Math 2: Equations you can't live without

Test Preparation
1

<p>I thought it'd be a great idea to put common and even uncommon equations/identities we may need for the Math 2. All help would be greatly appreciated (even if you've already taken the Math 2 SAT II!). I'll start with common and basic trig functions/identities and other functions I've seen lately.</p>

<p>For right triangles:
Pythagorean Theorem:
a^2 + b^2 = c^2</p>

<p>sin x = opposite side/hypotenus
cos x = adjacent side/hypotenus
tan x = opposite side/adjacent side</p>

<p>csc x = (sin x)^-1
sec x = (cos x)^-1
cot x = (tan x)^-1</p>

<p>Radians -> Degree conversion
2(pi) = 360 degrees</p>

<p>Other Trig Functions/Identities:</p>

<p>Pythagorean Identities
sin^2 x + cos^2 x = 1
tan^2 x + 1 = sec^2 x
cot^2 x + 1 = csc^2 x</p>

<p>Sum and Difference Formulas
sin(A + B) = (sin A)(cos B) + (cos A)(sin B)
sin(A - B) = (sin A)(cos B) - (cos A)(sin B)
cos(A + B) = (cos A)(cos B) - (sin A)(sin B)
cos(A - B) = (cos A)(cos B) + (sin A)(sin B)</p>

<p>tan(A + B) = [tan A + tan B]/[1 - (tan A)(tan B)]</p>

<p>tan(A - B) = [tan A - tan B]/[1 + (tan A)(tan B)]</p>

<p>Double Angle Formulas
sin 2A = 2(sin A)(cos A)
cos 2A = cos^2 A - sin^2 A
cos 2A = 2(cos^2 A) - 1
cos 2A = 1 - 2(sin^2 A)</p>

<p>tan 2A = [2 tan A]/[1 - tan^ 2 A]</p>

<p>Law of Sines</p>

<p>[sin A]/a = [sin B]/b = [sin C]/c
*A, B, C are angles and a, b, c are sides opposite of the respective angles</p>

<p>Law of Cosines
a^2 = b^2 + c^2 - 2bc (cos A)
b^2 = a^2 + c^2 - 2ac (cos B)
c^2 = a^2 + b^2 - 2ab (cos C)</p>

<p>Are of a triangle:</p>

<p>Area = (0.5) bc (sin A)
Area = (0.5) ac (sin B)
Area = (0.5) ab (sin C)</p>

<p>Conic Sections
Circle<a href="x%20-%20h">/u</a>^2 + (y - k)^2 = r^2</p>

<p>Ellipse
[(x - h)^2]/[a^2] + [(y - k)^2]/[b^2] = 1</p>

<p>Hyperbola
[(x - h)^2]/[a^2] - [(y - k)^2]/[b^2] = 1</p>

<p>Polar Graph
sin z = y/r
cos z = x/r
tan z = y/x</p>

<p>x = r (cos z)
y = r (sin z)
x^2 + y^2 = r^2</p>

<p>I hope this helps. Feel free to add anything, and sorry if anyone has already created a thread like this. Let's hope it'll help us all!</p>

2

<p>::huggles::
thank you so much krabble!
:-D</p>

3

<p>thank you!!!</p>

4

<p>this is awesome. =)</p>

5

<p>Does anyone know the method of graphing and using the equation of a hyperbola by only referring to this equation:</p>

<p>[(x - h)^2]/[a^2] + [(y - k)^2]/[b^2] = 1</p>

<p>There's another equation associate with the hyperbola, and it's when the major axis is perpendicular with the x-axis, but I never really learned this method when learning trig. Instead we only referred to the one equation (above) somehow, but now I forget. Can anyone give me an overview of how I can graph ANY hyperbola using only that top equation? Has anyone else learned hyperbolas by only referring to one equation? I believe it is the same idea that can be applied to ellipses as well. Help!</p>

6

<p>I'll bump this up incase anyone knows what I'm talking about and can answer my question!</p>

7

<p>ohh.... something about (h, k), widths and heights and drawing a rectangle and diagonal asymptotes. anyone?</p>

8

<p>Just remember soh cah toa...Sine Opposite (over) Hypotenuse, Cosine Adjacent (over) Hypotenuse, Tangent Opposite (over) Adjacent...</p>

<p>Everything else is easy after that.</p>

<p>Thank you so much! THis is a really neat thread.</p>

9

<p>Trig and Analytic Geometry should not be taken in 18 weeks. I remember the trig. Not the conic sections though.</p>

10

<p>Crap! We have to memorize all the sin/cos/tan identities?! Noo!</p>

11

<p>put them in your calculator</p>

12

<p>Krabble88, it is an ellipse formula what you ask. But if you mean
[(x - h)^2]/[a^2] - [(y - k)^2]/[b^2] = 1
[(y - k)^2]/[b^2] - [(x - h)^2]/[a^2] = 1
Then you graph this hyperbola by putting (h,k) as the center and putting points that are "a" units away from the center on the "x" axis and "b" units away from the center on the "y" axis. Then you draw lines through those points so that the intersection of four lines (horizontal and vertical lines) make a rectangle. Draw diagonals of the rectangle and put the parts of the hyperbola up and down of the rectangle if x part of the equation is subtracted from y part. If it is other way around, then put the parts of the hyperbola to the sides of the rectangle.</p>

<p>I hope it helps. It is hard to describe sth that should be seen.</p>

13

<p>I have taken the exam and these were the most useful ones in my opinion:</p>

<p>sin^2 x + cos^2 x = 1</p>

<p>Pythagorean Theorem:
a^2 + b^2 = c^2</p>

<p>sin x = opposite side/hypotenus
cos x = adjacent side/hypotenus
tan x = opposite side/adjacent side</p>

<p>csc x = (sin x)^-1
sec x = (cos x)^-1
cot x = (tan x)^-1</p>

<p>Law of Sines</p>

<p>[sin A]/a = [sin B]/b = [sin C]/c</p>

<p>BTW, I never used polar graphs in math iic (I didn't use them even in AIME)</p>

14

<p>has integration ever helped anyone on math iic?</p>

15

<p>Integration didn't help me on math iic. But you may use integration in one or two questions if you don't see the straight-forward solution. But it takes too much time anyway.</p>

16

<p>was anybody able to finish every single prob? Is it possible?</p>

17

<p>I'm just curious, but do many logic questions come up on the 2c test?</p>

18

<p>Logic problems do come up, but there is usually only one, two at max.</p>

19

<p>Can someone give and explain the equation for finding the distance between a point and a line? It's not a major topic on the Math 2 test, but it came up on a practice test (Barron's), and I was wondering if anyone could explain it to me for my own references.</p>

<p>Also, do you think the difficulty of Barron's practice tests are due to the fact that they cover more obscure topics, the problems require more mathematical steps, etc.?</p>

<p>Thanks!</p>

20

<p>Krabble88, the distance between two points can be found by the formula= square root of ([y2-y1]^2/[x2-x1]^2)
In fact, the logic of this formula is very simple. Just make a right triangle using those two points and you will see the distance can be found by using the Pythagorean Theorem.
Barron's questions require more mathematical steps and you need to "see the way to the answer" most of the time. (Most of the questions are tricky and take more time to solve than real questions.)(My friends bring me the book to solve some questions)
Spazallan;
I have finished every single problem in math iic in 40 minutes, and I solved all of the questions again in 20 minutes. Moreover, I hadn't studied anything before the exam. I mean it is possible. (By the way, I am an international student who loves mathematics, I enter all the math competitions that are available. I had 144/150 in AMC 10, and had certificate of distinction in AIME. I was placed first in Turkey two times, and placed third once in different math competitions. Therefore, for me, time was no problem at all, but if you are not a math person, you need to do a lot of practice tests)</p>