arctan function.
Of course, all three graphs look different from the originals, since they are inverse functions, and therefore have reverse (x,y) coordinates.

Here are links to the three inverse functions. This should help make their domains and ranges pretty clear.

http://www.math.rutgers.edu/~greenfie/mill_courses/math151a/gifstuff/arcsin.gif

http://www.intmath.com/Analytic-trigonometry/arccosx.gif

http://upload.wikimedia.org/wikipedia/commons/f/f6/Arctan_plot_real.png


arcsin (x)
Domain: [-1, 1]
Range: [-π/2, π/2]

arccos (x)
Domain: [-1,1]
Range: [0, π]

arctan (x)
Domain: [all real numbers]
Range: [-π/2,π/2 ]

Use of inverse functions: We can plug in coordinate points along the unit circle, and the inverse function gives the angle. It gives angles in quadrants where that particular function (sin, cos, tan) is positive.

Then we took the quiz, which brought us to the end of class.

According to the tags on the side, Collin has only done one scribe post, so now it's his turn for next class.

Friday February 26th Class

Today in class we started off by taking the quiz.
After the quiz we went over some questions on the homework. We did problems 27 and 55.
27.


To graph these two functions we first looked at a standard sin graph.
http://fooplot.com/index.php
Then we took the f(x) function and altered the graph by moving the Y values of 1 to -2, and the Y values of -1 to 2. (Red)
We added on to that axis the g(x) function onto the same graph. This transformed the original sin graph by changing all the Y values of 1 to 4 and Y values of -1 to -4. (Blue)
file:///Users/student/Desktop/sin.tiff
The period is still 2π, and the symmetry is odd, as it is for all sin graphs. This means sin(-x)=-sin(x)
The transformation of these two functions changed the amplitude of the sin function.

55.
We then changed this to to make it simpler to graph.
This graph looked like: file:///Users/student/Desktop/cos.tiff
The period was 4π and the amplitude was changed, along with the graph being condensed. This graph originally (before being transformed) had was even, being symmetrical across the Y axis as all cos graphs originally are.

After reviewing the homework we moved on and looked at the graphs of the trig functions. We noticed that the sin graph and cos graph were very similar, but just shifted over to the right a bit. Both had a domain of -1 to 1 and a range of all real numbers.
We spent most of the rest of class playing with these graphs. We also looked at the graphs of the other trig functions as well and looking at the periods and amplitudes of the graphs.
We also looked at the tan graph which looks like:file:///Users/student/Desktop/tan%20graph.tiff

The csc graph which looks like:file:///Users/student/Desktop/csc%20graph.tiff

And the sec graph which looks like:file:///Users/student/Desktop/sec%20graph.tiff

But we did not get into why they are the way they are. That will be learned later and is in Dan's scribe post of class on 3/2.

scribe post, petra, trigonometry

We started the class with a warm-up quiz on quizlet. Then we worked on a few problems: example:
We went over hw problems we didn't know.
83)a which is so the answers are:
19) 0"> from the function value we get that
We decided that it's in the IV quadrant. We calculated hypotenuse using Pythagorean theorem. The answer is:




27) 2x-y=0, quadrant III we made a slope -> y=2x and found the sides of the triangle. Then using the sides of the triangle found all six trigonometric functions.
Then we went on the site for a Ferris Wheel. http://maine.edc.org/file.php/1/AssessmentResources/FerrisWheelUnitCircle32_L.html
We tried to make a graph where height was on y-axis and time on x-axis.
Then we used a function
We found it's domain: all real numbers
and range: [-1, 1]
Then we made a graph where f (θ) was on y axis and θ was on x axis.
We then used GeoGebra and made a graph for g(θ) = cos (θ), where we used different values for θ in radians.
At the end we used a function:
to define amplitude of sine and cosine curves -> amplitude = |a|
and to define a period of sine and cosine curves ->