Work Period

Today, as mentioned above, was a work period due to the fact Mr. Max was not here. We were given the opportunity to catch up on any homework given over the past couple weeks, finish up late assignments such as out charts or unit circles, or else continue working on our accelerated math. I am advising you, my class, to keep on top of your due exercises and especially accelerated math! There are a few ways to get free marks in this class, and you CANT afford not to get them. These free marks can account for a substantial amount of your final grade, and in some cases can even mean the difference between a pass and fail. So try you hardest on these sections of this course. Secondly, I would like to strongly advise that if you don’t understand the material as it is being taught, take a little time to stay after school and go to a tutoring session. In most cases you will learn more there than you would ever learn in one class. You have nothing to loose and everything to gain, and besides you cannot get less intelligent! I tell you this cause I am speaking from experience, and I only wish I would have known these things the first time around. I believe that if you do these things and try your hardest, this course will go smoothly for you and you will be satisfied with the results. There will be a test this Thursday, so be aware that we will be having our first homework check in the next few days!!

So my question is: An angle is 100 degrees. Equivalent to the correct radian. To go about doing this question you have to first convert degrees into radians.To do this is you can make a little converting chart. You put 100 degrees at the top left side then diagnol to it you put 180 degrees, and at the right top corner you put pi. You put 180 in the bottom because one pi is only 180 degrees. Once you have this all written out the degrees will cancel out and then you times the 100 to pi and put it over 180. Then you have to reduce the fraction. So, the answer would be 5pi over 9 radians.

Graphing Trigonometric Functions

Today in class we learned how to graph trigonometric (circular) fuctions. This involves properties of the Unit Circle and using the six trig. ratios. This can be done by using programs such as Graphmatica or by putting in trig. values for points aong the Unit Circle.
For example: y= sin(x)+1
How I would approach this question without aid of a graphing calculator or any other programs is by finding exact values for Sin (-1,.5,1) and then add one to the values, since it is Sin(x)+1 {the +1 simply means that the function moves 1 spot upon the y-axis} If you have a chart it makes it that much easier, but if you don't than you have to check values using triangle values on the Unit Circle. Then it is just a matter of plotting your values on your graph and drawing your line.
Now you have to determine the amplitude, period, domain, range and zeros of the function.
Amplitude=1 (the distance up or down it moves from the middle of te function)
Period=2pi
Domain= X set Reals
Range= [0,2]
zeros=.....-5pi/2, -pi/2, 3pi/2, 7pi/2
This stuff makes sense to me......
Exercise 6: Questions 1-16 are assigned

My question is in from Jan/08 #1 Part 1
Solve for theta over the interval (0,2pi), to decimal places
2csc2θ +cscθ −10=0
*the first step to solving this equation is to factor it so that there is 2csctheat+5 in one bracket and csctheta-2 in the other bracket
*to solve 2csctheta+5, you get the csctheta by itself which will give you -5/2 on the other side. Then you flip the fraction (-2/5) to solve for sine, and the reference angle to 3 decimal points is 0.411.Theta then equals 3.553, 5.871 because sine is positive in quadrants 3 and 4.
*in the next bracket get csc theta by itself, making the other side equal to 2 or 2/1. Flip the fraction to solve the reference angle for sine (1/2). You will notice that sin=1/2 at pi/6. sin is positive in quadrants 1 and 2 in this question making your answer pi/6, 5pi/6.

Why not take your world for granted?

Here's a reason....nothing to do with math, and everything to do with the unparalleled ingenuity of the human spirit. Sit down for a few minutes and check this out...

http://www.freetech4teachers.com/2009/09/moving-windmills-inspiring-story-of.html

Solving Trig equations over Real Numbers

In today's class, in addition to solving trig equations over a set domain, we also learned the process of solving trig equations over real numbers.

In any case the answer over all reals will be what ever radian you get, plus 2 pie multiplied by K, where K is any integer

For home work we were assigned Exercise 5 questions 1 through 12.

Our test is on Thursday Oct 1st, one week from today.

At first this stuff had me completely lost but the more Mr.max explained it the easier it got.

Math Exam Q,

I'll be doing Question 2 (Solve the equation 3tan2θ−5tanθ−2=0 on the interval [0, 2π] .
State your answers correct to 3 decimal places.) from the June 2007 exam.

1st and 2nd Degree Trig Equations

Here are the notes we did in class for solving 1st and 2nd degree trig equations. We did these on Wednesday. At the end of class I had an idea what was going on, but when I went home to do the work my mind totally went blank. So I went on the link for the Blackboard Learning System on the blog and it helped alot. If you had no idea what went on in class on wednesday or you just blanked like me this is a good site to go on because it explains step by step how to solve the 1st and second degree trig equations. Also it explains about exact values. We are supposed to get going on our Accelerated math and get the first four objectives done before the test next week.

I will be doing my question from Jan/08 Booklet 1, long answer, question 1

Exam Question

I will be doing Question 2, on exam January 2007.

Website

I came across this website while searching on Google and it has notes/examples on all the things we have been learning about the Unit Circle

http://www.analyzemath.com/Trigonometry.html#angles_trigonometry

Courtney 's Question

#42 on june 2006 long answer

Will's Question

I will be doing question 16 part 2 June 08.

This question asks us to find the value of sine at (-4&pi/3). What worked best for me was to remember that its just looking for the sine value at that point on the circle, and if its negative instead of going counter clockwise you go clock wise (from 0&pi). This brought me to 120°. Once I found my point I looked up my sine ratio which is Y/1 so all I needed to do was find the Y value of this 30° 60° 90° triangle and since it was above the X axis I knew it was positive. the values of this triangle are hyp=1 X=1/2 and Y=&radic 3/2. This gave me the correct answer of &radic 3/2.

My question

I'll be doing january 2008, booklet 2, question 11.

My Question

I call dibs on June, 2007, booklet 2, question 26.

June 2009 Exam, Part 2, Qu. #13

13. The value of csc(5 pi/3):
a. -2
b. -2/√3
c. 2
d. 2/√3

Solution:
1. Look on unit circle and find 5pi/3
2. It's in quadrant 4 at 300 degrees

3. csc= 1/y

4. At that place on the unit circle y= -√3/2
5. Since csc= 1/y divide 1 by -√3/2 (multiply numerators then denminators)
6. You should get 2/-√3 Which is the answer to the question according to the site (Answer: b)
7. If you choose to go a step further and rationalize the denominator, multiply both the numerator and demonenator by -√3
8. This will give you -2√3/3

Jun. '09 Prt 2 #13

I'll be doing the June 2009 Part 2 #13 question.

Blog/ Work/ Review Period

Today we were given the period to find a trig. question from previous exams to solve and post to share with everyone, work on our exercises, work on our chart, and our unit circle and try some accelerated math. I'm personally pretty happy we were able to just review and catch up this block, I'm still a bit confused with the unit circle and need to start memorizing it for the test coming up next week Thursday, October 1. It's slowly making more sense to me, but I'm thinking I'll need to look at this stuff at home more than I have been. I also suggest looking back to some of the first lectures we've gotten because when I did, I actually understood what was going on and it's great review.

"Happiness does not come from doing easy work but from the afterglow of satisfaction that comes after the achievement of a difficult task that demanded our best."
-Theodore I. Rubin

Matt's Question

Question 12 on the January 2006 Pre-Calc Exam (Booklet 2)
If P(θ) is a point on the unit circle, then the coordinates of P (5/6π) are:

a)√3/2, 1/2

b) -√3/2, 1/2

c) √3/2, -1/2

d) -√3/2, -1/2

It is just a matter of finding where on the Unit Circle is 5/6π and using the corresponding triangle to determine the coordinates.

If you are unfamiliar with the Unit Circle and degrees is more your style than you can convert radians to degrees. (5/6π) (180)= 150 degrees

So now we know where our triangle is on the Unit Circle:

Since we are at 150 degrees, a 30-60-90 triangle is generated. Now you just plug in your sides of the triangle by setting your hypotenuse at 1. Now you know the coordinates are:

x=-√3/2 y=1/2 or -√3/2, 1/2

The answer than is b)

I call January 2007 # 20 in the multiple choice

A is the answer!

so 1 over x is what we want

Thus we get

But It would seem as if they don't want the rationalized answer.

The answer would be A because the x value is negative

My Question

Im explaining question 12 from the June 2009 Student Booklet Part 2.



Question
12. On the unit circle shown, the coordinates of P are (a, b).
The value of sec(theta) is:
a) a
b) b
c) 1/a
d) 1/b

Solution
The ratio for secant(theta) is hypotenuse/adjacent which would be 1/x on a unit circle.
The x value for the point is a so by substituting a in for x you get the solution: c) 1/a

Trig Equations

Today, we learned more about solving trig equations. We learned that you can graph the trig equations either on a unit circle, or on a grid. A good applet to show this is linked here. http://ronblond.com/M12/Trig.Solve.APPLET/index.html . Another program you can use for graphing is called Graphmatica. You may notice that for the graph of y=tan(x), the lines are not connected. The spaces in between the lines are called the asymptotes. Where the asymptotes are, the tangent ratio is 1/0 which has no solution, leading to no graph on that value. I am very confused by most of this stuff still. The assignment for Monday is to finish the angle measurement chart.

Exact Values for Trig Functions

Today we were to make a chart of the exact values of all 6 trig functions for up to 360 degrees on the unit circle. I didn't really get how to do it at all, but then i found out that for the chart you are only ever going to be using 2 circles. a 45:45:90 triangle or a 30:60:90 triangle, depending which angle you're on. If its 45, 135, 225, or 315 degrees the triangle you will use is a 45:45:90 triangle, all the other angles are the 30:60:90 triangle.

Radians & Degrees