Since we started exam review, we're doing ch. 1-6 (and 13)...
I'm going to review the rational root theorem, one of the things I've remembered:
ex. x^4+2x^3-7x^2-8x+12
factors of p(constant)=+/-12, +/-1, +/-6, +/-2, +/-3, +/-4
factors of q(leading coeff.)=+/-1
Find p/q=+/-12, +/-1, +/-6, +/-2, +/-3, +/-4
Plug all possibilities into table on calculator and find all where y is 0.
You should get 1, 2, -2, and -3.
Do cythetic division with any two.
1 1 2 -7 -8 12
1 3 -4 -12
1 3 -4 -12 0
(x^3+3x^2-4x-12)(x-1)
2 1 3 -4 -12
2 10 12
1 5 6 0
(x^2+5x+6)(x-1)(x-2)
Solve by factoring x^2+5x+6=
(x+2)(x+3)
x=-2, -3, 1, 2
Put in point form:
(-2, 0)(-3, 0)(1, 0)(2, 0)
It's pretty simple stuff, but it's also easy to forget.
Saturday, May 8, 2010
Sunday, May 2, 2010
Reflection 5/2
Ah, the trig exam tomorrow and tuesday. Hopefully I'll do good and I've been studying a lot. Anyways, I'm going to explain some of the stuff from chapter 11 on polar form and rectangular forms.
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.
z= sqrt of 5 cis 63.435
z= sqrt of 5 cos 63.435 + sqrt of 5 sin 63.435 i
z= negative sqrt of 5 cis 243.435
z- negative sqrt of 5 cos 243.435 + negative sqrt of 5 sin 243.435 i
It is now in polar form.
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.
z= sqrt of 5 cis 63.435
z= sqrt of 5 cos 63.435 + sqrt of 5 sin 63.435 i
z= negative sqrt of 5 cis 243.435
z- negative sqrt of 5 cos 243.435 + negative sqrt of 5 sin 243.435 i
It is now in polar form.
reflection
Trig test tomorrow..... aaaaaaaah. Hopefully I do good, anyways I'll review identites that we learned in chapter seven. The main thing to identities is memorizing all the formulas, if you don't then you wont be able to work the problem. To solve or simplify an identity you just replace what the have with something in the formulas.**Keep in mind you are trying to make the problem smaller not larger.
1. simplify:
(1-sinx)(1+sinx)
*Multiply the 2
1-sinx+sinx-sinx^2
=1-sin^2
*Use a formula
= cos^2x s^2+c^2=1
c^2=1-s^2
The answer would be 1.
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
**For this problem you are just trying to get the same thing that you started off the problem with to show that it is correct. Solve the problem like you normally would and the answer should be what it equals up to.
1. simplify:
(1-sinx)(1+sinx)
*Multiply the 2
1-sinx+sinx-sinx^2
=1-sin^2
*Use a formula
= cos^2x s^2+c^2=1
c^2=1-s^2
The answer would be 1.
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
**For this problem you are just trying to get the same thing that you started off the problem with to show that it is correct. Solve the problem like you normally would and the answer should be what it equals up to.
Last Reflection
Now thinking way back to logs. Because this will probably help me since im not to great with them.
logb^x=a
all you would do is switch the x and the a. (Kinda like an inverse)
b^a=x <---This is exponential form.
log3^2=7
3^7=2
This helps you solve problems sometimes.
But before you solve you have to change it to exponential form.
Example:
logx^4=2
x^2=4
x= +/- 2
For natural logs-
e^x=75
change e to ln(natural log), then switch the x and the 75
ln75=x
Plug into your calculator.
ln 75=4.317
logb^x=a
all you would do is switch the x and the a. (Kinda like an inverse)
b^a=x <---This is exponential form.
log3^2=7
3^7=2
This helps you solve problems sometimes.
But before you solve you have to change it to exponential form.
Example:
logx^4=2
x^2=4
x= +/- 2
For natural logs-
e^x=75
change e to ln(natural log), then switch the x and the 75
ln75=x
Plug into your calculator.
ln 75=4.317
Reflection May 2
Today, i will explain Law of Sines.
ok here we go:
You are given that in triangle ABC:
B = 63 degrees, b = 5, and c = 4
Find angle C.
to solve for this you will use law of sines....
Set it up as:
sinC/4 = sin63degrees/5, because the formula is sinA/a = sinB/b.
Now cross multiply:
4sin63 = 5sinC
-divide by 5 to get sinC by itself:
sinC = 4sin63/5
-take inverse:
C = sin^-1 (4sin63/5) = 45.464 degrees
So angle C is 45.464 degrees, yay.
Now one with law of cosines:
You are given that in triangle ABC:
C = 120 degrees, b = 40.8, and a = 10.5
Find side c.
-follow the law of cosines formula, and plug in:
x^2 = 40.8^2 + 10.5^2 - 2(40.8)(10.5) cos120
-type into calculator with a square root symbol over all this, b/c its x^2
you get that c = 46.939 from your calculator.
ok here we go:
You are given that in triangle ABC:
B = 63 degrees, b = 5, and c = 4
Find angle C.
to solve for this you will use law of sines....
Set it up as:
sinC/4 = sin63degrees/5, because the formula is sinA/a = sinB/b.
Now cross multiply:
4sin63 = 5sinC
-divide by 5 to get sinC by itself:
sinC = 4sin63/5
-take inverse:
C = sin^-1 (4sin63/5) = 45.464 degrees
So angle C is 45.464 degrees, yay.
Now one with law of cosines:
You are given that in triangle ABC:
C = 120 degrees, b = 40.8, and a = 10.5
Find side c.
-follow the law of cosines formula, and plug in:
x^2 = 40.8^2 + 10.5^2 - 2(40.8)(10.5) cos120
-type into calculator with a square root symbol over all this, b/c its x^2
you get that c = 46.939 from your calculator.
reflection
Here are some identities:
Cos(α +/- β)=cos α cos β -/+ sin α sin β
sin(α +/- β)=sin α cos β -/+ cos α sin β
sin x + sin y= 2 sin x + y/2 cos x-y/2
sin x - sin y= 2 cos x + y/2 sin x-y/2
cos x + cos y= 2 cos x + y/2 cos x-y/2
cos x - cos y= 2 sin x + y/2 sin x-y/2
tan (α + β)=tan α + tan β/1-tan α tan β
tan (α - β)=tan α - tan β/1+tan α tan β
sin2α=2sin α cos α
cos 2α=cos^2 α –sin^2 α = 1-2 sin^2 α= 2 cos^2 α -1
tan 2α = 2tan α /1-tan^2 α
sin α/2= +/- √1-cos α/2
cos α/2= +/- √1+ cos α/2
tan α/2= +/- √1-cos α or 1 + cos α
=sin α/1+cos α
=1-cos α/sin α
Does anyone have any helpful hints on this test so i dont fail?!?!?!
Cos(α +/- β)=cos α cos β -/+ sin α sin β
sin(α +/- β)=sin α cos β -/+ cos α sin β
sin x + sin y= 2 sin x + y/2 cos x-y/2
sin x - sin y= 2 cos x + y/2 sin x-y/2
cos x + cos y= 2 cos x + y/2 cos x-y/2
cos x - cos y= 2 sin x + y/2 sin x-y/2
tan (α + β)=tan α + tan β/1-tan α tan β
tan (α - β)=tan α - tan β/1+tan α tan β
sin2α=2sin α cos α
cos 2α=cos^2 α –sin^2 α = 1-2 sin^2 α= 2 cos^2 α -1
tan 2α = 2tan α /1-tan^2 α
sin α/2= +/- √1-cos α/2
cos α/2= +/- √1+ cos α/2
tan α/2= +/- √1-cos α or 1 + cos α
=sin α/1+cos α
=1-cos α/sin α
Does anyone have any helpful hints on this test so i dont fail?!?!?!
reflection 5/2
This has been a super long day, ive been studying math formulas all day and frustration and aggervation is kicking in, i feel very overwelmed, but i know i was taught all of this and hat i hopefully will do good on it the next two days.im not going to sit here and just type out formulas ill learn much better by keep on studying them right now. I will have better blogs in the future, i just think its more importat to concentrate on doing my math packets and studying my formulas.
Is there any helpful hints for remembering identites better!!!!!!!!!!!!!!!!
Is there any helpful hints for remembering identites better!!!!!!!!!!!!!!!!
reflection march 2
So, this week was just like any other week. It went by pretty fast because softball left thursday to go to sulphur, and we didn't win:(..there's always next year. But anyway, we just looked over our tests and asked questions about the ones that we didn't know how to do and the ones that we are unsure about. This reflecton will be a little review over things that we learned in the past few weeks, before we started to review for the trig exam, which is tomorrow, and tuesday. That test is going to be the biggest thest i think i've ever taken. Im so scared:/...so anyway, heres a review on determinants.....
EXAMPLES:
a b c
d e f
g h i
*remember to delete the row, delete the column while alternating signs.
+,-+,-+,-
*pick row or column with the most 0's or 1's.
1. cross product - finds a pependicular vector
2. if uxv=0 then they are parallel.
3. /uxv/=/u/ /v/sin theta
u=(2,3,4) v=(1,0,5)
4. finds the area or volume, etc. of the object.
/uxv/
EXAMPLES:
a b c
d e f
g h i
*remember to delete the row, delete the column while alternating signs.
+,-+,-+,-
*pick row or column with the most 0's or 1's.
1. cross product - finds a pependicular vector
2. if uxv=0 then they are parallel.
3. /uxv/=/u/ /v/sin theta
u=(2,3,4) v=(1,0,5)
4. finds the area or volume, etc. of the object.
/uxv/
Reflection 5/2
okayyyyy, i'm super upset that i missed the field trip on friday, (haha NOT). But i am upset becasue i didn't get to study as much as i wanted to. I think i'm going to do okay on the exam though, because i understand all of the study guides from all of the previous chapters. This trig exam is going to be beast, but as long as i remember all of the identities, i should do okay, because i understand everything else that involves trig. So i've been stuck in my room all day reading over the study guides! Sounds boring, but as long as i get a good grade then it was worth it. So i'm just gonna give examples to review over before the exam to refresh myself and others.
_______________________________________________________________
Identities:
(Reciprocal relationships)
cscθ=1/sinθ
secθ=1/cosθ
cotθ=1/tanθ
(Relationships with negatives)
sin(-θ)=-sinθ
csc(-θ)=-cscθ
tan(-θ)=-tanθ
cos(-θ)=-cosθ
sec(-θ)=-secθ
cot(-θ)=-cotθ
(Pythagorean relationships)
sin^2θ+cos^2θ=1
1+tan^2θ=sec^2θ
1+cot^2θ=csc^2θ
(Cofunction relationships)
sinθ=cos(90-θ)
tanθ=cot(90-θ)
secθ=csc(90-θ)
cosθ=sin(90-θ)
cotθ=tan(90-θ)
cscθ=sec(90-θ)
EXAMPLE:
1) csc^2x(1-cos^2x)
1/sin^2x(sin^2x)=1
*both of the sin^2x cancel and you are left with 1.
2) cosθcot(90-θ)
cosθtanθ
cosθsinθ/cosθ=sinθ
*the cosθ cancel out and you are left with sinθ.
_______________________________________________________________
Formulas from chapter 10:
(Sum and difference formulas for cos & sin)
cos(a+-b)=cosacosb-+sinasinb
sin(a+-b)=sinacosb+-cosasinb
(Rewriting a sum or difference as a product)
sinx+siny=2sin x+y/2 cos x-y/2
sinx-siny=2cos x+y/2 sin x-y/2
cosx+cosy=2cos x+y/2 cos x-y/2
cosx-cosy=2sin x+y/2 sin x-y/2
(Sum formula for tangent)
tan(a+b)= tana+tanb/1-tanatanb
(Difference formula for tangent)
tan(a-b)= tana-tanb/1+tanatanb
(Double-angle and Half-angle)
sin2a=2sinacosa
cos2a=cos^2a-sin^2a...1-2sin^2a...2cos^2a-1
tan2a= 2tana/1-tan^2a
sin a/2= +-sqrt 1-cosa/2
cos a/2= +-sqrt 1+cosa/2
tan a/2= +-sqrt 1-cosa/1+cosa...sina/1+cosa...1-cosa/sina
EXAMPLE:
1) cos75cos15+sin75sin15
=cos(a+b)=cos60--> 1/2
2) sin75cos15+cos75sin15
=sin(a-b)=sin60--> sqrt3/2
_______________________________________________________________
So pretty much my whole blog i just typed is what i need to study the most. That is probably the case with everyone else too. All of the other trig stuff is easy. But if anyone else can give me examples of the hard identity ones, i would greatly appreciate it :) And if anyone knows that there is some more confusing problems on the exam that you get. I could study that too, haha. THANKSS :)
_______________________________________________________________
Identities:
(Reciprocal relationships)
cscθ=1/sinθ
secθ=1/cosθ
cotθ=1/tanθ
(Relationships with negatives)
sin(-θ)=-sinθ
csc(-θ)=-cscθ
tan(-θ)=-tanθ
cos(-θ)=-cosθ
sec(-θ)=-secθ
cot(-θ)=-cotθ
(Pythagorean relationships)
sin^2θ+cos^2θ=1
1+tan^2θ=sec^2θ
1+cot^2θ=csc^2θ
(Cofunction relationships)
sinθ=cos(90-θ)
tanθ=cot(90-θ)
secθ=csc(90-θ)
cosθ=sin(90-θ)
cotθ=tan(90-θ)
cscθ=sec(90-θ)
EXAMPLE:
1) csc^2x(1-cos^2x)
1/sin^2x(sin^2x)=1
*both of the sin^2x cancel and you are left with 1.
2) cosθcot(90-θ)
cosθtanθ
cosθsinθ/cosθ=sinθ
*the cosθ cancel out and you are left with sinθ.
_______________________________________________________________
Formulas from chapter 10:
(Sum and difference formulas for cos & sin)
cos(a+-b)=cosacosb-+sinasinb
sin(a+-b)=sinacosb+-cosasinb
(Rewriting a sum or difference as a product)
sinx+siny=2sin x+y/2 cos x-y/2
sinx-siny=2cos x+y/2 sin x-y/2
cosx+cosy=2cos x+y/2 cos x-y/2
cosx-cosy=2sin x+y/2 sin x-y/2
(Sum formula for tangent)
tan(a+b)= tana+tanb/1-tanatanb
(Difference formula for tangent)
tan(a-b)= tana-tanb/1+tanatanb
(Double-angle and Half-angle)
sin2a=2sinacosa
cos2a=cos^2a-sin^2a...1-2sin^2a...2cos^2a-1
tan2a= 2tana/1-tan^2a
sin a/2= +-sqrt 1-cosa/2
cos a/2= +-sqrt 1+cosa/2
tan a/2= +-sqrt 1-cosa/1+cosa...sina/1+cosa...1-cosa/sina
EXAMPLE:
1) cos75cos15+sin75sin15
=cos(a+b)=cos60--> 1/2
2) sin75cos15+cos75sin15
=sin(a-b)=sin60--> sqrt3/2
_______________________________________________________________
So pretty much my whole blog i just typed is what i need to study the most. That is probably the case with everyone else too. All of the other trig stuff is easy. But if anyone else can give me examples of the hard identity ones, i would greatly appreciate it :) And if anyone knows that there is some more confusing problems on the exam that you get. I could study that too, haha. THANKSS :)
Reflection
Trig test tomorrow..... aaaaaaaah. Hopefully I do good, anyways I'll review identites that we learned in chapter seven. The main thing to identities is memorizing all the formulas, if you don't then you wont be able to work the problem. To solve or simplify an identity you just replace what the have with something in the formulas.**Keep in mind you are trying to make the problem smaller not larger.
1. simplify:
(1-sinx)(1+sinx)
*Multiply the 2
1-sinx+sinx-sinx^2
=1-sin^2
*Use a formula
= cos^2x s^2+c^2=1
c^2=1-s^2
The answer would be 1.
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
**For this problem you are just trying to get the same thing that you started off the problem with to show that it is correct. Solve the problem like you normally would and the answer should be what it equals up to.
----------------------------
Something thats confusing me is the whole graphing thing with the amplitude and period and phase shift and all that. Help pleaseeeeee?
1. simplify:
(1-sinx)(1+sinx)
*Multiply the 2
1-sinx+sinx-sinx^2
=1-sin^2
*Use a formula
= cos^2x s^2+c^2=1
c^2=1-s^2
The answer would be 1.
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
**For this problem you are just trying to get the same thing that you started off the problem with to show that it is correct. Solve the problem like you normally would and the answer should be what it equals up to.
----------------------------
Something thats confusing me is the whole graphing thing with the amplitude and period and phase shift and all that. Help pleaseeeeee?
Reflection
I'm going to review how you solve the area of a non right triangle. This is what you use to get the area of a non-right triangle: A= 1/2 (leg) (leg) Sin (angle between).
Example: For Area say you have a triangle with:
A side length of 4
A side length of 5
And an 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.
Example: 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
Example: For Area say you have a triangle with:
A side length of 4
A side length of 5
And an 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.
Example: 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
5/2
We are having a math party at chauvins right now and I decided to take a break. Now Im doing my blog and Im going to give the formulas for the chapter 9 limacon, cardiod, and rose curve.
Limacon:
r=a+b sin (theta)
r=a+b cos (theta)
Cardiod:
r=a+-b sin theta
r=a+-bcos theta
rose curve
r=a sin (ntheta)
r=a cos (ntheta)
Archimedes
r=a theta+b
Logarithmic
r=ab^theta
Limacon:
r=a+b sin (theta)
r=a+b cos (theta)
Cardiod:
r=a+-b sin theta
r=a+-bcos theta
rose curve
r=a sin (ntheta)
r=a cos (ntheta)
Archimedes
r=a theta+b
Logarithmic
r=ab^theta
reflection 5/2
well, May is here
time for the last few weeks of school, then summer :D
just can't wait for it
trying to get away from a whole mess that came up last week, and im an idiot on completely different levels -.-
anyways.........diverging from that statement.........back to math.........
well, i can't really think of much, and the trig exam is coming up, i guess i'll just put up the trig chart
0 30 45 60 90
sin 0=0 π/6=1/2 π/4=√2/2 π/3=√3/2 π/2=1
cos "=1 "=√3/2 "=√2/2 "=1/2 "=0
sec "=1 "=2√3/3 "=√2 "=2 "=Ø
csc "=Ø "=2 "=√2 "=2√3/3 "=1
tan "=0 "=√3/3 "=1 "=√3 "=Ø
cot "=Ø "=√3 "=1 "=√3/3 "=0
thats about all ive got
deff gonna need this for the trig exam
hope we all do good on it
good luck everybody
time for the last few weeks of school, then summer :D
just can't wait for it
trying to get away from a whole mess that came up last week, and im an idiot on completely different levels -.-
anyways.........diverging from that statement.........back to math.........
well, i can't really think of much, and the trig exam is coming up, i guess i'll just put up the trig chart
0 30 45 60 90
sin 0=0 π/6=1/2 π/4=√2/2 π/3=√3/2 π/2=1
cos "=1 "=√3/2 "=√2/2 "=1/2 "=0
sec "=1 "=2√3/3 "=√2 "=2 "=Ø
csc "=Ø "=2 "=√2 "=2√3/3 "=1
tan "=0 "=√3/3 "=1 "=√3 "=Ø
cot "=Ø "=√3 "=1 "=√3/3 "=0
thats about all ive got
deff gonna need this for the trig exam
hope we all do good on it
good luck everybody
REFLECTION 5/2
Okay this week I'll go over some things from Chapter 11 to help review for that trig exam. Here's a few examples:
Ex. 1) Convert (12,13) to polar
*First, when you convert to polar you have to get two answers. the answers should be in the form (r,theta) *where r is a number and theta is a degree
*First to get r, you have to use this formula: r = sqrt of x^2 + y^2
*So you get r = +/- sqrt of 12^2 + 13^2
= +/-sqrt of 313
*Now to get the angle you take the inverse of tangent
*So you get tan^-1(y/x) >> tan^-1(13/12)
*and your reference angle for that is approx. 47.291 degrees (tangent is positive, so you're looking for the angles in the 1st and 3rd quadrants) So you add 180 degrees to 47.291 and you get 227.291
*47.291 and 227.291 are your two angles. now you have to figure out which angle goes with which sqrt of 313 (positive or negative)
first plot the point (12,13)..it ends up in the 1st quadrant. sooo the positive sqrt of 313 goes with the angle in the 1st quadrant which is 47.291
*So your final answers should be: (sqrt of 313,47.291)(-sqrt of 313,227.291)
Ex. 2) Convert (3,0 degrees) to rectangular
*to convert to rectangular you have to use these two formulas:
x = rcos(theta) y = rsin(theta)
(where r is the number 3 and theta is 0 degrees..and you're finding a coordinate (x,y))
*So first let's find the x. So you get x = 3cos(0)
= 3(1)
So x = 3
*Now find the y. So you get y = 3sin(0)
= 3(0)
So y = 0
**So your answer is (3,0)
Ex. 3) If z1 = 5 + 4i and z2 = 2 - 5i ...Find (z1)(z2)
*So all (z1)(z2) means is that you have to multiply the two together
*so you get (5+4i)(2-5i)
All you have to do is FOIL it out
So you get 10-25i+8i-20i^2
= 10-17i+20
So your answer is 30-17i
Ex. 4) Convert r = sin(theta) to rectangular
*okay when you have an equation like this and you're asked to convert, the first thing you have to do is use identities to get rid of the number in front of theta..but since there is no number there, you skip that step.
*Next you have to replace "sin" with "y/r" in the equation
*So you get >> r = y/r
now you have to get rid of the fraction, so you multiply both sides by "r".
*So now you have > r^2 = y
*Now you have to replace "r" with "sqrt of x^2+y^2"
So you get (sqrt of x^2+y^2)^2 = y ..then solve for y
so the square root and ^2 cancel out ^^
**So your final answer would be y = x^2 + y^2
***CAN SOMEONE PLEEEAAAAASSSEEEE HELP ME WITH NUMBER 5 ON THE CHAPTER 9 WORD PROBLEM TEST ASAP? (: ..I completely forgot how to do it.
It says:
Find the area of a 60 sided figure inscribed in a circle with radius 2cm
Ex. 1) Convert (12,13) to polar
*First, when you convert to polar you have to get two answers. the answers should be in the form (r,theta) *where r is a number and theta is a degree
*First to get r, you have to use this formula: r = sqrt of x^2 + y^2
*So you get r = +/- sqrt of 12^2 + 13^2
= +/-sqrt of 313
*Now to get the angle you take the inverse of tangent
*So you get tan^-1(y/x) >> tan^-1(13/12)
*and your reference angle for that is approx. 47.291 degrees (tangent is positive, so you're looking for the angles in the 1st and 3rd quadrants) So you add 180 degrees to 47.291 and you get 227.291
*47.291 and 227.291 are your two angles. now you have to figure out which angle goes with which sqrt of 313 (positive or negative)
first plot the point (12,13)..it ends up in the 1st quadrant. sooo the positive sqrt of 313 goes with the angle in the 1st quadrant which is 47.291
*So your final answers should be: (sqrt of 313,47.291)(-sqrt of 313,227.291)
Ex. 2) Convert (3,0 degrees) to rectangular
*to convert to rectangular you have to use these two formulas:
x = rcos(theta) y = rsin(theta)
(where r is the number 3 and theta is 0 degrees..and you're finding a coordinate (x,y))
*So first let's find the x. So you get x = 3cos(0)
= 3(1)
So x = 3
*Now find the y. So you get y = 3sin(0)
= 3(0)
So y = 0
**So your answer is (3,0)
Ex. 3) If z1 = 5 + 4i and z2 = 2 - 5i ...Find (z1)(z2)
*So all (z1)(z2) means is that you have to multiply the two together
*so you get (5+4i)(2-5i)
All you have to do is FOIL it out
So you get 10-25i+8i-20i^2
= 10-17i+20
So your answer is 30-17i
Ex. 4) Convert r = sin(theta) to rectangular
*okay when you have an equation like this and you're asked to convert, the first thing you have to do is use identities to get rid of the number in front of theta..but since there is no number there, you skip that step.
*Next you have to replace "sin" with "y/r" in the equation
*So you get >> r = y/r
now you have to get rid of the fraction, so you multiply both sides by "r".
*So now you have > r^2 = y
*Now you have to replace "r" with "sqrt of x^2+y^2"
So you get (sqrt of x^2+y^2)^2 = y ..then solve for y
so the square root and ^2 cancel out ^^
**So your final answer would be y = x^2 + y^2
***CAN SOMEONE PLEEEAAAAASSSEEEE HELP ME WITH NUMBER 5 ON THE CHAPTER 9 WORD PROBLEM TEST ASAP? (: ..I completely forgot how to do it.
It says:
Find the area of a 60 sided figure inscribed in a circle with radius 2cm
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