Relfection #17

This week we've been working on study guides. I'll reflect about something we learned a couple weeks ago. This might help you on your study guides. This is how you find the area of a triangle when given two lengths and angle.

For Area say you have a triangle with:

A side length of 4
A side length of 5
And angle of 30 degrees

Then plug it all in: A=1/2 (4) (5) Sin 30 degrees

A= 10 Sin 30 degrees which aproximately =5.

For another triangle:

A side length of 3
A side length of 8
And an angle of 60 degrees

Then plug it all in: A=1/2 (3) (8) Sin 60 degrees

A=12 Sin 60 degrees which approximately = 10.392

Reflection #17

This week we worked on study guides. From chapters past, one thing I'm going to reflect on is logs: change of base:

**Only use when there is no other way to change the base of a log or solve for x in an exponential
First, write as an exponential if it is not already
Second, take the log of both sides (base 10)
Third, move exponent (x) to front
Forth, solve for variable by division
Fifth, write as log fraction (simplify further if whole number only, plug in calc.)

Example:
log base 3 of 7
3^x=7
log of 3^x=log of 7
x log of 3=log of 7
x=log of 7/log of 3

5^x=10
log of 5^10=log of 10
**Note: log of 10 = 1
x log of 5=1
x=1/log of 5

Now, I don't understand from the study guides...
Chapter 8 - #4, #9, #10, #11
Chapter 7 - #16, #17, #21
Any help with these problems??

Reflection 17

Last week we really didnt do too much. It was a pretty easy, but busy week. We worked on our study guides and did our bridge project. Unfortunatly some people had to do it twice, and we still got the wrong glue, thanks to who ever gave us that. Anyways, lets review about right angles of triangles to help with our mid term exam that is coming up on wednesday. Right Triangle, here we go!!!! The hypotenuse is always opposite a right angle. The area of a right triangle is 1/2 the base x height. To deal with traingles remember the saying soh cah toa. This saying means sin is equal to the opposite leg divided by the hypotenuse, cos is equal to the adjacent leg over the hypotenuse, and tan is equal to the opposite leg divided by the adjacent leg. You always want your calculater in degrees when trying to solve these problems. When you want the degree of a triangles angle you have to hit the second button first so that u get the inverse, also knowing the trig chart will help on some of those problems, so just a hint, KNOW THE TRIG CHART< IT WILL HELP!!!!!!!!!!!!!! Finding the area of a non right traingle is pretty easy all you have to do is A=1/2(leg)(leg)sin(angle). We also learned the law of sin and we learned the law of cosine, i was absent those days so anyone that wants to show me an example prblem and help me i would appreciate that, especially the law of cosine, i dont even think i have the notes.

Reflection 17

I did not know what to reflect about from this week since we did study guides and our bridge contest. So I decided to reflect about something we learned a couple weeks ago. This might help you on your study guides. This is how you find the area of a triangle when given two lengths and angle.

For Area say you have a triangle with:

A side length of 4
A side length of 5
And angle of 30 degrees

Then plug it all in: A=1/2 (4) (5) Sin 30 degrees

A= 10 Sin 30 degrees which aproximately =5.

For another triangle:

A side length of 3
A side length of 8
And an angle of 60 degrees

Then plug it all in: A=1/2 (3) (8) Sin 60 degrees

A=12 Sin 60 degrees which aproximately = 10.392

This formula can only be done when you have two given lengths and an angle.

Reflection

This week was mostly study guides in advanced math so nothing really new. I do however need some help with simplifying trig functions so if anyone would help me with an example that would be nice. So far i think that tan is equal to sin over cos and i know that sin squared plus cos squared always equals one. And that tan is equal to one over sin and cot is equal to one over cos. Could somebody please post a few more examples before the exam.

Reflection 17!

So this is going to be a little harder than usual because we didn't really learn anything new this week. Our bridge held up to 75 pounds, how cool? :) Congrats to the winning bridge, 20 points sounds like a great Christmas present! Keenen, we'll alll miss you. Don't forget to keep in touch, and any chance that you get to come visit PLEASE DO! We loooove you! Good luck wherever you go, :/

I'll just go through my study guides and try to explain a few things from a while back?
First, i'll start with synthetic division.
Synthetic division is strictly used to find zeroes or roots of polynomials.

Equation: (x^5+x^4+2x^3+5x^2-3x+6) / (x+3)
First you'll set x+3=0
x=-3

this gives you what to synthetically divide by(not sure if that's the correct way to say that! lol) :

-3|1 1 2 5 -3 6




1st: Bring your first number down.

-3|1 1 2 5 -3 6
------------------
1

2nd: multiply the number at the bottom by the number being divided and add.

-3|1 1 2 5 -3 6
| -3 6 -24 57 -162
---------------------------
1 -2 8 -19 54 -156

Remainder for this equation would be:
R= -156.

All in all,
these are the steps in words...

• Bring down the 1.
• Multiply it by the -3 to get -3.
• Add 1+-3 to get -2.
• Multiply -2 times -3 to get 6.
• Add 2 plus 6 to get 8.
• Multiply 8 times -3 to get -24.
• Add 5 plus -24 to get -19.
• Multiply -19 times -3 to get 57.
• Add -3 plus 57 to get 54.
• Multiply 54 times -3 to get -162.
• Add 6 plus -162 to get -156.

Second, the six trig functions are:

sin=y/r
cos=x/r
tan=y/x
cot=x/y
csc=r/y
sec=r/x

ALSO:
degrees to radians= D * pi/180
radians to degrees= R * 180/pi(in this case, pi's cancel leaving you with a number in DEGREES)

and finally, exponential equations?

in an equation:
(b^2/a)^-2 TIMES (a^2/b)^-3

you must first distribute your exponents to all parts of your fraction, in the case of exponents, you must multiply the exponential values together, i.e.,

(b^-4/a^-2) TIMES (a^-6/b^-3)

in order to remove all negative exponents from the equation, put a 1 over each and use the sandwich method!

(1/b^4)/(1/a^2) TIMES (1/a^6)/(1/b^3)

multiply the top by the bottom(BREAD) and the two inner fractions by each other(MEAT, or peanut butter- whichever you prefer. lol)

a^2/b^4 TIMES b^3/a^6

cancel what you can:

a^2 and a^6= a^4
b^4 and b^3= b


FINAL EQUATION:

1/ba^4

---------------------------------------------------------------
WORD PROBLEMS STILL GET ME, PLEASE PLEASE PLEASE HELPPP!

Reflection 17

So, our bridge won! 20 bonus points i hope! So this week what we did was our study guides and bridges, which we had to stay up till 430 in the morning! So these are the six trig functions and the unit circle:

Six trig functions:
sin: y/r
cos: x/r
tan: y/x
cot: x/y
csc: r/y
sec: r/x

Unit Circle:
90(pi/2): (0,1)
180(pi): (-1,0)
270(3pi/2): (0,-1)
360(2pi): (1,0)

And here are a couple 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

and thats all the time i have folks! peace!

reflection 9

The thing we learned this week was identities. In identities you pretty much just keep canceling things out. The thing you have to remember in identities is first you find an identity then you do some type of math to make the problem smaller. In this you have to memorize all of the identites it's the only way you will be able to work the problems.

These are the identites you will use a lot:

Pythagorean Relationships:

sin^2 x + cos^2 x = 1

1 + tan^2 x = sec^2 x

1 + cot^2 x = csc^2 x

Reciprocal Relationships:

csc x = 1/sin x

sec x = 1/cos x

cot x = 1/tan x

Ill give an example.

simplify: tanA x cosA
tan=sin/cos now its sinA/cosA x cosA and that equals sinAcosA/cosA then your cos cancels, your answer is sinA.

Identities are chap and you have to be all careful and stuff if you wanna do it right. just remember to do identities before your math stuff and you straight.

I forgot like graphing. pretty much everything about it. aha.

Reflection 17 (and a shout out to keenan)

To start, I dont care that this is not math....Keenan I'ma miss you dude. Riverside aint gone be the same wit you gone. It was tight havin you in some classes while it lasted. So good luck in life brother.

Ok so now to the math. We worked on study guides this week and there was a lot of remembering needed to do some of the early chapter tests for the study guide. One thing I remeber how to do, and its from way back in chapter 1, is how to work the i chart.

-When you get a number like say, i^3224, its easy to solve for it. All you have to do is divide the exponent (3224) by 4. In this case that will give you 86.

-Now use the i chart (plug the number you got when you divided by 4 (86) into the i chart:

.25 = i
.5 = -1
.75 = -i
1or any whole number = 1

-So because 86 is a whole number, i^3224 = 1. Yay!



Okay so another thing that was a great deal of fun at the beginning of the year that I remember well is logs. I'm going to show you some log stuff.

-To put a log into exponetial form use this handy dandy formula:
logb(x)=a ---logarithmic form
b^a=x ---exponential form

-Guess what....here's an example with numbers instead of letters:

logx(4)=2
-Put into exponetial form:
x^2=4

Reflection 17

this week was mainly focused on review for the exam and the bridges. congrats to those who held over 20 lbs by the way. I basically got a fresh review of chapters 1 thru 9 pounding in my head right about now.
IDENTITIES
csc=1/sin x
sec= 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*-x)
tan x=cot (90*-x)
sec x=csc (90*-x)
cos x=sin (90*-x)
cot x=tan (90*-x)
csc x=sec (90*-x)
tan x=sinx/cosx
cot x=cosx/sinx

if i am correct these should be all the identities we need to know

My Last Reflection...

This week we just worked on our study guides and review for our exam we learned how to use the identities to simplify equations, graph trig functions, and solve trig functions all by using identities for them.

The most common used Identities

Pythagorean Relationships:
sin^2 x + cos^2 x = 1
1 + tan^2 x = sec^2 x
1 + cot^2 x = csc^2 x

Reciprocal Relationships:
csc x = 1/sin x
sec x = 1/cos x
cot x = 1/tan x

Example:
1/cos x - sin x/1 (sin x/cos x)
1/cos x - sin^2 x/cos x
1-sin^2 x/cos x
cos^2 x/cos x
= cos x

reflection #17

this week was spent mostly as an exam review, and the bridge projects... which were awesome

one of the things i retained from chapter one (after looking back at it) was finding an intersection. this is how you do it...

eg: y=2x2-8x+5

step 1. does it open up or down?

if its positive it opens up, and if its negative it opens down

step 2. how many x-intercepts does it have?

b2-4ac discriminate: positive-2x intercepts negative-no x intercept 0-1 x intercept

(-8)2 -4(2)(5)=24 so there are 2 x intercepts

step 3. find the intercepts

complete the square to find

2x2-8x+5

2x2-8x =-5

x2-4x+4=-5/2+4

(x-2)2=3/2 [shortcut]

x-2= [square root of] 3/2

x-2= [square root] 3/ [square root] 2 X [square root] 2/ [square root]2

x=2+/- [square root] 6/2

put in point form

(2+ [square root]6/2, 0) (2- [square root] 6/2, 0)

step 4. find the y-intercept

plug in y

2(0)2-8(0)+5

=5

(0,5)

step 5. find the axis of symmetry

x=-b/2a

=-(-8)/2(2)

=2

step 6. find the vertex

x=2

2(2)2-8(2)+5

=-3

vertex= (2, -3)

reflection 17

this week we learned identities. there are many different identities you can use to make the problem look like its smaller. but you always have to remember to do the identities first and not the algebra. the algebra comes second. here are some examples on identities:.....

1. simplify:

(1-sinx)(1+sinx)
1-sinx+sinx-sinx^2
=1-sin^2
= cos^2x s^2+c^2=1
c^2=1-s^2


2. prove:

cotA(1+tan^2A)/tanA=csc^2A

cotA(sec^2A)/tanA

=1/tan9sec^2A)/tanA

=(sec^2A/tanA)/(tanA/1) = sec^2A/tan^2A

=(1/cos^2A)/(sin^2A/cos^2A) = cos^2A/ cos^2Asin^2A = 1/sin^2A


= csc^2A


my biggest problem with these identities is that i can never remember the identities and i never know what to do with the algebra part. it all looks so easy when we do it in class, but when im on my on doing homework or the packets im clueless and i get so confused.

Reflection

This reflection is a recap of Identities.Here is a list of the identities.

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

remember you must always start with an identity. DO NOT START WITH ALGEBRA.

Example:
Simplify: sec x - sin x tan x

1/cos x - sin x/1 (sin x/cos x)
1/cos x - sin^2 x/cos x
1-sin^2 x/cos x
cos^2 x/cos x
= cos x


Question. What is the formula for the sum and product to the roots of an equation?

Reflection 17.

The thing we learned this week was identities. In identities you pretty much just keep canceling things out. The thing you have to remember in identities is first you find an identity then you do some type of math(factoring, sandwich, ect.) to make the problem smaller. In this you have to memorize all of the identites it's the only way you will be able to work the problems.

These are the identites you will use a lot:

Pythagorean Relationships:
sin^2 x + cos^2 x = 1
1 + tan^2 x = sec^2 x
1 + cot^2 x = csc^2 x

Reciprocal Relationships:
csc x = 1/sin x
sec x = 1/cos x
cot x = 1/tan x

Examples**

simplify: tanA x cosA
tan=sin/cos now its sinA/cosA x cosA and that equals sinAcosA/cosA then your cos cancels, your answer is sinA.

here is another one: (1-sinA)(1+sinA)
foil it out with algebra because there is no identities for that
1 - sinA + sinA - sin^2A=1-sin^2A
Now you have to use a identitiy: sin^2 + cos^2 = 1.
solve for 1-sin^2 and you get cos^2 so the answer is really cos^2A.

Identites can be really confusing but you have to take it slow and do one step at a time. And remember that you always find an identity before trying to work it with math!
-----------------------------------

I kinda forgot all the stuff about conics. Like hyperbolas and cirles and ellipses. If someone could explain all that to me again that would be nice :)

Reflection 17

The law of sines states that in any triangle, the ratio between each angle and the side opposite of that angle is the same for all angles and opposite sides.

Look at figure one. The ratio between side a and angle a1 is the same as the ratio between side b and angle b1. Likewise the ratio between side b and angle b1 is the same as the ratio between side c and angle c1.




Another words:

(length of side a)/(sin a1) = (length of side b)/(sin b1) = (length of side c)/(sin c1)

In order to figure out the sin of a1, we need to divide the triangle into two right triangles as shown in figure two.

sin a1 = opposite side/hypotenuse = side h/(side b)

Likewise

sin b1 = opposite side/hypotenuse = side h/(side a)

Problem:

In figure one, side b is 2 inches long, side a is 3 inches long and angle a1 is 35. What is angle b1?

Solution:

According to the law of sines

side a/sin a1 = side b/sin b1

Multiply both sides of the equation by sine b1

sin b1 * side a/sin a1 = side b

Multiply both sides of the equation by sine a1

sin b1 * side a = side b * sin a1

Divide both sides of the equation by side a

sin b1 = side b * sin a1/side a

In order to solve for sin a1, we have to solve for the length of side h in figure two first.

We have the length of b and the angle a1. Side b, the angle a1 and side h form a right triangle with side b being the hypotenuse.

In this right triangle, side h is opposite angle a1.

sin a1 = side h/side b

multiply both sides of the equation by side b

sin a1 * side b = side h

We know the angle a1 and we know the length of side b. Therefore we can solve for the length of side h.

sin 35o * 2 = side h

sin 35o =0.574

0.574 * 2 = 1.148.

Side h is 1.148 inches long.

Now we have the following data:

side h = 1.148 inches

side b = 2

side a = 3

angle a1 = 35o

We can plug this data into the law of sines and solve for b1.

sin b1/side b = sin a1/side a

sin b1 = side b * sin a1/side a

sin b1 = 2 * 0.574/3 = 0.383

sin b1 = 0.383

arcsin 0.383 = b1

b1 = 22.5o

Problem 2:

In figure two, side h = 1.148 inches, side a = 3 inches and side b = 2 inches. What is the angle between side a and side b?

Solution:

The angle between side a and side b is angle c1 in figure one.

this week we learned how to use the identities to help simplify equations, graph trig functions, and solve trig functions using identities. this week i understood how to graph trig functions the most.
here is examples:
y=3cos(piex-2)+1
1. amp=3
2.perieod=2pie/pie=2 p/4=2/4=1/2
1.0
2.1/2
3.1
4.3/2
5.2
3. phase shift=2
1. 0+2=2
2.1/2+2=5/2
3.1+2=3
4.3/2+2=7/2
5.2+2=4
y=3cos(piex-2)+1
***then you graph your equation with your amp., phase shift, and period.
4. vertical shift=1
***then after you find your vertical shift you move your graph that many spaces, in this case you move the graph one up.

what i didn't understand this week was how you simplify equations by using the identities. i don't understand how you use the identities and how you solve them.

Reflection #17

Okay, this week went by really fast. I guess it is because we have christmas break coming up and exams next week, but i don't know. We did not really learn anything new. We are just working on study guides for review for our exam. My blog today is going to be on review for the exam if anyone needs to refresh their memory. I'm going to give examples of how to find the zeros of an equation, the trig chart, and some identities to help you on the exam.
_________________________________________________________________________

Identities:


Reciprocal Relationships

csc x = 1/sin x
sec x = 1/cos x
cot x = 1/tan x

Relationships with Negatives


sin (-x) = -sin xcos (-x) = cos x
csc (-x) = -csc xsec (-x) = sec x
tan (-x) = -tan xcot (-x) = -cot x

Pythagorean Relationships

sin^2 x + cos^2 x = 1
1 + tan^2 x = sec^2 x
1 + cot^2 x = csc^2 x
_____________________________________________________________________

Triangle Formulas:


SOHCAHTOA
sin = opposite leg/ hypotenuse
cos = adjacent leg / hypotenuse
tan = opposite leg / adjacent leg


area of a right triangle = 1/2 (b)(h)
area of an isoceles triangle = bh
area of a non-right triangle = 1/2 (leg) (leg) (sin (angle between those 2 legs)

_____________________________________________________________________

Six trig functions:
sin: y/r
cos: x/r
tan: y/x
cot: x/y
csc: r/y
sec: r/x

Unit Circle:
90(pi/2): (0,1)
180(pi): (-1,0)
270(3pi/2): (0,-1)
360(2pi):(1,0)
_____________________________________________________________________

Now the only thing that i did not understand through chapters 1-9, are the identities. I know what they are, but i just do not know how to put them in context with the problem. Can someone please give me an example of one? And i also do not get how to apply the trig chart sometimes. It gets a little confusing, but hopefully i can study and pass my exam on wednesday. I need example problems.