Day 3: Trigonometric Equations & Quadrantal Angles

Hello Everyone! HAPPY HALLOWEEN!!!!! This is Alyana scribing for today. Today we didn't really learn anything new. Anyways, this is what we discuss in class.

The trigonometric functions cosine and sine may be defined on the unit circle as follows. If (x,y) is a point of the unit circle, and if the ray from the origin (0,0) to (x,y) makes an angle theta from the positive x-axis,
then cos(t) = x
sin(t) = y

The equation x2+ y2 = 1 gives the relation
cos2(t) + sin2(t) = 1

P(t)=(x,y) radius is equal to 1 = (cos(t), sin(t)) radius is greater than 1.




Sin(t)= O/H => Sin(t)= Y/1=Y
Cos(t)= A/H => Cos(t)= X/1=X
Tan(t)= O/A => Tan(t)= Y/X= Sin(t) divide by Cos(t)

The standard position starts at (1,0).

Positive distance is measured in counterclockwise direction &
Negative distance is measured in a clockwise direction.

Review of solving Trigonometric Equations:

1. Solve the trigonometric function(sin(t), cos(t) or tan(t))
2. Find the related angle by using the inverse trig function on the positive guy only.
3. Find the quadrants based on the sign of the trig guy (use the C.A.S.T. rule)
4. Find the solution in each quadrant determined in step 3.
5. Write your solution.
   

QUADRANTAL ANGLES <;3
An angle in standard position with terminal side lying on x-axis or y-axis is called as Quadrantal Angle. The terminal side is on an axis: 0 degree, 90 degrees, 270 degrees, 360 degrees ....

I'm not good at blogging
Sorry... oh well... hope this helped you guys a little :D

Hall of Famer for October

Please cast your votes (under comments) for the Hall of Famer of October.
Thank you and good luck.
Mr.P

The winners are:
1. Gian - 3 bonus points
2. Angelo - 2 bonus points
3. Abby, Christian - 1 bonus point each

Thanx and good luck in November.

Hey peeps i'm tired............. it's 3 am and i'm hungry
I don't know why i'm doing this so late should of done it earlier but you know psssssssht tumblr :p
so anyways it's Halloween and I hope the teacher gives us candy ;) if not then

I forgot to say my name it's Angelo and i'll be showing you how to do Trigonometric Eqautions
YAAAAAAAAAAAAAAAAAAAAAAAAAY

Trigonometric Equations

First thing you need to know is the Cast ruleIt is very important that you know this rule or you won't be able to answer the question

The Equation we are doing is the first example in the booklet and the equation is:


*Before we move onto the first step you need to know that this equation is gonna be Q1 and Q2. If you don't know why it is in these quadrants, it's because Q1 and Q2 are the only two that have a positive sin.*

The first step in solving this equation is to multiply this equation

* Do not know what happen but the picture should say Sin*-1 (0.78615) ok lettuce move on.

The results by multiplying should be
This is your reference angle it also represents Quadrant 1

The next step is to figure out for Quadrant 2

You must subtract 51.827 degrees from 180 degrees to get your answer
*For Quadrant 3 you must add to 180*
*For Quadrant 4 you must subtract from 360*

OK After solving that you have your answers for the Quadrants you needed to find out for. In this case I rounded both answers to the nearest tenth


YAAAAAAAAAAAAAAAAAAAAAY The problem is solved
I'm going back to sleep goodnight :D

Hi my name is Ryley and i well be your scribe for today.


Ambiguous Case Triangles

Today im going to show you how to solve an Ambiguous Triangle using sine law.


First thing your going to want to do is draw out the angle
you are given in this case its 30 degrees.


next your going to want to measure out the b length in this case its 10 cm

and lastly to make it a triangle take the a value and connect it with the c value so that it lines up.



to solve for the other angles you now need to use sine law

here i am solving for SinB



once you have 2 of the angles you can subtract them from 180 degrees to get your last angle



now with the last angle you can get the last side using sin law


and there you go the triangle is solved

A=30 degrees
B=90 degrees
C=60 degrees
a=5cm
b=10cm
c=8.66cm

hope this helped you learn more about Ambiguous Triangles

sine law

Hi everybody my name is Breena and i am from precal 30s. Yesterday we learned about "the sine law" so today I am going to blog about it for you.

Yesterday we learned about solving for the missing angle in a triangle...

Example #1:

First find your angles and you sides, be sure that you label them it well make things much easier.

Then do the Pythagoras theory..

Then do cosine....

Example#2:

The sine law is a relationship between sides
and angles in any kind of triangle. In this example let angle ABC be any
triangle where a,b & c represent the measures of the opposite sides.

Sorry i kinda had some computer problems but i hope this blog well help you guys out! :)

Points and Exact Values of Angles

Hi everyone! I'm Hannah and I'll scribe for yesterday's class. :)

Yesterday, we learned about finding the coordinates of a point of an angle. We've answered a few examples on our booklet and I've listed them below together with its solution. :)

When solving for the point of an angle, we should identify the cos and sin of the given angle. We can get the cos through the SOHCAHTOA. For cos, we should identify the adjacent over hypotenuse. For the sin, we should identify the opposite over hypotenuse.

In our first example, the angle of 30 degrees has a cos of square root of 3 over 2 and 1/2. We should also consider the CAST rule. Since 30 degrees is in quadrant 1, wherein all are positive, the final value of the point is (square root of 3/ 2 , 1/2).

We also learned about finding the exact value of a given angle. Below is an example from our booklet.

The given angle is tangent of 300 degrees. In getting the answer, we will use the SOHCAHTOA again. Since the given is tangent, we will be looking for its opposite over adjacent. The angle 300 degrees is a related angle, with 60 degrees as its reference angle. In the special triangle of 60 degrees, the opposite is square root of 3, and its adjacent is 1. Using the CAST RULE, the 300degrees is in the fourth quadrant, where cos is positive and the rest are negative. Since the given is tangent in quadrant IV, the answer should be negative. So our final answer, the exact value of tan300degrees is -square root of 3.

So yeah, I'm not really good at this so I just hope you learned something. Thanks :)

Special Right Triangles

Hey everyone, This is Alex here and I'll be scribing for yesterdays class. I apologize for the late scribe I seemed to be having some technical difficulties with the blog. But anyways here goes,
Yesterday we covered the topic of Special right Triangles and below is some info on them.

Special Angles: Most angles on the unit circle on based on reference angles of either 30°, 45°, or 60°.

Below is an Example of an Isosceles right triangle and how to solve for the missing hypotenuse which will aid us in find the value of SinΘ, CosΘ, and Tan Θ.



As you can see, by using the Pythagorean Theorem you can calculate the missing hypotenuse and use in calculating SinΘ for other given angles.

We can also use this method in solving for angles in a 30° and 60° Triangle.

Once we know the Sin, Cos and Tan values of 30°, 45°, 60°, and 90° we then already know the values of all the other Reference Angles. These Values are basically the same, we must however remember the C.A.S.T rule in order to determine whether or not the value of Sin, Cos, or Tan is negative or positive.

Below is a diagram of the C.A.S.T rule and an example of how it is used.

The C.A.S.T rule tells us which values are positive in the unit circle. For example if a 45° triangle
was located in Point A of the unit circle then all the values of Sin, Cos, and Tan would be Positive.

Alright then now that we this we can solve the missing values in the Tables of our Trig booklets, and seeing as we've already done then congratulations, you now know how to solve the Sin, Cos and tan values of the unit circle :)

Hope you enjoyed my scribe, bye everyone.

Reference & Related Angles

Hey everyone this is Gian, and I'm gonna be your scribe for today. Today we started the class of by going over some of the questions that we had for home work last class. Then we started a new lesson on Reference and Related Angles.

• Reference Angles are acute angles that is less than 90 degrees. They are formed in quadrant 1 and is in between the terminal arm of the angle and the x-axis. 

• Related Angles are angles formed between the positive x-axis and the terminal arm. So that the acute angle formed between the terminal arm and the x-axis is equal to the reference angle.  

We learned how to find reference angles, and the rules to find them. First you must find the related angle. The first thing that you must determine what quadrant the terminal arm finishes in. 

• If the angle is between 0 and 90 (degrees) the terminal finishes in quadrant 1.
• If the angle is between 90 (degrees) and 180 (degrees) the terminal finishes in quadrant 2. 
•  If the angle is between 180 (degrees) and 270 (degrees) the terminal finishes in quadrant 3.
• If the angle is between 270 (degrees) and 360 (degrees) the terminal finishes in quadrant 4.  

For example: 

 So what I did here was first determine my angle which is 98 (degrees) because the angle lands on quadrant II, then I subtracted 180 (degrees) which gives me the reference angle of 98 (degrees). 

* Never forget about the cast rule, and make sure to pay attention to the signs.

Related Angle Formulas:

•  If the angle terminates in Quadrant I:  = 
•  If the angle terminates in Quadrant II:  = 180 - r
•  If the angle terminates in Quadrant III:  = 180 + r
•  If the angle terminates in Quadrant IV:  = 360 - r

Day 5 (I think) - Trig Ratios and The C.A.S.T rule

Hey guys, this is Naavi & I’m here to explain what we did on Friday. I was supposed to do it on thursday, but I couldn’t get into the blog. x)

On Friday, we learned about the trigonometric ratios, and the C.A.S.T rule.

Trigonmetric Ratios:

The value of ‘r’ is known as the radius vector, it’s length is always positive and is the hypotenuse on a triangle.

The other 2 points/sides are ‘x’ and ‘y’. If ‘x’ and ‘y’ is on the terminal arm of an angle in std position, that means that:

Sin o (theta) = opposite/hypotenuse (y/r)

Cos o (theta) = Adjacent/Hypotenuse (x/r)

Tan o (theta) = opposite/adjacent (y/x)

Make sense? I hope so. xD

The C.A.S.T Rule

The C.A.S.T rule is pretty simple, basically, it tells us if the trig ratio is positive or negative on the four quadrants.

And yeah, I think that’s all we learned on Friday. I hope this made sense to you guys, I know I’m not very good at explaining things. x)

Day 3-Trigonometry-Graphs of Sine and Cosine

Hey whatsup everybody it's me Emmanuel from your Pre-Cal30S class. Today I'm going to talk about what we learned in class yesterday from the beginning of our Trigonometry Unit. We learned about Graphing Sine and Cosine values, Comparing Graphs of Sine and Cosine Values, and naming the Five Key Points
of Sine and Cosine.

The basic equations of Graphing Sine and Cosine are y= a sin bx, y= a cos bx

The intermediate equations for Transformation of Graphing Sine and Cosine are y= a sinb(x-h)+k , y= a cosb (x-h)+k , h
represents the horizontal movement of the graph and k represents the vertical movement of the graph.

Amplitude- The amplitude is the distance from the x-axis to the highest point or lowest point for a sin x or cos x function, and is
unlimited for a tan x function.

A period is the length of a cycle which is measured in degrees or radians.

The periods of a sin x or cos x function is 360°

The period for a tan x function is 180°


II-these brackets represent the absolute value of a function. Absolute Value means that whichever number is between the brackets becomes positive.

The domain of a function is a set of numbers with all first x-coordinates of the ordered pairs in a function.

The range of a function is a set of numbers with all first y coordinates of the ordered pairs in a function.

The maximum value is the highest point of a function and the minimum value is the lowest point on the curve of the function, they are represented by
y-values.

For the basic graph of y=sin x


Domain: (-∞,∞)

Range: [-1, 1]

Period: 360°

Maximum Value: y = 1

Minimum Value: y = -1

X-intercept: x = c180°


For the basic graph of y=cos x


Domain: (-∞,∞)

Range: [-1,1]

Period: 360°

Maximum Value: y = 1

Minimum Value: y = -1

X-intercept: x = c180°-90°

Day *insert number of posts* - Quadratic Functions Review

Hey everyone, it's Christian from Pre Cal 30S. I am here to talk about what we did in class today (or yesterday I guess).

We had a review period today because we will have a test pretty soon. Some of us finished the Quadratic Functions Assignment 3 sheet and most of us worked on the review.

Someone asked a question from the review. It was from question 2G and it was asking to "describe the transformation with respect to y = x^2". This means we have to compare a given quadratic function to the basic function which is y = x^2.

Example:

This is the equation that needs to be analyzed.

This is the basic function. See the difference? Now it's all about knowing "what does what"

See the circles? They represent the a value the -h value and the k value.

The a value represents the shape of the parabola(narror or wide), and the direction of the opening. If the number is larger than 1, the parabola is narrower, and if it's less than one but larger than 0, the parabola is wide. Also the a value is negative which makes the parabola face down. The parabola is narrower and faces down.

The -h value represents the shifts on the x-axis(left and right). Notice that I wrote -h. It means that whatever the h value is, you have to get the opposite value to get the true value. In this case, the vertex of the parabola is moved 1 value to the left(negative side)

The k value represents the shifts on the y-axis(up and down). In this case, the vertex is found 3 values down.

SO the parabola is NARROWER, FACING DOWN, and the VERTEX is found at (-1, -3)

The other thing that we discussed today was another question in the review sheet. Question 2M states "find the new equation if the parabola is moved 2 units upward and 3 units to the left". All we have to do is imagine how it would affect on the quadratic function. Since it said 2 units up and 3 units left, we have to change the h and k value.

Here is an example:




Here is the original function and all we have to do is follow what the question is asking us to do. It stated 2 units upward and 3 units to the left

So here's what I did. Note that I added 3, it's because the h value is always opposite of what it is written so instead of -3, I put +3.

And here is our final answer. I underlined the h and k values so you guys can see the difference.

SO THAT'S WHAT WE DID IN CLASS TODAY. I HOPE THIS HELPED YOU GUYS OUT. I CAN'T EXPLAIN MATH UP-CLOSE SO I DON'T KNOW HOW I DID OVER THE INTERNET. I HOPE WE ALL DO WELL ON THE TEST. I HOPE YOU GUYS !!VOTE FOR ME!! AT THE END OF OCTOBER. EPIC NINJAS!!! PEACE!

~LOOK BEHIND YOU~

JK =)

Sup

Hi guys, it's A.J. from Mr.P's 30s pre-cal class. Today in class we finished the last page of our unit one booklet and started are unit one review. Wow all we did was a review so what do I talk about. Kay well I'll review Transformations of Quadratic functions 3 and Applications example 7.

"During the summer months Terry makes and sells necklaces on the beach. Last summer he sold necklaces for $10 each. His sales averaged 20 per day. Considering a price increase, he took a small survey and found that for every dollar increase he would lose two sales per day. If the material for each necklace costs $6, what should the selling price be to maximize profit."

Step one
find the equation and let one value equal "x"
20 = necklaces
2 = lost from price increase
x = number of the one dollar increase
10 = how much the necklaces are
6 = money for the material
THUS
(20-2x)(10-6+x)
or
(20-2x)(4+x)

Step 2
First
Outside
Inside
Last

Thus getting
80+20x-8x-2x²

Then change it around
-2x²+20x-8x+80
Then combine the alike terms
-2x²+12x+80
That's out equation

Step 3
Solve for X





















Step 4
Place X in the equation

-2(3)²+12(3)+80
3²= 9
9 . 2- = -18

12 . 3 = 36

-18+36+80=98

Step 5
Add the sentence
Terry made a maximum profit of $98

yeah. so I hope this helps guys sorry I just came home from watching volleyball. Oh yes don't forget to vote for the people in September. I have alot of grammar mistakes. I'm going to say "no don't read it" tomorrow