Saturday, December 5, 2009
help on ch. 8 test?
I'm having trouble on the chapter 8 test, number 10 a. If anyone can help me out before the end of the weekend that would be great!! thanks (:
REFLECTION #16
Finally it's the weekend! So I thought this week wasn't too bad at all. It went by at a good pace, not too boring and not too stressful. Although I could have used a few extra days of review before that test we have to take on Monday. Well I'm sure studying all night will do me some good. Anyway starting off this week on Monday we reviewed what we learned before the holidays. We continued doing a bunch of problems with trig inverses in them and I'm glad to say that those are super easy for me now! I don't even have to think too much about how to do them anymore. Also on Monday we started learning the trig identities, (which reminds me, I still need to memorize those) And we worked a bunch of problems like that where you have to use the identities and then use algebra to solve the equation. Then Tuesday and Wednesday we continued working on trig problems. And Thursday and Friday we reviewed over everything for our test on Monday. (which I am not ready for)
So what I thought was the easiest thing we went over this week is trig inverses. I know I know they're so easy. Everyone should be able to do them by now. But something else I understand very well now is solving trig equations. It's pretty simple to me because it helps when I think of the equations as quadratics.
Here's an example:
Ex. 1.) 2tan^2theta = 3tantheta - 1
*Okay the first thing you want to do with this problem is move everything over to the left side of the equal sign. (so you subtract 3tantheta over and add 1 over. then set it equal to zero)
So you get >> 2tan^2theta - 3tantheta + 1 = 0
now you can look at it as if it's a quadratic like this to make it easier > 2x^2 - 3x + 1
Then you can factor the equation to get >> (2tan^2theta - 2tantheta) - (tantheta + 1) = 0
then from there you get >> 2tantheta (tantheta - 1)-1(tantheta - 1) = 0
(2tantheta - 1) (tantheta - 1) = 0
^^then solve whats in the parenthesis for theta
*so for the first one you get theta=tan^-1(1/2) and once you find the inverse of that the two angles are 26.565 degrees and 206.565 degrees
*then for the second equation after you set it equal to zero you get that theta=tan^-1(1) and once you find the inverse of that, the two angles you get are 45 degrees and 225 degrees.
*so your final answer is theta=26.565, 206.565, 45, 225 (all are in degrees)
**Now for what I didn't quite understand this week. Well I'm still having a little trouble with the graphs, but not drawing them. I'm just having trouble figuring out an equation when you are given a graph. If anyone can help me with that it would be greatly appreciated (:
So what I thought was the easiest thing we went over this week is trig inverses. I know I know they're so easy. Everyone should be able to do them by now. But something else I understand very well now is solving trig equations. It's pretty simple to me because it helps when I think of the equations as quadratics.
Here's an example:
Ex. 1.) 2tan^2theta = 3tantheta - 1
*Okay the first thing you want to do with this problem is move everything over to the left side of the equal sign. (so you subtract 3tantheta over and add 1 over. then set it equal to zero)
So you get >> 2tan^2theta - 3tantheta + 1 = 0
now you can look at it as if it's a quadratic like this to make it easier > 2x^2 - 3x + 1
Then you can factor the equation to get >> (2tan^2theta - 2tantheta) - (tantheta + 1) = 0
then from there you get >> 2tantheta (tantheta - 1)-1(tantheta - 1) = 0
(2tantheta - 1) (tantheta - 1) = 0
^^then solve whats in the parenthesis for theta
*so for the first one you get theta=tan^-1(1/2) and once you find the inverse of that the two angles are 26.565 degrees and 206.565 degrees
*then for the second equation after you set it equal to zero you get that theta=tan^-1(1) and once you find the inverse of that, the two angles you get are 45 degrees and 225 degrees.
*so your final answer is theta=26.565, 206.565, 45, 225 (all are in degrees)
**Now for what I didn't quite understand this week. Well I'm still having a little trouble with the graphs, but not drawing them. I'm just having trouble figuring out an equation when you are given a graph. If anyone can help me with that it would be greatly appreciated (:
Reflection 16
This week we learn about Idenities and Equations (Relationships among the Functions)
In this section, we investigated some of the relationshops among the trigonometric functions by using the recoprocals, Pythagorean, and Cofunction of that specific function.
Steps for solving
1. check Idenities
2.Use Algebra: combining, factoring, multiplacation, fraction, and sandwiching.
Simplify
SecX-SinX TanX
use and identity for secX and tanX
1/cosX-sinX(sinX/cosX)
the distribute sinx into ( sinx/cosx)
1-sin^2X/cosX
use the iden. and solve for cos^2X with 1-sin^2X
Cos^2 X/ CosX= Cosx
In this section, we investigated some of the relationshops among the trigonometric functions by using the recoprocals, Pythagorean, and Cofunction of that specific function.
Steps for solving
1. check Idenities
2.Use Algebra: combining, factoring, multiplacation, fraction, and sandwiching.
Simplify
SecX-SinX TanX
use and identity for secX and tanX
1/cosX-sinX(sinX/cosX)
the distribute sinx into ( sinx/cosx)
1-sin^2X/cosX
use the iden. and solve for cos^2X with 1-sin^2X
Cos^2 X/ CosX= Cosx
Wednesday, December 2, 2009
ignore these week 11/9
law of cosine, i kind of get this too. the law of cosine formula:
(opp. leg)^2= (adj leg)^2+ (other leg)^2 - 2(adj leg)(adj leg)cos(adj b/w)
long and scarey but not that bad after all....
lets work one: **key: % = alpha
hypotenuse= 5 base= 6 angle b/t= 36degrees oppoiste leg= X
plug it in:
X^2= 6^2 + 5^2 - 2(5)(6)cos %
square root to make X stand alone:
X= square root of (6^2 + 5^2 - 2(5)(6)cos %)
then work out the addition, subtraction, and multiplication.
X approx. = 3.530
this one has an angle in it.
isos. triangel. one side= 5 other side= 6 angle between= % base=7
after you work it out you have to solve for cos % which = 7^2 - 6^2 - 5^2/-2 (6) (5)
then you have the inverse of % and it equals approx. 78.463 degrees.
(opp. leg)^2= (adj leg)^2+ (other leg)^2 - 2(adj leg)(adj leg)cos(adj b/w)
long and scarey but not that bad after all....
lets work one: **key: % = alpha
hypotenuse= 5 base= 6 angle b/t= 36degrees oppoiste leg= X
plug it in:
X^2= 6^2 + 5^2 - 2(5)(6)cos %
square root to make X stand alone:
X= square root of (6^2 + 5^2 - 2(5)(6)cos %)
then work out the addition, subtraction, and multiplication.
X approx. = 3.530
this one has an angle in it.
isos. triangel. one side= 5 other side= 6 angle between= % base=7
after you work it out you have to solve for cos % which = 7^2 - 6^2 - 5^2/-2 (6) (5)
then you have the inverse of % and it equals approx. 78.463 degrees.
ignore these reflection week of 11/2
SOH CAH TOA!!!! and i get this finally i understand something soo well in advanced math! okay, im done. this is only used for right triangles. **key 0=theta
sin 0= opp. leg/ hypotenuse
cos 0= adj. leg/ hypotenuse
tan 0= opp. leg/ adj. leg
when given a right triangle with two sides equaling 553 meter and 100 meters.. and you are trying to find angle A it is adjacent to the right angle... refer back to SOH CAH TOA. visualize it: angle a is opposite a side and adjacent a side so you would use tangent.
tanA= 553/100
A=tan inverse (553/100) Remember you are trying to find an angle therefore you
A= 79.750 degrees always use the inverse of tan, sin, or cos.
now when given a triangle with a side and an angle you need to find both sides x and y.
right triangle. hypotenuse = 20ft angle A= 70 degrees
look at the side opp. the hypotenuse.
side y. we know that opposite is 20 ft and adj is x so plug it in to the formula
cos 70=x/20
x=20cos70 approx. = 6.840/ft
side x. we know that opp. side is y and hypotenuse is 20 ft. the angle is 70 degrees so lets plug it in...
sin 70= y/20
y= 20sin70 approx. = 18.79 ft
sin 0= opp. leg/ hypotenuse
cos 0= adj. leg/ hypotenuse
tan 0= opp. leg/ adj. leg
when given a right triangle with two sides equaling 553 meter and 100 meters.. and you are trying to find angle A it is adjacent to the right angle... refer back to SOH CAH TOA. visualize it: angle a is opposite a side and adjacent a side so you would use tangent.
tanA= 553/100
A=tan inverse (553/100) Remember you are trying to find an angle therefore you
A= 79.750 degrees always use the inverse of tan, sin, or cos.
now when given a triangle with a side and an angle you need to find both sides x and y.
right triangle. hypotenuse = 20ft angle A= 70 degrees
look at the side opp. the hypotenuse.
side y. we know that opposite is 20 ft and adj is x so plug it in to the formula
cos 70=x/20
x=20cos70 approx. = 6.840/ft
side x. we know that opp. side is y and hypotenuse is 20 ft. the angle is 70 degrees so lets plug it in...
sin 70= y/20
y= 20sin70 approx. = 18.79 ft
ignore these reflection week 10-21
ignore these they are the blogs a i missed.....
week of october 21
A sector of a circle has a arc length of 6cm and area of 75 cm squared. Find It s radius and the measure of its central angle.
follow the formulas ***key 0= theta
1. S=r0
2. K= 1/2r^20
K=1/2rs
you know that S= 6 K= 75cm2 and you need to find R and 0.
plug in what you know....
K=1/2r(6)
75=3(r)
r= 25 cm
S=RO
6cm=25theta
theta=6/25
all this was was plugging into formulas and being able to identify in a word problem.
week of october 21
A sector of a circle has a arc length of 6cm and area of 75 cm squared. Find It s radius and the measure of its central angle.
follow the formulas ***key 0= theta
1. S=r0
2. K= 1/2r^20
K=1/2rs
you know that S= 6 K= 75cm2 and you need to find R and 0.
plug in what you know....
K=1/2r(6)
75=3(r)
r= 25 cm
S=RO
6cm=25theta
theta=6/25
all this was was plugging into formulas and being able to identify in a word problem.
reflection
im posting my reflection now sense i get what we did in class yesterday and its still fresh in my memory.
solving difficult trigonometric fuctions isnt really that difficult after all, you just have to remember when you look at a problem think of it as algebra and if thats too hard replace the sin cos tan etc with x's... here's an example
solve.
solving difficult trigonometric fuctions isnt really that difficult after all, you just have to remember when you look at a problem think of it as algebra and if thats too hard replace the sin cos tan etc with x's... here's an example
solve.
sinXtanX=3sinX for 0 < or =" X" or =" 2">
now look at it, and think of Algebra.
first you set it equal to zero:
sinXtanX-3sinX=0
then you factor out what you can:
sinX(tanX-3)=0
then set each one to zero and work them out:
sinX=0
X=sin inverse (0)
X= 0, pi, 2pi
*refer to unit circle for the answers locations.
tanX-3=0
tanX=3
X= tan inverse (3)
X= 71.565/180 pi, 251.565/180 pi
* refer back to grid for the answers locations.
now what i could use a little help on:
identites and equations, i guess when i look at the problems and see all the sin's and cos's i start to freak out, is there anyway for me to look at it and keep my cool?
Monday, November 30, 2009
Reflection #15
Okay, so I didn't know we had to do two blogs over the holidays, so here's my second one. I didn't really like learning the curves of sine and cosines 2 weeks ago, so I'll try to explain something different...I think we're allowed to describe any old concept we learned throughout the year, so...
I'm going to describe how to solve for the angles in Chapter 8 using trigonometrical inverses. It's kind of like a review of a previous chapter we learned in the first nine weeks.
Example:
Find the values of x between 0 and 2pi for which sin x = .6
Therefore
x= sin^-1 (.6)
Now you have to determine which quadrants it is in.
Plug it in your calculator and you get the value of 36.870 degrees in the first quadrant. Make it negative and add 180 degrees and you have 143.130 degrees in the second quadrant.
Now since it is between 0 and 2pi, the answers have to be in radians.
Just convert the answers to radians.
36.87 times pi/180 = 1229/6000 pi
143.13 times pi/180 = 4771/6000 pi
Those are the two answers.
Now the only thing I'm really struggling with still is the curves of sine and cosine. I guess I'll still have to practice some more with the homework and do more of it. Some practice problems would be great. Thanks :)
Reflection #15
So, for this blog, I'm going to review conversions:
To convert degrees to radians: multiply by pi/180
ex. 120 degrees
120 times pi/180
(keep answer in pi form)
120 divided by 180
=2/3
your answer =2/3 pi or 2 pi/3
To convert radians to degrees: multiply by 180/pi
ex. 3 pi/4 or 3/4 pi
3 pi/4 times 180/pi
pi cancels out
3/4 times 180
3 times 180
=540
divided by 4
your answer =135 degrees
119.2
37.92
and so forth.
119.2/180 (plug in cal.) 149/225
then multiply pi =149 pi/225
To convert degrees decimals to minutes and seconds:
take the decimal of the degrees and multiply by 60 for minutes
68.33 .33 times 60 =19.8
take the decimal of the minutes and multiply by 60 for seconds
19.8 .8 times 60 =48
your answer =68 degrees 19 minutes 48 seconds
*if seconds have decimal, round to the 3rd decimal place
*if minutes don't have decimals, leave in degrees minutes
To convert degrees minutes seconds to decimals:
68 degrees 19 minutes 48 seconds
divide seconds (48) by 3600 =.013
divide minutes (19) by 60 =.317
add the two decimals =.33
your answer is 68.33
I think the only thing I didn't understand was the graphs we did right before the holidays. Care to explain anyone?
Sunday, November 29, 2009
Reflection
Well, i am soo glad we had this week off because i'm officially tired of school. My group worked on our bridge and finally got it finished, that project took longer than i thought it would have so thankfully it's finished! But since we didnt have school, we didnt learn anything new. I'll just give some examples of what i thought was the easiest thing we learned this year: logs.
EXAMPLES:
Condensing.
1.)logm + log7 + 4logn
= log7mn^4
2.)5loga + logd + log6
= log6da^5
3.)logn - 3logh -logy
= n/yh^3
4.)4logt - logc
= t^4/c
Expanding.
1.)log5gh^2
= log5 + 2logh +logg
2.)m^3b^7/f
= 3logm + 7logb - logf
Well i hope everyone had a great Thanksgiving! :)
EXAMPLES:
Condensing.
1.)logm + log7 + 4logn
= log7mn^4
2.)5loga + logd + log6
= log6da^5
3.)logn - 3logh -logy
= n/yh^3
4.)4logt - logc
= t^4/c
Expanding.
1.)log5gh^2
= log5 + 2logh +logg
2.)m^3b^7/f
= 3logm + 7logb - logf
Well i hope everyone had a great Thanksgiving! :)
ReFlection 15
Happy thanksgiving!!!!!!! Ive been enjoyin my week with just hanging out, deer hunting, asnd chilling at home. And the bridge is coming out just fine. Now i'm off to do sum comments
reflection 15
this is my thanksgiving blog. i hope everyone had a nice thanksgiving and that they got to spend it with their family and friends. my week of was really relaxing but im ready to go back to school to see everyone and get these 3 weeks over with so it can be christmas break:) my groups bridge project is coming along, we got the bottom part down and the river and everything but i bought the wrong craft sticks..sooo my mom has to go buy some more tomorrow. i don't really know what else to say because we were off this week and didn't learn anything.
reflection 14
so i had no idea that i had to do a super awesome blog. thats cool but ummmm happy late thanksgiving to you all. hope you had a nice one. but anyways heres some mathematical goodness.
you can find the area of triangles by using a cool formula.
the formula is
1/2(leg)(leg)sin(angle b/w)
so if you plug in some cool numbers such as
1/2(2)(5)sin(48)
you get 3.346
so thats your area of that triangle. woot woot. you know math and such.
we learned alot more stuff which im sorry to say i dont remember AT ALL at the moment. so if anyone wants to catch me up again i would really appreciate it:)
love, Chad.
relfection 15
This is my thanksgiving week blog, and i have no idea what to say, except i'll tell everyone happy thanksgiving again! haha. Good luck to the basketball teams whenever yall play, cuz yall'd probly pwn anyway! haha
SOHCAHTOA:
sin=opp/hyp
cos=adj/hyp
tan=opp/adj
never really understood that until this year!
but then again our gemoetry teachers were never really totally understandable, so thanks b-rob.
SOHCAHTOA:
sin=opp/hyp
cos=adj/hyp
tan=opp/adj
never really understood that until this year!
but then again our gemoetry teachers were never really totally understandable, so thanks b-rob.
reflection 14
ok so, i didnt realize that we were supposed to do two blogs, but i guess it gets us points right!
So i want to start off by wishing everyone a late but much needed HAPPY THANKSGIVING!! hope yall all got really fat! lol! j/k! And now, its baseball season, for me at least, but i wish it wasnt over yet, ya know for the seniors! :(
and now...for the math!!
line formula:
1) m=tan (alpha or fish)
m=slope
alpha or fish=angle of inclination
which is almost all i remember from last week! haha.
anyone wanna help w/ everything else???
So i want to start off by wishing everyone a late but much needed HAPPY THANKSGIVING!! hope yall all got really fat! lol! j/k! And now, its baseball season, for me at least, but i wish it wasnt over yet, ya know for the seniors! :(
and now...for the math!!
line formula:
1) m=tan (alpha or fish)
m=slope
alpha or fish=angle of inclination
which is almost all i remember from last week! haha.
anyone wanna help w/ everything else???
Reflection 15
Let me just start by saying I DO NOT want to go back to school at alll! I don't feel like waking up bookoo early and I don't feel like making my brain work..what can I say I'm just lazy. haha, but my holidays were really good and I hope everyones were good also. I didn't bring home my binder because I forgot we had to do blogs so thisis just something I remember off the top of my head. *Using SOHCAHTOA to solve right triangles
In trianlge ABC, C =90, A=28, B isn't given, and a=40. find b & c.draw your triangle and label it with the information from above.
**remember SOHCAHTOA
Pick an angle to help you solve the equation.What I do is figure out which side I want to find first. I'll find b first we know the adj. is 40 so in SOHCAHTOA what invovles opposite and adj.? It's tan. So you take the TAN of your angle and set it equal to the side your tying to solve for over what side you already have.
tan 28=40/b
b tan 28=40
b=40/tan 28
b=75.229
You would find the last side by doing the same thing. You would use SIN because it's opposite over hypot.Take SIN of your angle and set it equal to the side your tying to solve for over what side you already have.
sin 28=40/c
c sin 28=40
c=40/sin 28
c=85.202
------------------------------------
Since I didn't bring my book home I can't say I don't remember anything. But I know I understood pretty much everything we learned the week before school let out.
In trianlge ABC, C =90, A=28, B isn't given, and a=40. find b & c.draw your triangle and label it with the information from above.
**remember SOHCAHTOA
Pick an angle to help you solve the equation.What I do is figure out which side I want to find first. I'll find b first we know the adj. is 40 so in SOHCAHTOA what invovles opposite and adj.? It's tan. So you take the TAN of your angle and set it equal to the side your tying to solve for over what side you already have.
tan 28=40/b
b tan 28=40
b=40/tan 28
b=75.229
You would find the last side by doing the same thing. You would use SIN because it's opposite over hypot.Take SIN of your angle and set it equal to the side your tying to solve for over what side you already have.
sin 28=40/c
c sin 28=40
c=40/sin 28
c=85.202
------------------------------------
Since I didn't bring my book home I can't say I don't remember anything. But I know I understood pretty much everything we learned the week before school let out.
Reflection 14
The week before we left for thanksgiving I pretty much understood everything we learned really well because it was just a review of everything and it was similar to stuff we learned in previous chapters. One thing that was new though was finding the angle of inclination. It's similar to trig inverses. Once you know which fromula you use you do the inverse and you should get two answers.
for lines:m=tan where m=slope, alpha=angle of inclination
for conics:tan 2(alpha)=B/A-C
also for conics, if A=C then alpha=pi/4
example:find the angle of inclination...
2x+5y=15
m=-2/5 tan=-2/5 alpha=tan^-1(-2/5)
----------------------------------------
I pretty much understood everything...wooooo :)
for lines:m=tan where m=slope, alpha=angle of inclination
for conics:tan 2(alpha)=B/A-C
also for conics, if A=C then alpha=pi/4
example:find the angle of inclination...
2x+5y=15
m=-2/5 tan=-2/5 alpha=tan^-1(-2/5)
----------------------------------------
I pretty much understood everything...wooooo :)
REFLECTION #15
Why did this week have to go by soooo fast!!!??! I'm not ready to go back to school! But then again I think it's time for me to go back because I completely forgot what we learned in this class before we left for the holidays. ha my bad. So before I get into any details that are actually about math, I'd like to take a moment to say how BORING my holidays were. Let's see, what did I do? Hmm..I drove two hours a day for drivers e.d. which was sooo fun, as you can imagine. I went see a movie, and no it was not new moon. I cheered at the football game Friday, we lost :/ Annnnddddd I read a book. Fascinating I know. So now that I pretty much wasted my time saying all of this nonsense let me get to some advanced math stuff. Well let's see, I think I'll just reflect on some random things we learned all year that I can remember really good. I don't have my binder or anything to use for guidance, so when I say random I mean verrryyy random.
So the first thing that pops into my head is something that we learned not too long ago. RIGHT TRIANGLES!! yay. So here's an example of how to solve a right triangle. Since I'm not too creative at making up word problems, I'll just make up some numbers and go from there. sounds good to me.
Ex. 1.) Triangle ABC. Angle C is 36 degrees. Side b is 7. Find sides a and c. Then find the triangle's area. (*keep in mind this is a right triangle)
*Since this is a right triangle you know you can use SOHCAHTOA to solve it. You are asked to find sides a and c. Let's find side c first. Once you draw the picture you know that you will be using tangent as your trig function. (opposite over adjacent)
so you set up the equation >> tan 36 = c/7 (*multiply 7 to both sides)
side c is approximately 5.086
*Next you can find side a. You can use sine to do this.
Set up the equation >> sin36 = 5.086/a (*multiply a to both sides then divide by sin 36)
then you get that side a is approximately 8.653
*Now to find the area you just use the simple formula A = 1/2bh (*used only for right triangles)
Sooo... A = 1/2(5.086)(7)
Area is approximately 17.801
*Another thing I remember is how to find reference angles.
Heres an example:
Ex. 2.) Find the reference angle of sin 245.
*The first thing you have to do is figure out which quadrant it is located in. It is located in the 3rd quadrant. Next you notice that it is sine and sine relates to the y axis. The y axis is negative in the 3rd quadrant therefore sin will be negative. (-sin) Then to find the reference angle you subtract 180 from 245 and you get 65.
so your reference angle is -sin 65
okay I'm done
So the first thing that pops into my head is something that we learned not too long ago. RIGHT TRIANGLES!! yay. So here's an example of how to solve a right triangle. Since I'm not too creative at making up word problems, I'll just make up some numbers and go from there. sounds good to me.
Ex. 1.) Triangle ABC. Angle C is 36 degrees. Side b is 7. Find sides a and c. Then find the triangle's area. (*keep in mind this is a right triangle)
*Since this is a right triangle you know you can use SOHCAHTOA to solve it. You are asked to find sides a and c. Let's find side c first. Once you draw the picture you know that you will be using tangent as your trig function. (opposite over adjacent)
so you set up the equation >> tan 36 = c/7 (*multiply 7 to both sides)
side c is approximately 5.086
*Next you can find side a. You can use sine to do this.
Set up the equation >> sin36 = 5.086/a (*multiply a to both sides then divide by sin 36)
then you get that side a is approximately 8.653
*Now to find the area you just use the simple formula A = 1/2bh (*used only for right triangles)
Sooo... A = 1/2(5.086)(7)
Area is approximately 17.801
*Another thing I remember is how to find reference angles.
Heres an example:
Ex. 2.) Find the reference angle of sin 245.
*The first thing you have to do is figure out which quadrant it is located in. It is located in the 3rd quadrant. Next you notice that it is sine and sine relates to the y axis. The y axis is negative in the 3rd quadrant therefore sin will be negative. (-sin) Then to find the reference angle you subtract 180 from 245 and you get 65.
so your reference angle is -sin 65
okay I'm done
Reflection 15
Since we didn't have school this week, I am going to show some stuff we learned in Chapter 7 along with the chart and unit circle that we should not forget
In 7-1 to convert degrees to radians you have to multipy the degrees by pi/180. To convert radians to degrees you take the radian and multiply by 180/pi.
Ex: 225 degrees x pi/180 = 5/4 pi
3pi/4 x 180/pi=135 degrees
In 7-3 you have to know the unit circle and chart.
The Chart:
Sin=y/r Csc=r/y
Cos=x/r Sec=r/x
tan=y/x Cot=x/y
Ex:
Find all 6 trig functions of (-3,4).
Sin=4/5 Csc=5/4
Cos=-3/5 Sec=-5/3
tan=-4/3 Cot=-3/4
The unit circle:
90 degrees= pi/2 and (0,1)
180 degrees=pi and (-1,0)
270 degrees=3pi/2 and (0,-1)
360 degrees= 2pi and (1,0)
What is our test I think for Tuesday going to be on? What charts and formulas we have to know for it.
In 7-1 to convert degrees to radians you have to multipy the degrees by pi/180. To convert radians to degrees you take the radian and multiply by 180/pi.
Ex: 225 degrees x pi/180 = 5/4 pi
3pi/4 x 180/pi=135 degrees
In 7-3 you have to know the unit circle and chart.
The Chart:
Sin=y/r Csc=r/y
Cos=x/r Sec=r/x
tan=y/x Cot=x/y
Ex:
Find all 6 trig functions of (-3,4).
Sin=4/5 Csc=5/4
Cos=-3/5 Sec=-5/3
tan=-4/3 Cot=-3/4
The unit circle:
90 degrees= pi/2 and (0,1)
180 degrees=pi and (-1,0)
270 degrees=3pi/2 and (0,-1)
360 degrees= 2pi and (1,0)
What is our test I think for Tuesday going to be on? What charts and formulas we have to know for it.
Reflection 14
This is some of the area stuff we just learned the past week before the Holidays.
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.
The law of sines and cosines still confuse me, I had trouble on the test with those.
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.
The law of sines and cosines still confuse me, I had trouble on the test with those.
Reflection 15
Even though we did not have school this week you are to reflect back on an older concept. This is in preparation for your mid-term. As we talked about in class, pick anything from the year and write your reflection based on that.
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