Reflection

So this week went by so slow! But thankfully it is finally spring break :)
We got b-rob back as our teacher this week, so we had to catch up on math. I was a little rusty but we started a new chapter. We learned stuff like vectors from Chapter 12. It was definitely easier than i thought it would be. We also learned how to draw vectors. Also, we started to do some review on trig for our trig exam coming up.

Vectors:
u = (1,2) v = (4,3)

1. u + v
(1+4) (2+3)
= (5,5)

2. u - v
(1-4) (2-3)
= (-3, -1)

3. 3u + v
(3+4) (6+3)
= (7, 9)

I did have some trouble with the stuff we learned that related to physics, so if anyone wants to help me, that'll be great! HAPPY EASTER :)

Comments post

We finally learned something new. Vectors....
vectors are usually recognized as U and V. Or they will have these: <> around them. vectors are weird and are given in point form (I think)

to add vectors you simply add the points i.e.
U(3,7) V(2,3)

U+V=(5,10)
Subraction- same principle
U-V= (1,4)
an equation
2U+V=(8,17)

Reflection

State convention was pretty fun...I'm glad we had the time together as a club, and now B-Rob's back to teach us more math...fun. So here's some of the stuff I remember from way back when.

LAW OF COSINES

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

You can use an angle to orient yourself like SOHCAHTOA


Example:

a is 5 cm
b is 6 cm
angle C is 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



TRIG INVERSES

you have to find 2 angles usually with trig inverses.
first you need to determine which quadrants your angles are going to be in
(find out where that function is in its sign--ex: sine would be positive in the upper two quadrants, I and II)
then you plug in the positive inverse value into your calculator to get a reference angle, if there is not one given already

quadrant rules:
I to II---make it negative and add 180 degrees
I to III---add 180 degrees
I to IV---make it negative and add 360 degrees
II to IV---add 180 degrees

Example:

cos^-1 (-1/2)

Cosine is negative in the II and III quadrants
Your reference angle is given as 60 degrees.
I to II---make 60 negative and add 180 degrees = 120 degrees
I to III---add 180 degrees to 60 degrees = 240 degrees

Reflection

State was pretty fun. B-Rob is finally back and that means we can finally learn something new. Springfest was poo poo as always. Had a long, pretty fun weekend. Can't wait till easter break. K bye ;)

Reflection

Graphing ellipse

Formula:
(x-h)^2/(length of x/2)^2+(y-k)^2/(length of y/2)^2=1

1. center: (h,k)

2. major axis: larger denominator

3. vertex: on major axis; +/- square root of small denominator

4. find focus: smaller # or denominator^2=larger # or denominator^2-focus^2

5. length of major axis 2 square root of larger denominator

6. length of minor axis 2 square root of smaller denominator

7.then graph...

Reflection March 28

Some older stuff to reflect back on:

Give the polar coordinates for (3,4)
first you have to use the formula r = square root of (x^2 + y^2)
so for this problem r = square root of (3^2 + 4^2)
which gives you square root of (9 + 16) = (25)
so now you have your x = + or - 5
Now pllug your y/x into the inverse of tangent:
a = tan^-1(4/3) = 53.130 degrees.
Now find where tangent is positive, quadrants I and III, so your other angle is going to be 233.
130 because you ad 180 to 53.
130 to move it to the third quadrant.
Now you have to graph the two to figure out which goes with positive 5 and which goes with negative 5.
In this case, it's:
(5 , 53.130 degrees) and (-5 , 233.130 degrees)
Also, changing from polar to rectangular is easy:
Give the rectangular coordinates for (3,30 degrees)
use your formulas x = rcosa and y = rsinax = 3(square root of (3)/2) and y = 3(1/2)solve and you end up with:
(3 square root of 3/2, 3/2)

reflection

Solving anything bigger than a quadratic usuing quadratic form. To use quadratic form you have to have 3 terms only. The first term must equal the 2nd exponentx2, and the last term must be a constant. The first thing you do is make g=x^exponent/2 so that you would get g^2+g+#. The second thing is to do the quadratic formula, factor, or complete the square. The the last thing is to plug back in for g. (Whenever you do step three you are basically just plugging back into step one. g=x^2)

An example:
x^4-4x^2-12=0

1. g=x^4/2
g=x^2
g^2-4g-12

2. (g^2-6g)+(2g-12)
g(g-6)+2(g-6)
(g+2)(g-6)
g=-2 g=6

Reflection #32

One thing we'll always need in this class...the trig chart.
So, let's review

0 = 0 degrees
pi/6 = 30 degrees
pi/4 = 45 degrees
pi/3 = 60 degrees
pi/2 = 90 degrees

sin 0 = 0
sin 30 = 1/2
sin 45 = square root of 2 over 2
sin 60 = square root of 3 over 2
sin 90 = 1
cos 0 = 1
cos 30 = square root of 3 over 2
cos 45 = square root of 2 over 2
cos 60 = 1/2
cos 90 = 0
tan 0 = 0
tan 30 = square root of 3 over 3
tan 45 = 1
tan 60 = square root of 3
tan 90 = undefined
cot 0 = undefined
cot 30 = square root of 3
cot 45 = 1
cot 60 = square root of 3 over 3
cot 90 = 0
sec 0 = 1
sec 30 = 2 square root of 3 over 3
sec 45 = square root of 2
sec 60 = 2
sec 90 = undefined
csc 0 = undefined
csc 30 = 2
csc 45 = square root of 2
csc 60 = 2 square root of 3 over 3
csc 90 = 1

Now, examples:
**a=alpha

4 sin pi/6 cos pi/6
=2 sin 2a
=2 sin 2(pi/6)
=2 sin(pi/3)
=2(square root of 3 over 3)
=square root of 3

cos^2 x/2 - sin^2 x/2
=cos 2a
=cos 2(x/2)
=cos x

And don't forget it...

golb

Im doing my blog on SOCAHTOA because its easy and i am getting very lazy. So here it is ladies and gentlemen.

SOCAHTOA



Sin=opposite leg/hypotenuse

Cos=adjacent leg/hypotenuse

tan=opposite leg/adjacent leg


By following the chart you will get what you need. Meaning that for Sin you are trying to find the adjacent leg, Cos the opposite leg, and tan the hypotenuse.


Hint: If you have an isoceles triangle all you have to do is split it down the middle to make two right triangles. Then you split the length and they have the same angles on both side of the bottom.

Reflection 32.

This week is ACT prep week so hopefullllllly we don't learn anything new.. :). Anyways I'll explain formulas for a line and when to use them.

Formula for a line:
1) m=tan alpha
m=slope
alpha=angle of inclination

Examples:

Find the angle of inclination of the line that passes through (4,8) and (7,1).
First find the slope by plugging in the two points into the slope formula, y-y/x-x.

8-1/4-7 = -7/3
*Plug into formula.
-7/3 = tan alpha
alpha = tan^-1(-7/3)
*Find quadrants: II & IV
*Plug inverse into calculator.
66.801 degrees
*Make negative and add 180.
113.199 degrees
*Make negative and add 360.
293.199 degrees
*Angle of Inclination = 113.199 degrees, 293.199 degrees

REFLECTION #32

Okay this week I'm going to review some more things from trig, since we have that huge exam coming up. Here are just a few examples of things in trig that we learned.

Ex. 1.)
**Find the reference angle for the following:
A.) sin 210 degrees
*The first thing you have to do is figure out which quadrant 210 is in.
*it's in the 3rd quadrant. you're dealing with sine, which deals with the y axis, so in the the 3rd quadrant sine would be negative...So that means you have -sin so far
*Now to find the angle. All you have to do is subtract 180 degrees from 210 and you get 30 degrees
*So your reference angle is -sin 30

B.) tan 695 degrees
*The first thing you notice is that 695 can't be found in the quadrants because it's bigger than 360 degrees...Soo what you have to do first is subtract 360 from it to get a new angle.
*So you now have tan 335 degrees
*So you find 335 in the quadrants, and it's in the 4th quadrant
*you're dealing with tangent (which is y/x). And in the 4th quadrant the y axis is negative and the x axis is positive. (so that means > -y/x >> negative) So therefore tangent would be negative (-tan)
*Then you subtract 180 degrees from 335 and you get 155. *reference angles can only be between 0 and 90 degrees, so you have to subtract 180 from 155. You get -25, but pay no attention to the negative
*So your reference angle is -tan 25

Ex. 2.) sin^-1(-sqrt of 3/2) >> Find the inverse
*to find the inverse first you have to find a reference angle.
*So you look on your trig chart and find sin(-sqrt of 3/2) and the angle is 60 degrees.
*you always start off in the first quadrant with your reference angle. and for inverses you have to find two angles.
*Now look at sin^-1(-sqrt of 3/2) Is it negative or positive? negative. does it relate to the x or y axis? y axis. So you ask yourself, "where is sine negative?" you would check off the 3rd and 4th quadrants because that is where sine is negative
*To find your angles you first want to move from the first quadrant to the 3rd quadrant. to do this you take your reference angle, 60 degrees, and add 180 to it. So your first angle is 240 degrees.
*Now you want to move from your first quadrant to the 4th quadrant. to do that you take 60 degrees, make it negative and add 360 degrees to it. So your other angle is 300 degrees.

Alright that's it for this week. Can anyone tell me what you guys did in class on Friday? thank youuuuuu (:

Reflection 3/28

Im doing my blog on SOCAHTOA because we are going to need this when brob gets back. So I thought I would freshen everybody up on the whole concept.

SOCAHTOA



Sin=opposite leg/hypotenuse

Cos=adjacent leg/hypotenuse

tan=opposite leg/adjacent leg


By following the chart you will get what you need. Meaning that for Sin you are trying to find the adjacent leg, Cos the opposite leg, and tan the hypotenuse.


Hint: If you have an isoceles triangle all you have to do is split it down the middle to make two right triangles. Then you split the length and they have the same angles on both side of the bottom.