REFLECTION #31

Alright this week I thought I'd review some basic trig concepts that we learned right in the begining of trig, just to refresh everyone's memories. ha
So here's some examples:

Ex. 1.) Find the negative coterminal angle to 1150 degrees.
*Remember, to find coterminal angles all you have to do is either add or subtract 360 degrees.
*Since they're asking for a negative coterminal angle, all you're going to do is subtract 360 degrees from 1150 until you get a negative number
*So you subtract 360 once and you get 790, subtract 360 again and get 430, subtract 360 again and you get 70, then finally subtract 360 one more time and you get -290. So -290 is your negative coterminal angle.

Ex. 2.) Convert 3pi/4 to degrees.
*When converting to degrees all you have to do is multiply what you're given by 180/pi.
*So you get 3pi/4 * 180/pi. the pi's cancel out and you multiply 3*180 then divide by 4 and you get 135.
*So 3pi/4 is equal to 135 degrees.

Ex. 3.) Convert 80 degrees to radians.
*When converting to radians all you have to do is divide the degree you're given by 180, get a fraction for it, and add the pi back in.
*So you get 80/180 which reduces to 4/9, then you add the pi in
*So 80 degrees in radians is 4pi/9

Ex. 4.) Put in degrees only > 17degrees10'26"
*Okay for this one, all the numbers aren't degrees so you have to make them into degrees in order to solve the problem
*So to put 10' in degrees you make it 10/60
*And to put 26" in degrees you make it 26/3600
(*^^I don't actually know why you do that, but that's what I have in my notes soooo..ha yeah)
*Now since all the numbers are degrees you add them all up in your calculator
So you add 17 + 10/60 + 26/3600 and you get 17.174
*So your answer is 17.174 degrees

Ex. 5.) (3,4) Find all 6 trig functions.
*The first thing that should pop into your mind when you're asked to find all 6 trig functions is a triangle..because you're going to have to draw one ha
*So when drawing your triangle, first it's going to be in the 1st quadrant. the point you're given is (3,4) so you're going to go to the right 3 times and up 4 and put your dot. then you draw a line from the dot to the middle of your graph "0". then draw a line straight down from the dot to the x axis
*Now you have to find the hypotenuse of the triangle so you're going to use c^2 = a^2 + b^2
*So plugging in 3 and 4 into the formula you get that c is 5. *Orrrr you could've just known that a 3,4,5 triangle is a perfect triangle, but whatever works for you (;
*Now you find your trig functions. *keep in mind that "r" is the hypotenuse
1. sin (y/r) = 4/5
2. cos (x/r) = 3/5
3. tan (y/x) = 4/3
4. csc (r/y) = 5/4
5. sec (r/x) = 5/3
6. cot (x/y) = 3/4
And thereeee youuuu gooo!

Well I say that's enough reviewing for the week, so I'm done here (:

Comments

Here is an example of the law of cosines

(opp leg)^2 = (adj leg)^2 + (other adj leg)^2 - 2(adj leg)(adj leg) cos (angle b/w)

a = 5 cm
b = 6 cm
angle C = 36 degrees
find c

c^2 = 5^2 + 6^2 - 2(5)(6) cos(36 degrees)
c = sqrt of ((6^2 + 5^2 - 2(5)(6) cos 36 degrees))
c = 3.53

here's chapter 1 review. The things that I found the easiest are quadratic formula, and completing the square. Here’s an example of quadratic formula to show how it works:3x^2-4x+1=0-b+-square root b^2-4ac/2a--4+-square root (-4) ^2-4(3) (1)/2a4+- square root 4/64+-2/66/6=1 (1,0)2/6=1/3 (1/3,0)Step 1: take the equation and plug it into the formula.Step 2: plug everything under the square root into your calculator.Step 3: simplify the square root then simplify the fraction and plug in as a point.The thing I had the most trouble with is Graphing Parabolas. I understand everything till you have to solve for the vertex. In the example solving for the vertex, the original problem is y=2x^2-8x+5. The example it has (2,-3), 2(2)^2-8(2)+5=-3 so where does the 2 come from? Does it matter what number you plug in?As the school year goes on and we learn more it will get tougher but we will all get through it.

Reflection

Conics:The steps to find the intersection of a line and a circle are: solve the linear equation for y, next substitute in circle equation, after this you solve for x, and last you plug x value in to get y value **If your x value is imaginary, then there is no point of intersection.

Say you have:
x^2+y^2+12y+16x-5=0

First you rewrite the problem in order with x's in front and y's in back, or vice versa, and you would get this:
x^2+16x__+y^2+12y__=5

Next you would fill in the blanks with the number that belongs, for this you divide the x and y by 2 and then square it. For this problem you would use 16x and 12y, and you would get 64 and 36. So the answer would be:
x^2+16x+64+y^2+12y+36=5

After this you add the new numbers to the other side of the problem and you would get:
x^2+16x+64+y^2+12y+36=5+64+36
or
x^2+16x+64+y^2+12y+36=105

Then you factor out the x's and y's:
(x+8)^2+(y+6)^2=105

The answer is:
Center=(-8,-6) Radius=square root of 105

Reflection

Here are a few examples of some easy logs. I can't wait for the easter break. AHHHHHHHHHHHHH

log 7 of 49 = 2 log x of 36 = 2
7^2 =49 x^2 = 36
x = 6

log 3 of 27 = 3 log x of 64 = 3
3^3 =27 x^3 = 64
x = 4

log 2 of 32 = 5 log x of 100 = 2
2^5 =32 x^2 = 100
x = 10

Reflection #30

Let's review logs:

To exponential form:
log(x)4=2
take the base (x), raise it to = ... (2), and set it equal to log of ... (4)
x^2=4
solve for x
x=+/- 2

To log form when base 10:
10^x=75
take number raised (10) and set as base, take the = ... (75) and set as log of ..., then take raised to ... (x) and set = to
log 75=x
then plug into calculator and solve for x
x=1.875

To natural log form:
e^x=75
put e as base of ln and put 75 as ln of ... then set x equal and solve for x by plugging into calculator
ln 75=x
x=4.317

Domain and Range:

Exponential:
D=(-infinity, infinity)
R=(o, infinity)

Log and ln:
D=(0, infinity)
R=(-infinity, infinity)

Any more examples?

reflection

Law of Sines: sin(opp. angle)/Leg=sin(Opp. angle)/Leg. Set up a proportion.

Example. triangle ABC where A=36 degrees a=3 and B=56 degrees. find b

sin36/3=sin56/x 3sin56/sin36= x Plug into your calculator and done.

Law of Cosines:

(opposite leg)^2=(adjacent leg)^2+(other opposite leg)^2-2(leg)(leg)cos(angle in between)

example:
for a triangle with C=36 degrees a=5 b=6

c^2=5^2+6^2-2(5)(6)cos36
c=Square root of(25+36-2(5)(6)cos36)
c= 3.53

Are of non-right triangle:
1/2(leg)(leg)sin(angle between)

Example: Triangle ABC has sides a=5 b=3 and C=40 degrees

Your formula will be (1/2)(5)(3)(sin(40)) type that in your calculator and you will get the answer.

Reflection.

I'm gonna go over like the first thing we ever learned in trig and it's like the simplest. Converting degrees into radians and vice versa. To convert degrees to radians you multiply the degrees by Pi/180. Your final answer will have Pi in it. To convert radians to degrees multiply by 180/Pi. Pi will cancel out and you will be left with just the number.

Example: Degrees---->Radians

225 x Pi/180= 225/180Pi= 5/4Pi


Example: Radians---->Degrees

3/4Pi x 180/Pi= 135 Degrees


**Also converting Decimals to degrees
-Convert 6.5 to degrees

*Alright the first thing you want to do is type this in your calculator:
6.5 X (180/pi) =
-hit enter and you get 372.4225
Since 372 is in front of the decimal, that's going to be your degrees.
The next thing you do on your calculator is subtract 372 and you just get .4225. Then you multiply that by 60 to get your minutes.
Then you get 25.3540. Since 25 is in front of the decimal, your minutes will be 25'. Now you subtract 25 and you get .3540, then you multiply that by 60 and you get 21.241>>these are your seconds. (and you leave the decimal)

Answer:
372 degrees 25' 21.241"

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What is on the exam exactly? Will it just be questions about the book like short answer or something? Then will we have to know that vocab we took in class?

Reflection 3/14

okayy guysss, today is pi dayyy. yayyyy :), so we took chapter tests all week, and our exam is on tuesdayyyy. so here's time for some review before the exam.
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IDENTITIES:

csc x = 1/sin x
sec x = 1/cos x
cot x = 1/tan x
sin (-x) = -sin x
cos (-x) = cos x
csc (-x) = -csc x
sec (-x) = sec x
tan (-x) = -tan x
cot (-x) = -cot x
sin^2 x + cos^2 x = 1
1 + tan^2 x = sec^2 x
1 + cot^2 x = csc^2 x
sin x = cos (90 degrees - x)
tan x = cot (90 degrees - x)
sec x = csc (90 degrees - x)
cos x = sin (90 degrees - x)
cot x = tan (90 degrees - x)
csc x = sec (90 degrees - x)
tan x = sin x/cos x
cot x = cos x/sin x

^^you might need to review identities before we take that chapter test, and remember**use identities first, then algebra.
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SYMMETRY:

y=x^3=4x

a.) x-axis
(-y)=x^3+4x
y=-x^3-4x Reflect; No

b.) y-axis
y=(-x^3)+4(-x)
y=-x^3-4x No

c.) y=x
x=y^3+4y
y^3+4y=x
y(y^2+4)=x No

d.) origin
(-y)=(-x)^3+4(-x)
-y=-x^3-4x
y=x^3+4x Yes

^^ways to find the symmetry of a problem, that was on one of our recent chapter tests we just took in class.
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TRIANGLES:

SOHCAHTOA:
sin (angle)=Opposite leg/Hypotenuse
cos (angle)=Adjacent leg/Hypotenuse
tan (angle)=Opposite leg/Adjacent leg

^^you might also need to know this again if we take those chapter tests also.
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and thats about it, if the exam is on flatland, we just study the questions we took in class? or is it on the book, like the whole book?

3/14

This week we have our exam on Flatland and still are doing chapter tests so here is something from chapter 11.

11-2 Imaginary Numbers are no longer "imaginary"

Rectangular form is defined as
a + bi

Polar form is defined as
z = r cos theta + r sin theta i
abbreviated z = r cis theta

Example: Express 2 cis 50degrees in rectangular form

2 cos 50 + 2 sin 50 i


-1-2i in polar form
radius = +- sqrt of ((-1)^2 + (-2)^2)) = +- sqrt of (5)
theta = tan^-1(-2/-1)
theta = tan^-1(1)

if you do this, tangent is positive in the first and third quadrants, so it comes out to be 63.435 and 243.435

Since the 63 is positive for cosine, we can put it with the positive sqrt of 5.
And the 243 is negative for cosine, we can put it with the negative sqrt of 5.

reflection march 14

This week was like the first full week we had in a while. All we did was do the chapter tests, and im not surprised that i don't remember anything. I really need to start reviewing the chapters in my binder so i can actually know the stuff that will be on the chapter test the next day. And i need to start looking over the flatland stuff, like the vocabulary words and the questions and everything. Then after our exam on tuesday, were going to be getting back to the chapter tests. ugh...ha.

reflection for 3/14

all week, we did chapter tests from the beginning of the year....theres pretty much nothing new happenin in advanced math.....and the exams on flatland? that was a good review week wasnt it....

well, since we didnt learn anything new, ill just post somethin that gave me trouble in the chapter tests during the week

convert to radians: 120º
120 X π/180=120/180 π
=12/18 π
=2/3 π

man, i had to go waaaaay back into my notebook to figure out how to do this one
o well, at least the basketball team won the state championship

Reflection

So this week in class we did our review test, and i remember most of the stuff. I like how working the problems are kinda simular. Also, the formulas are basically easy to memorize.

EXAMPLES: Circles

Standard Form: (x-h)^2+(y-k)^2=r^2
center: (h,k) radius: r

Find the center and radius of each cirlce.
1.) (x-3)^2+(y+7)^2=19
center: (3,-7) radius: squareroot of 19

Find the intersection of the circle.
1.) x^2+4^2-25 and y=2x-2
a) y=2x-2
b) x^2=(2x-2)^2=25
c) x^2+4x^2-8x+4=25
5x^2-8x+4=25
5x^2-8x-21=0
5x^2-15x+7x-21=0
5x(x-3)+7(x-3)=0
(x-3)(5x+7)
x=3 x=-7/5
y=2(x)-2
2(3)-2=4
y=2(-7/5)-2=-24/5

Final Answer: (3,4) (-7/5,-24/5)

Write in Standard form.
1.) Center: (4,3) Radius: 2
(x-4)^2+(y-3)^2=4

I also thought the hyperbolas were mostly simple, but i'm still trying to understand them. But i thought the cirlces were most easiest to understand.