Some of us didn't understand problems 41 and 43 from the previous hw. Could you please post them here? Thanks.
Petra
Friday, April 30, 2010
Thursday, April 29, 2010
Tuesday, April 27, 2010
Scribe post, 4/27/10
We started with a warm up where we had to
1. Solve on [0,2π)
sin2x-cosx=0
2. Find the exact value of cos105˚
3. Use the figure to find tan2u
We went over how to figure out if our answer is right when we finally get an answer.
For number 1
sin2x-cosx=0
2sinxcosx-cosx
cosx(2sinx-1)=0
so cosx = 0 and sinx = 1/2
for number 2
cos(105˚)
we then went over the Law of Sines which is and
we started deriving the law of sines
then we proved that they are all equal for all triangles
We also talked about the added bonus which is
we tried this on a random triangle from GeoGebra where (using AAS therom)
A=59.85˚
C=26.05˚
c=4.78 units
so B =180˚-A-C
B=94.1
so a = 9.41
b=10.86
for the last 10 minutes of class we did our homework
1. Solve on [0,2π)
sin2x-cosx=0
2. Find the exact value of cos105˚
3. Use the figure to find tan2u
We went over how to figure out if our answer is right when we finally get an answer.
For number 1
sin2x-cosx=0
2sinxcosx-cosx
cosx(2sinx-1)=0
so cosx = 0 and sinx = 1/2
for number 2
cos(105˚)
we then went over the Law of Sines which is and
we started deriving the law of sines
then we proved that they are all equal for all triangles
We also talked about the added bonus which is
we tried this on a random triangle from GeoGebra where (using AAS therom)
A=59.85˚
C=26.05˚
c=4.78 units
so B =180˚-A-C
B=94.1
so a = 9.41
b=10.86
for the last 10 minutes of class we did our homework
Friday, April 16, 2010
April 16 scribe post
Mr. O' Brien reminded us to spend a solid hour on math homework, so that on quizzes we would not be stuck on the problems by not knowing what to do in the beginning as the first step.
We began class with a warm-up: 
We were asked to attempt to graph this without technology. So...
This is what we would have gotten if it originally was a positive tan. However, the negative makes it more difficult.
Therefore...
This makes us able to graph this equation by hand.
After the warm up we went over homework problems. We covered #33,77,29,75,
While doing this we went over the identities that we needed. #75 is the most important question on the homework, and there will probably be a question like this on the unit test. Make sure that you know how to do this problem!!! Remember to draw a triangle to help you get through it. If you don't know how to do this problem yet, then go to the download and try to work through it until you think you have more of a handle on it, then try and do it by yourself until you know that you could do this problem by yourself on a quiz or test.
While still working on this problem 75 we started to learn to put the sin, cos, and tan answers into the calculator up on the board. Then, we checked all three of these on our own using our own calculator. The sin, cos, and tan were:
inverse sin(-4/5)
inverse cos(-3/5)
inverse tan(4/3)
Put them all into your calculator and you can see for yourself that they are correct, and that the inverse sin and the inverse tan equal the same value.
The last thing we did in class we all took out our quiz and the homework's it related to Mr. O'Brien came around to people individually to help us understand what we missed.
Our Goals were:
Correct Each..
Quiz questions
either on quiz or in notebook using HW #3,4,5
Class was only an hour because we have the Bunny Bowl later on today.
Collin will be the scribe for next class.
We began class with a warm-up:
We were asked to attempt to graph this without technology. So...
Therefore...
This makes us able to graph this equation by hand.
After the warm up we went over homework problems. We covered #33,77,29,75,
While doing this we went over the identities that we needed. #75 is the most important question on the homework, and there will probably be a question like this on the unit test. Make sure that you know how to do this problem!!! Remember to draw a triangle to help you get through it. If you don't know how to do this problem yet, then go to the download and try to work through it until you think you have more of a handle on it, then try and do it by yourself until you know that you could do this problem by yourself on a quiz or test.
While still working on this problem 75 we started to learn to put the sin, cos, and tan answers into the calculator up on the board. Then, we checked all three of these on our own using our own calculator. The sin, cos, and tan were:
inverse sin(-4/5)
inverse cos(-3/5)
inverse tan(4/3)
Put them all into your calculator and you can see for yourself that they are correct, and that the inverse sin and the inverse tan equal the same value.
The last thing we did in class we all took out our quiz and the homework's it related to Mr. O'Brien came around to people individually to help us understand what we missed.
Our Goals were:
Correct Each..
Quiz questions
either on quiz or in notebook using HW #3,4,5
Class was only an hour because we have the Bunny Bowl later on today.
Collin will be the scribe for next class.
Thursday, April 15, 2010
April 14 Scribe Post
Class begin with more work on the trigonometric identities packet
sin2x = 2sinxcosx
cos2x = (cos^2)x - (sin^2)x
this can also be written as:
1-(sin^2)x - (sin^2)x
thus,
cos2x = 1- 2(sin^2)x
written solely in terms of cosine,
cos2x = 2(cos^2)x - 1
To derive the identity for tan2x:
tan(x+x)
= (tanx + tanx)/(1-tanxtanx)
tan2x = (2tanx)/(1-(tan^2)x)
Next, we derived identities for sin(x/2) and cos(x/2)
cosx = 1- 2(sin^2)(x/2)
2(sin^2)(x/2) = 1 - cosx
sin(x/2) = +/- sqrt((1-cosa)/(2))
sin(x/2) = +/- sqrt((1-cosa)/(2))
cosx = 2(cos^2)(x/2) - 1
2(cos^2)(x/2) = 1 + cosx
cos(x/2) = +/- sqrt((1+cosx)/(2))
cos(x/2) = +/- sqrt((1+cosx)/(2))
Finally, we derived an identity for tan(x/2)
(+/- sqrt((1-cosx)/(2)))/(+/- sqrt((1+cosx)/(2)))
= +/- sqrt((1 - cosx)/(1 + cosx)) * (sqrt(1-cosx))/(sqrt(1-cosx))
= +/- sqrt(((1 - cosx)^2)/(1-(cos^2)x))
= +/- sqrt((1-cosx)^2/(sin^2)x)
= (1-cosx)/(sinx)
tan(x/2) = (1-cosx)/(sinx)
This concluded the work we did on the trigonometric identities packet [sorry if it's hard to visualize, but I couldn't get the online equation editor's html code to show up on the blog for some reason].
HW Questions We Reviewed:
13.
cos2x - cosx = 0
cos2x = cosx
(cos^2)x - (sin^2)x = cosx
cos^2(x) - (1 - (cos^2)x) - cosx = 0
2(cos^2)x - cosx - 1 = 0
(2cosx + 1)(cosx - 1) = 0
2cosx + 1 = 0
x = 2π/3, 4π/3
or
cosx - 1 = 0
x = 0
49.
sinu = 5/13, π/2 <> cos u = -12/13
sin(u/2) = sqrt((1-cosu)/2) = sqrt((1+12/13)/2) = (5sqrt(26))/26
cos(u/2) = sqrt((1+cosu)/2) = sqrt((1- 12/13)/2) = (sqrt(26)/26
tan(u/2) = (sinu)/(1+cosu) = (5/13)/(1-12/13) = 5
19.
6sinxcosx = 3(2sinxcosx) = 3sin2x
25.
tanu = 3/4, o <> sinu = 3/5 and cosu = 4/5
sin2u = 2sinucosu = 2(3/5)(4/5) = 24/25
cos2u = (cos^2)u - (sin^2)u = 16/25 - 9/25 = 7/25
tan2u = (2tanu)/(1-(tan^2)u) = (2(3/4))/(1-(9/16)) = (3/2)(16/7) = 24/7
11.
4sinxcosx = 1
2sin2x = 1
sin2x = 1/2
2x = π/6 + 2πk
x = π/12 + πk
x = π/12, 13π/12
or
2x = 5π/6 + 2πk
x = 5π/12 + πk
x = 5π/12, 17π/12
23.
sinu = -4/5, π <> cosu = -3/5
sin2u = 2sinucosu = 2(-4/5)(-3/5) = 24/25
cos2u = (cos^2)u - (sin^2)u = 9/25 - 16/25 = -7/25
tan2u = (2tanu)/(1 - (tan^2)u) = (2(4/3))/(1 - (16/9)) = (8/3)(-9/7) = -24/7
We finished off class with the Chapter 5 Quiz #2
Scribe for next class will be Nate
sin2x = 2sinxcosx
cos2x = (cos^2)x - (sin^2)x
this can also be written as:
1-(sin^2)x - (sin^2)x
thus,
cos2x = 1- 2(sin^2)x
written solely in terms of cosine,
cos2x = 2(cos^2)x - 1
To derive the identity for tan2x:
tan(x+x)
= (tanx + tanx)/(1-tanxtanx)
tan2x = (2tanx)/(1-(tan^2)x)
Next, we derived identities for sin(x/2) and cos(x/2)
cosx = 1- 2(sin^2)(x/2)
2(sin^2)(x/2) = 1 - cosx
sin(x/2) = +/- sqrt((1-cosa)/(2))
sin(x/2) = +/- sqrt((1-cosa)/(2))
cosx = 2(cos^2)(x/2) - 1
2(cos^2)(x/2) = 1 + cosx
cos(x/2) = +/- sqrt((1+cosx)/(2))
cos(x/2) = +/- sqrt((1+cosx)/(2))
Finally, we derived an identity for tan(x/2)
(+/- sqrt((1-cosx)/(2)))/(+/- sqrt((1+cosx)/(2)))
= +/- sqrt((1 - cosx)/(1 + cosx)) * (sqrt(1-cosx))/(sqrt(1-cosx))
= +/- sqrt(((1 - cosx)^2)/(1-(cos^2)x))
= +/- sqrt((1-cosx)^2/(sin^2)x)
= (1-cosx)/(sinx)
tan(x/2) = (1-cosx)/(sinx)
This concluded the work we did on the trigonometric identities packet [sorry if it's hard to visualize, but I couldn't get the online equation editor's html code to show up on the blog for some reason].
HW Questions We Reviewed:
13.
cos2x - cosx = 0
cos2x = cosx
(cos^2)x - (sin^2)x = cosx
cos^2(x) - (1 - (cos^2)x) - cosx = 0
2(cos^2)x - cosx - 1 = 0
(2cosx + 1)(cosx - 1) = 0
2cosx + 1 = 0
x = 2π/3, 4π/3
or
cosx - 1 = 0
x = 0
49.
sinu = 5/13, π/2 <> cos u = -12/13
sin(u/2) = sqrt((1-cosu)/2) = sqrt((1+12/13)/2) = (5sqrt(26))/26
cos(u/2) = sqrt((1+cosu)/2) = sqrt((1- 12/13)/2) = (sqrt(26)/26
tan(u/2) = (sinu)/(1+cosu) = (5/13)/(1-12/13) = 5
19.
6sinxcosx = 3(2sinxcosx) = 3sin2x
25.
tanu = 3/4, o <> sinu = 3/5 and cosu = 4/5
sin2u = 2sinucosu = 2(3/5)(4/5) = 24/25
cos2u = (cos^2)u - (sin^2)u = 16/25 - 9/25 = 7/25
tan2u = (2tanu)/(1-(tan^2)u) = (2(3/4))/(1-(9/16)) = (3/2)(16/7) = 24/7
11.
4sinxcosx = 1
2sin2x = 1
sin2x = 1/2
2x = π/6 + 2πk
x = π/12 + πk
x = π/12, 13π/12
or
2x = 5π/6 + 2πk
x = 5π/12 + πk
x = 5π/12, 17π/12
23.
sinu = -4/5, π <> cosu = -3/5
sin2u = 2sinucosu = 2(-4/5)(-3/5) = 24/25
cos2u = (cos^2)u - (sin^2)u = 9/25 - 16/25 = -7/25
tan2u = (2tanu)/(1 - (tan^2)u) = (2(4/3))/(1 - (16/9)) = (8/3)(-9/7) = -24/7
We finished off class with the Chapter 5 Quiz #2
Scribe for next class will be Nate
Monday, April 12, 2010
April 12 Scribe Post
We began class with a warm up. In our warm up we did part a of the Sum, Difference, Double, and Half Angle Identities.
First, we discovered how we could write a sin graph in terms of cosine, which was
sinx=cos(x-pi/2)
this can also be cos (pi/2-x). it is complementary.
To write cos in terms of sin:
cosx= sin (x+pi/2)
These relate to the confunction identites, but are not the same.
Next, we figured out how to write cos(-a) in terms of cosa. We know that cos is even, so cos(-a) does not equal -cosa.
We then looked at #1 on the same sheet.
We drew in three angles. a,b, and a-b. We then labeled four points. Point A was (1,0), point B was where angle b intersected the circle (X1 ,Y1,), point C was where angle a-b intersected the circle (X2, Y2,), and then point D is where angle a intersects the circle (X3, Y3,).
Then, on #2, we figured out that by using pythagorean theorem and the equation of a circle, that for our points B, C, and D, x^2 + y^2 = 1.
#3 asks why X1= cosb and Y1 = sinb, which is because cos =x value on the unit circle and sin = y value on the unit circle. From that, you can derive the values.
They are points on the circle with points
#4 asks why arcs AC and CD are the same length, which is because they have the same angle and the same radius, so the arc length is the same.
COSINE
In #5 we must use the distance formula, which in our case, is square root ((X2-1)^2+Y2^2). On its own, the distance formula is: d= square root((X2-X1)^2+(Y2-Y1)^2).
Since one of our points has an X value of 1, and Y of zero, we were able to insert these into the first distance equation which is catered to our situation (but does not work if you aren ot on a unit circle!)
Since we are trying to find a formula for cos(a-b), we could put this in for the X2 value.
So, if we want to see if AC and BC line segments are the same length, we would put these two equal to each other, getting:
X2^2 - 2X2 + 1 +Y2^2 = X3^2-2X1X3+ X1^2 + Y3^2-2Y1Y3 + Y1^2.
We can take out all our value that equal one, which would be all the X^2 + Y^2 or just 1. We have 2 of these on either side, so we can -2 from each side.
We then have
-2X2= -2X1X3-2Y1Y3
After dividing each by -2, we get
X2= X1X3+Y1Y3,
which is the same as
cos(a-b)= cosa*cosb+sina*sinb
This is called the "mother" of all identities.
We can also say that cos(a+b) is the same as cosa*cos(-b) + sina*sin(-b) (by using the mother of all identities) and can then be written as
cosa*cos-sina*sinb
SIN
Now we want to get the identity for sin (we just got the identity for cos.)
Since we know that we can rewrite sin in terms of cos, we can write sin(a-b) as cos((a-b)-pi/2).
We then want to regroup the terms inside the cos function. This gets us
cos(a-(b+pi/2)).
Then, using the mother if all identities again, we will get
cosa*cos(b+ pi/2) + sina*sin(b+ pi/2).
Since we know how to break apart a cos we can turn the cos(b+pi/2) to cosb*cos(pi/2) - sinb * sin(pi/2),
so our whole function is
cosa* cosb*[cos(pi/2) - sinb * sin(pi/2)] + sina*cosb
We can reduce the cos and sin inside the [] to 0-1 because of the location on the unit circle. So the value in the [] is just -1.
So, this = cosa (-sinb) + sina* cosb
which =
sina*cosb - cosa*sinb.
SIN(A-B)= SINACOSB - COSASINB
To find sin (a+b), we will use both sin(a-b) and cos(a+b) as reference to reduce the process.
SO,
sin (a+b)=
sin(a-(-b))=
sinacos(-b)-cosasin(-b)=
sina*cosb +cosa*sinb
TANGENT
tan(a-b) can be found by remembering tan=sin(a-b)/cos(a-b), so
sin(a-b)/cos(a-b)=
sinacosb-cosasinb/ cosacosb+sinasinb.
This would be fine, but we want to find this in terms of tan, instead of sin and cos.
SO, we must use a FuFoo.
sinacosb-cosasinb/ cosacosb+sinasinb * (1/cosacosb/1/cosacosb) (this is the FuFoo).
Some values cancel out, so we are left with
tana-tanb/ 1+ tana tanb.
SO, TAN (A-B)= TANA-TANB/ 1+TANA TANB
To find tan (a+b), we will use the same trick as in finding sin(a+b) and cos (a+b), replacing the b with a -b. So
Tan (a+b)= tana +tanb / 1-tana tanb
After this, we went over a few problems from the homework (p. 404).
Problems we went over: 19, 29, 37, 47, 53, 65, 71:
19.
13pi/12
sin (13pi/12)
Change to "nicer" values, so...
sin (3pi/12 + 10pi/12), these reduce to
sin (pi/4 + 5pi/6)
sinpi/4 cos 5pi/6 + sin 5pi/6 cos pi/4=
(sqrt. 2)/2 * (sqrt. 3)/2 + (sqrt. 2)/2 1/2=
(sqrt. 2 - sqrt. 6)/4
29. tan 2x + tan x/ a+tan 2x tanx
By using the mother identity, we can find that
= tan(2x+x), which = tan3x.
37.
sin (u+v) = sinu cosv +sinv cosu
sinu = 5/13
cosv=-3/5
u=arcsin 5/13
v=arccos -3/5
u and v are both in quad. II, and we must turn arcsin into pi - arcsin so it will be in quad. II instead of quad. I.
we will then have:
(5/13)(-3/5) + (4/5)(-12/13)= -63/65
47.
tan (u-v) = tanu -tanv/ 1+ tanutanv
sin u = -7/25 cos v = -4/5
= (7/24 - 3/4)/[ 1 + (7/24) (3/4)]
= (-11/24) / (39/32)
= 44/117
53.
cos(arccosx+arcsinx)=
cos(arccosx) cos(arcsinx)-sin(arccosx)(arcsinx)
= x (sqrt 1-x^2)- (sqrt 1-x^2)x
= 0
65. cos (3pi/2 - X)
cos 3pi/2 cosx+ sin 3pi/2 sinx
= 0(cosx)+ -1(sinx)
= -sinx
71. cos (x+ pi/4) - cos (x-pi/4)= 1
Simplifies to
1 = -2 sin(x) (sqrt.2)/2
1= -sqrt. 2 sinx
sinx= 1/(sqrt. 2)
x= 5pi/4, 7pi/4
After we went over the homework questions we began to look at sin2a and cos2a.
Sin2a:
= sin (a+a)
= sinacosa + sinacosa
= 2sinacosa
Cos2a:
= cos (a+a)
= cosasina cosasina
= cos^2a - sin^2a
= 1-sin^2a - sin^2a
= 1- 2sin^2a
= cos^2a - (1-cos^2a)
= 2 cos^2a -1
Homework Quiz next Class!
Scribe for next class is Kyle.
First, we discovered how we could write a sin graph in terms of cosine, which was
sinx=cos(x-pi/2)
this can also be cos (pi/2-x). it is complementary.
To write cos in terms of sin:
cosx= sin (x+pi/2)
These relate to the confunction identites, but are not the same.
Next, we figured out how to write cos(-a) in terms of cosa. We know that cos is even, so cos(-a) does not equal -cosa.
We then looked at #1 on the same sheet.
We drew in three angles. a,b, and a-b. We then labeled four points. Point A was (1,0), point B was where angle b intersected the circle (X1 ,Y1,), point C was where angle a-b intersected the circle (X2, Y2,), and then point D is where angle a intersects the circle (X3, Y3,).
Then, on #2, we figured out that by using pythagorean theorem and the equation of a circle, that for our points B, C, and D, x^2 + y^2 = 1.
#3 asks why X1= cosb and Y1 = sinb, which is because cos =x value on the unit circle and sin = y value on the unit circle. From that, you can derive the values.
They are points on the circle with points
#4 asks why arcs AC and CD are the same length, which is because they have the same angle and the same radius, so the arc length is the same.
COSINE
In #5 we must use the distance formula, which in our case, is square root ((X2-1)^2+Y2^2). On its own, the distance formula is: d= square root((X2-X1)^2+(Y2-Y1)^2).
Since one of our points has an X value of 1, and Y of zero, we were able to insert these into the first distance equation which is catered to our situation (but does not work if you aren ot on a unit circle!)
Since we are trying to find a formula for cos(a-b), we could put this in for the X2 value.
So, if we want to see if AC and BC line segments are the same length, we would put these two equal to each other, getting:
X2^2 - 2X2 + 1 +Y2^2 = X3^2-2X1X3+ X1^2 + Y3^2-2Y1Y3 + Y1^2.
We can take out all our value that equal one, which would be all the X^2 + Y^2 or just 1. We have 2 of these on either side, so we can -2 from each side.
We then have
-2X2= -2X1X3-2Y1Y3
After dividing each by -2, we get
X2= X1X3+Y1Y3,
which is the same as
cos(a-b)= cosa*cosb+sina*sinb
This is called the "mother" of all identities.
We can also say that cos(a+b) is the same as cosa*cos(-b) + sina*sin(-b) (by using the mother of all identities) and can then be written as
cosa*cos-sina*sinb
SIN
Now we want to get the identity for sin (we just got the identity for cos.)
Since we know that we can rewrite sin in terms of cos, we can write sin(a-b) as cos((a-b)-pi/2).
We then want to regroup the terms inside the cos function. This gets us
cos(a-(b+pi/2)).
Then, using the mother if all identities again, we will get
cosa*cos(b+ pi/2) + sina*sin(b+ pi/2).
Since we know how to break apart a cos we can turn the cos(b+pi/2) to cosb*cos(pi/2) - sinb * sin(pi/2),
so our whole function is
cosa* cosb*[cos(pi/2) - sinb * sin(pi/2)] + sina*cosb
We can reduce the cos and sin inside the [] to 0-1 because of the location on the unit circle. So the value in the [] is just -1.
So, this = cosa (-sinb) + sina* cosb
which =
sina*cosb - cosa*sinb.
SIN(A-B)= SINACOSB - COSASINB
To find sin (a+b), we will use both sin(a-b) and cos(a+b) as reference to reduce the process.
SO,
sin (a+b)=
sin(a-(-b))=
sinacos(-b)-cosasin(-b)=
sina*cosb +cosa*sinb
TANGENT
tan(a-b) can be found by remembering tan=sin(a-b)/cos(a-b), so
sin(a-b)/cos(a-b)=
sinacosb-cosasinb/ cosacosb+sinasinb.
This would be fine, but we want to find this in terms of tan, instead of sin and cos.
SO, we must use a FuFoo.
sinacosb-cosasinb/ cosacosb+sinasinb * (1/cosacosb/1/cosacosb) (this is the FuFoo).
Some values cancel out, so we are left with
tana-tanb/ 1+ tana tanb.
SO, TAN (A-B)= TANA-TANB/ 1+TANA TANB
To find tan (a+b), we will use the same trick as in finding sin(a+b) and cos (a+b), replacing the b with a -b. So
Tan (a+b)= tana +tanb / 1-tana tanb
After this, we went over a few problems from the homework (p. 404).
Problems we went over: 19, 29, 37, 47, 53, 65, 71:
19.
13pi/12
sin (13pi/12)
Change to "nicer" values, so...
sin (3pi/12 + 10pi/12), these reduce to
sin (pi/4 + 5pi/6)
sinpi/4 cos 5pi/6 + sin 5pi/6 cos pi/4=
(sqrt. 2)/2 * (sqrt. 3)/2 + (sqrt. 2)/2 1/2=
(sqrt. 2 - sqrt. 6)/4
29. tan 2x + tan x/ a+tan 2x tanx
By using the mother identity, we can find that
= tan(2x+x), which = tan3x.
37.
sin (u+v) = sinu cosv +sinv cosu
sinu = 5/13
cosv=-3/5
u=arcsin 5/13
v=arccos -3/5
u and v are both in quad. II, and we must turn arcsin into pi - arcsin so it will be in quad. II instead of quad. I.
we will then have:
(5/13)(-3/5) + (4/5)(-12/13)= -63/65
47.
tan (u-v) = tanu -tanv/ 1+ tanutanv
sin u = -7/25 cos v = -4/5
= (7/24 - 3/4)/[ 1 + (7/24) (3/4)]
= (-11/24) / (39/32)
= 44/117
53.
cos(arccosx+arcsinx)=
cos(arccosx) cos(arcsinx)-sin(arccosx)(arcsinx)
= x (sqrt 1-x^2)- (sqrt 1-x^2)x
= 0
65. cos (3pi/2 - X)
cos 3pi/2 cosx+ sin 3pi/2 sinx
= 0(cosx)+ -1(sinx)
= -sinx
71. cos (x+ pi/4) - cos (x-pi/4)= 1
Simplifies to
1 = -2 sin(x) (sqrt.2)/2
1= -sqrt. 2 sinx
sinx= 1/(sqrt. 2)
x= 5pi/4, 7pi/4
After we went over the homework questions we began to look at sin2a and cos2a.
Sin2a:
= sin (a+a)
= sinacosa + sinacosa
= 2sinacosa
Cos2a:
= cos (a+a)
= cosasina cosasina
= cos^2a - sin^2a
= 1-sin^2a - sin^2a
= 1- 2sin^2a
= cos^2a - (1-cos^2a)
= 2 cos^2a -1
Homework Quiz next Class!
Scribe for next class is Kyle.
April 8th
At the start of today's class Mr. O'Brien gave us two warm up problems to do, to be solved by hand and then checked using technology. Here's the first problem:
After the warm up problems we went over questions people had from the homework. Number 73 is below:
Next, Mr. OB handed back our first quiz on unit 6! In general, I think most people did fairly well. Remember the two identities on the back are very similar because they use the same identities. If you understand one you should be able to figure out the other. Also, if you still have questions about the information we've covered so far, look over the sections in the textbook or come in to see Mr. OB!
Finally, we took notes on a new set of identities. These are called the sum, difference, double and half angle identities.
Cos(a-b)=Cos(a)Cos(b)+Sin(a)Sin(b)
Cos(a+b)=Cos(a)Cos(b)-Sin(a)Sin(b)
Sin(a-b)=Sin(a)Cos(b)-Cos(a)Sin(b)
Sin(a+b)=Sin(a)Cos(b)+Cos(a)Sin(b)
You can also find these identities on the front inside cover of your textbook!
Homework p. 404/5, 19, 29, 33, 37, 47, 53, 63, 65, 71
The scribe for next class is Sophie.
Thursday, April 8, 2010
Homework question
So I understand problem 55 up until you come out with sin x = {3/4, 1/3}. However, I simply did arcsin(3/4) and arcsin(1/3) and came out with x?{.8481,.3398}. However, the book showed two other answers that I should get, and I understand that they are also points on the arcsin graphs. However, I don't understand why we use two of them on each graph. Shouldn't the arcsin function only have one value for x? And even if it did use more than the one, then how would I know which to use when there are infinite points on the arcsin graph? Furthermore, how would I calculate the other value? It gives ~2.2935 as one of the values, and if I enter in sin(2.2935), i get ~.75. But I don't know how to find that without having it in the book. Thanks!
-Molly
Tuesday, April 6, 2010
Quarter 3 projects
Please take the time to look at the one or two quarter projects from your classmates:
http://math-ob.wikispaces.com/Pre-AP+Calculus+Quarter+3+Projects+2009-10
While I was impressed with them all, Annie's in R3 and Andy's in W2 are particularly well done.
http://math-ob.wikispaces.com/Pre-AP+Calculus+Quarter+3+Projects+2009-10
While I was impressed with them all, Annie's in R3 and Andy's in W2 are particularly well done.
Friday, April 2, 2010
We spent a good deal of class working on the warmup problems. Basically, it was the application of the Fundamental Trigonometric Identities. The primary benefit of the exercise was that it gave us a chance on making our proof format more acceptable. Format work included using the equals sign in a transitive manner, and making sure everything was thoroughly explained. If you’re not evidently applying a property, it’s best to leave a side note for thorough explanation. For example,
If cotx + csc= cos/(sin^2), you couldn’t make the leap to “cos/(sin^2)=cos/(sin^2)” without first explaining your manipulation of cotx and cscx.
On concept that came up several times throughout the warm-up was the technique of multiplying by fufoos to get lowest common denominator. This can be particularly useful with complicated fractions.
After we had learned to write proofs in a more explanatory manner, we started work on the homework problems.
While working on the homework, there was some reviewing of how to derive the other pythagorean identities. For those who need a reminder, just divide the values on both sides by (sin^2)x or (cos^2)x. Basically, working on the homework solidified strategies we had previously gone over, like multiplying by fufoos.
Then, we worked on Solving Trig Functions. Instead of trial and error, we worked with technology to solve the equations. Note that some of the solutions to these trig functions looked similar to the format used in our Francois worksheet.
At the end of the class, we briefly talked about future algebraic solving.
If cotx + csc= cos/(sin^2), you couldn’t make the leap to “cos/(sin^2)=cos/(sin^2)” without first explaining your manipulation of cotx and cscx.
On concept that came up several times throughout the warm-up was the technique of multiplying by fufoos to get lowest common denominator. This can be particularly useful with complicated fractions.
After we had learned to write proofs in a more explanatory manner, we started work on the homework problems.
While working on the homework, there was some reviewing of how to derive the other pythagorean identities. For those who need a reminder, just divide the values on both sides by (sin^2)x or (cos^2)x. Basically, working on the homework solidified strategies we had previously gone over, like multiplying by fufoos.
Then, we worked on Solving Trig Functions. Instead of trial and error, we worked with technology to solve the equations. Note that some of the solutions to these trig functions looked similar to the format used in our Francois worksheet.
At the end of the class, we briefly talked about future algebraic solving.
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