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.
