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.

Friday February 26th Class

Today in class we started off by taking the quiz.
After the quiz we went over some questions on the homework. We did problems 27 and 55.
27.


To graph these two functions we first looked at a standard sin graph.
http://fooplot.com/index.php
Then we took the f(x) function and altered the graph by moving the Y values of 1 to -2, and the Y values of -1 to 2. (Red)
We added on to that axis the g(x) function onto the same graph. This transformed the original sin graph by changing all the Y values of 1 to 4 and Y values of -1 to -4. (Blue)
file:///Users/student/Desktop/sin.tiff
The period is still 2π, and the symmetry is odd, as it is for all sin graphs. This means sin(-x)=-sin(x)
The transformation of these two functions changed the amplitude of the sin function.

55.
We then changed this to to make it simpler to graph.
This graph looked like: file:///Users/student/Desktop/cos.tiff
The period was 4π and the amplitude was changed, along with the graph being condensed. This graph originally (before being transformed) had was even, being symmetrical across the Y axis as all cos graphs originally are.

After reviewing the homework we moved on and looked at the graphs of the trig functions. We noticed that the sin graph and cos graph were very similar, but just shifted over to the right a bit. Both had a domain of -1 to 1 and a range of all real numbers.
We spent most of the rest of class playing with these graphs. We also looked at the graphs of the other trig functions as well and looking at the periods and amplitudes of the graphs.
We also looked at the tan graph which looks like:file:///Users/student/Desktop/tan%20graph.tiff

The csc graph which looks like:file:///Users/student/Desktop/csc%20graph.tiff

And the sec graph which looks like:file:///Users/student/Desktop/sec%20graph.tiff

But we did not get into why they are the way they are. That will be learned later and is in Dan's scribe post of class on 3/2.

We started class with a radical function warm up. After we did this problem, but before we went over it, we went over the quiz for the first half of class, and learned some ways to use our calculators to solve for i or find a complex conjugate. Next we went over the homework problems from two nights ago, on page 179. We went over problems 43 and 47, finding the roots over rational numbers, real numbers, and complex numbers for #43. For #47, we solved by factoring by grouping, then solving for zero. We learned how to factor a number that is the sum of two squares. So (x^2 +25) becomes (x+ 5i) (x-5i), which is the difference between two squares. When a complex number is a root, another root will be the complex conjugate. So since 5i is a root, - 5i is too. After going over those two problems we went back to the warm up. To help solve this problem, we went to http://calc101.com/webMathematica/long-divide.jsp This website can do long division, and it will show you the steps as well. We then went over how to find a slant asymptote. To find the slant asymptote, you can use the form of the radical function where there is and x + c outside of the fraction. To find the y-intercept it is easiest to use the original, unfactored form. All the x values will cancel out, and you will be left with the c values, which have no x value. This is the y intercept. We then talked about a removable asymptote, which is when a factor cancels out of a radical function, but should still be in the domain. For example: (x^2 + 1)(x + 1) ( x-3)/(x-2) (x-1) (x-3) The (x-3) will cancel out, but x cannot equal 3, so you must draw an open circle at this point on the graph.
There is no quiz on Friday, but the Unit 2 test is on Tuesday.