## Saturday, November 5, 2011

### Sum and difference formulas for tangent

10 - 2 Formulas for tan(a+ b)

The sum and difference formulas for tangent are valid for values in which tan a, tan b, and tan(a +b) are defined.

 tan(A + B) =(tan A + tan B)/(1 - tan A tan B) tan(A - B) =(tan A - tan B)/(1 + tan A tan B)

We can also use the tangent formula to find the angle between two lines. We will get two cases which are supplementary to each other. To find the angle in between two lines, we need to know the slope of both lines. The equation looks like:

Sample problems

1) Find the exact value of tan 105o

Solution:
Use one of the above formulas. Find a pair of numbers that you know the exact value of that add up to 105. Try 45 and 60! Use the addition formula!

2) Find the two supplementary angles formed by the lines

y = 2x -5

and y = -3x + 2

Solution:

let m1 = 2 and m2= -3

One angle is 45o and the other is 135o.

3) If the tan x = -7/24 and cot y = 3/4, x is in quadrant II and y is in quadrant III, find each of the following:

a) tan(x + y)

b) tan(x - y)

Solution:

Since the tangent and cotangent functions are reciprocals, tan y = 4/3

4) Verify tan(x - p/2) = -cot x

Solution:

Since the tangent is undefined at p
/2, we must change to sine and cosine.

5) Let sin x = 3/5 and sin y = 5/13 and both angles are in quadrant I, find tan(x + y).

Solution:

Since sin x = 3/5, cos x = 4/5 and tan x = 3/4.

Since sin y = 5/13, cos y = 12/13 and tan y = 5/12

Now let's take a look at the double and half-angle formulas!!

## Thursday, November 3, 2011

### Sum and difference formulas for sine and cosine

10 - 1 Formulas for Cos(a + b) and sin(a + b)

Formulas for Cos(a+ b)

We will prove the sum formula for the cosine in class. I will state it here and use it to prove other formulas later. The two main purposes for these formulas are:

1) finding exact values of other trig expressions

2) simplifying expressions to find other identities

 Cos(a ± b) = cos a cos b ∓ sin a sin b sin (a ± b) = sin a cos b ± sin b cos a

All of these formulas should be memorized. Plus you need to recall the trig values for 0, 30, 45, 60 and 90!

Demo: sin(A + B) (Manipula Math)

Sample problems

1) Find the exact value of cos 15o

Solution:

Think of two angles that you know the value of, that either add or subtract and give you 15. There are many. One that comes to mind is 45 and 30.

let a = 45 and b = 30

Use the above formula for the difference of the cosine

cos 15 = cos(45 - 30) = cos 45 cos 30 + sin 45 sin 30

2) Find the exact value of sin 50o cos 10o + sin 10o cos 50o

Solution:

Match the problem to one of the formulas. Do you see that it is the sum formula for the sine function?

= sin (50 + 10) = sin 60o
= √(3)/2

3) Suppose sin a = 4/5 and sin b = 5/13, where both a and b are in the first quadrant. Find cos (a - b)

Solution:

We need the difference formula for cosine

Cos(a - b) = cos a cos b + sin a sin b

We know the sin a and sin b, we need to find the cos a and cos b. We can use basic trig to find it. First, draw a picture for each.

Look at the diagrams. Do you see how we got the missing values? That's right, using x, y and r and their relationship to the sine and cosine.

cos a = x/r = 3/5

cos b = x/r = 12/13

Put the values in the formula:

Cos(a - b) = (3/5)(12/13) + (4/5)(5/13)

= (36/65) + (20/65)

= 56/65

4) Verify that cos(x - p ) = - cos x

Solution:

Use the difference formula for cosine:

Cos(a - b) = cos a cos b + sin a sin b

with a = x and b = p

cos(x - p) = cos x cos p + sin x sin p

remembering that cos p = -1 and sin p = 0, we have:

= -1(cos x) + 0(sin x)

= - cos x

That should give you some background in the add/subtract formulas. Let's take a look at the tangent formulas!

## Thursday, October 6, 2011

### Hyperbolas

6-4 Hyperbolas

Definition of the hyperbola

A hyperbola is the set of all points P(x, y) in the plane such that | PF1 - PF2| = 2a
Again F1 and F2 are focus points. This time the difference of these distances remain a constant at 2a. The explanation is similar to that of the ellipse. Since the ellipse is the sum of the distances and the hyperbola is the difference of the distances, the equations are very similar. They differ only in the sign and the longest side for a hyperbola is c. (Remember for the ellipse it was a)

Drawing a Hyperbola (Manipula Math)

The equation of the above hyperbola would have the form:

The hyperbola opens left and right. Notice it comes in two parts. Different than an ellipse which is a closed figure. Hyperbolas can also open up and down. I am sure you can guess at the equation!

This hyperbola has the form:

To get the correct shape of the hyperbola, we need to find the asymptotes of the hyperbola. The asymptotes are lines that are approached but not touched or crossed. These asymptotes are boundaries of the hyperbola. This is one difference between a hyperbola and a parabola. For the hyperbolas that open right/left, the asymptotes are: y = (b/a) and y = - (b/a) and for hyperbolas opening up/down, the asymptotes are: y = (a/b) and y = - (a/b)

To form the asymptotes easily on the graph, all we need do is form a rectangle using a and b.

Sample problems

1) Graph the hyperbola x2/16 - y2/4 = 1 Find the vertices, foci and equations of the asymptotic lines.

Solution:

This hyperbola opens right/left because it is in the form x - y. a2
= 16, b
2 = 4, c2
= 16 + 4 = 20. Therefore, a = 4, b = 2 and c = 4.5

Vertices: (4, 0) and (-4, 0)

Foci: (4.5, 0) and (-4.5, 0)

Equations of asymptotic lines: y = .5x and y = - .5x

To graph the hyperbola, go 2 units up/down from center point and 4 units left/right from center point.

2) Graph the hyperbola y2/25 - x2/9 = 1. Give the vertices, foci and equations of asymptotic lines.

Solution:

This hyperbola opens up and down because it is in the form y - x. a2
= 25, b
2 = 9 and c2
= 25 + 9 = 34. Thus, a = 5, b = 3, c = 5.8

Vertices: (0, 5) and (0, -5)
Foci: (0, 5.8) and (0, -5.8)
Equations of asymptotic lines: y = (5/3)x and y = (-5/3)x
The box is formed by going 5 units up/down from center and 3 units left/right from center.

3) Find an equation of a hyperbola with center at the origin, one vertex at (7, 0) and a focus at (12, 0).

Solution:

The vertex and focus are on the x-axis, so the hyperbola opens right/left. a = 7, c = 12. That makes a2
= 49, c
2 = 144 and

b2
= 144 - 49 = 95. Therefore the equation is:

x2/49
- y
2/95 = 1

Translations of Hyperbolas

If the hyperbola opens right/left the translation is: (x-h)2/a2 - (y-k)2/b2 = 1
with the equations of the asymptotic lines as:
y - k = + (b/a)(x - h)

If the hyperbola opens up/down the translation is: (y-k)2/a2 - (x-h)2/b2 = 1
with the equations of the asymptotic lines as:
y - k = + (a/b)(x - h)

Sample Problems

1) Graph the equation: (y-2)2/36 - (x-1)2/25 = 1. Find the center, vertices, foci and the equations of the asymptotic lines.

Solution:

Since it is y - x it opens up/down. a2= 36, b2 = 25 and

c2
= 36 + 25 = 61. Thus, a = 6, b = 5 and c = 7.8

Center: (1, 2)
Vertices: (1, 8) and (1, -4) ( six units up and down from center)
Foci: (1, 9.8) and (1, -5.8) ( 7.8 units up/down from center)
Equations of asymptotic lines: y - 2 = (+6/5)(x - 1)
The box is formed by going 6 units up/down and 5 units right/left from center.

2) Find the equation of a hyperbola with center (1, 1), vertex (3, 1) and focus at (5, 1).

Solution:

The vertex and foci are on the same horizontal line. This makes the hyperbola open right/left. a = 2 (distance from vertex to center), c = 4 (distance from focus to center).
Thus a
2
= 4, c
2 = 16 and b2
= 16 - 4 = 12.

The equation is: (x-1)2/4 - (y-1)2/12 = 1

## Wednesday, September 7, 2011

### Evaluating and Graphing Sine and Cosine

7-4 Evaluating and Graphing Sine and Cosine

Sines and Cosines of Special Angles

30o, 45o and 60o angles are used many times in mathematics. I strongly urge you to memorize, or at least be able to derive the sine and cosine of these special angles.

In a 30-60-90 triangle, the sides are in ratio of 1: √3 :2

Look at the triangle below:

Sin 30o = y/r = 1/2, while the Cos 30o = x/r = √3/ 2

Sin 60o = y/r = √3/ 2, while the Cos 60o = x/r = 1/2

In a 45-45-90 triangle, the sides are in ratio of 1: 1 : √2

Study the triangle below:

Sin 45o = x/r = √2/ 2, while Cos 45o = √2/ 2

The wise person will memorize the following chart:

 Degrees Radians Sin q Cos q 0 0 0 1 30 π/6 1/2 45 π/4 60 π/3 1/2 90 π/2 1 0

The graph of Sine and Cosine Functions

y = Sin x

Demonstration of Sine Graph (Manipula Math)

Notice that this graph is a periodic graph. It repeats the same graph every 2 pi units. It is increasing from 0 to half pi, decreasing from half pi to negative 1.5 pi and increasing to 2 pi. Then the repeat starts. This matches what happens to the Sine function in the quadrants. Positive in first and second and negative in the third and fourth. Maximum value for the graph is 1 and the minimum value is -1.

y = Cos x

Demonstration of Cosine Graph (Manipula Math)

This graph is similar to the previous shape. It is also a periodic graph with the cycle being 2 pi. It also matches the signs of the quadrants with quad one being positive, quads two and three, negative and quad 4 back to positive. The difference in these two graphs is the starting point for the Cosine graph. It starts at the maximum value. The Sine curve started at the origin point.

An easy way to remember these graphs is to know their 5 important points. The zeros, maximum and minimum points.

The Sine curve has zeros at the beginning, middle and end of a cycle. The maximum happens at the 1/4 mark and the minimum appears at the 3/4 mark.

The Cosine curve begins and ends with the maximum. It has a minimum at the middle point. Zeros appear at the 1/4 and 3/4 mark of the cycle.

Reference Angles

All angles can be referenced back to an angle in the first quadrant. This is true because the trig functions are periodic. Study each of the quadrant formulas below to find the reference angles.

To find the reference angle a, simply use the chart above to locate the angle q.

Example: If
q =120,then you are in quadrant II. Thus,
use the formula 180 - 120 to get a reference angle of 60.

Example: If
q = 195, then you are in quadrant III.
Thus, use the formula 195 - 180 to get a reference angle of 15.

Example: If
q = 300, then you are in quadrant IV.
Thus, use the formula 360 - 300 to get a reference angle of 60.

Relating this idea of reference angles and Sine and Cosine is easy. Determine the reference angle as we did above and put the correct sign on each function. From previous sections the Sine function is positive in quadrants I and II and negative in quadrants III and IV. The Cosine function is positive in quadrants I and IV, while negative in quadrants II and III.

Examples

Sin 135o =
Sin ( 180o - 135o) = Sin 45o

Cos 310o =
Cos (360o - 310o) = Cos 50o

Sin 210o =
Sin (210o - 180o) = - Sin 30o (Sin is

Cos 112o =
Cos (180o - 112o) = - Cos 68o (Cos is

Using your TI-82 to evaluate Sine and Cosine

1) To calculate in degrees:

Procedure 1

a) Press mode button and highlight degrees. Press enter.

b) Type in problem as example: Sin 45. Press enter.

Note:
Most trig answers are round to 4 decimal places. You can set your calculator to fixed mode by: press mode. Press down arrow and
highlight 4. Press enter. Press clear. Now enter the problem: Sin 45

Procedure 2. (Use the degree button)

a) Enter problem: Sin 45

b)Press 2nd function. Press "angle" button. You get a menu with the degree symbol as choice 1. Either press 1 or enter.

c) Your display should now look like this: Sin 45o Press enter.

Answer is .7071 if you are still in fixed mode!

Note:
Procedure 2 works regardless of the mode you are in.

2) To calculate in radian measure.

Procedure 1

a) Press mode button. Press down arrow twice. Highlight radian. Press enter. You are now in rads. All problems will now be calculated in rads.

b) Enter a problem. Example Cos 5. Press enter.

Answer (rounded to 4 decimal places) is .2837

Procedure 2. (Using the rad button)

a) Enter a problem such as Sin -2 Press
2nd function. Press angle button. Press down twice or 3.

b) Your display should look like: sin -2r
Press enter.

Note: The use of the radian button will override the setting in mode. Just like the degree button does.

## Tuesday, September 6, 2011

### Polar Coordinates

11-1: Polar Coordinates

Converting from polar to rectangular:

1) x = r cos 0

2) y = r sin 0

Converting from rectangular to polar:

3) r2 = x2 + y2

4) tan 0 = (y/x)

Polar graphing calculator! (Manipula Math)

Example problems

1) Change (3, 4) to polar coordinates.

Solution:

Using property 3 from above, find r. 32 + 42 = 25 and take the square root. Therefore, r = 5

Using property 4 from above, tan 0 = (4/3). Use your calculator set to degree mode, the answer is: 53.1 degrees. (rounded to nearest tenth. Therefore, the point is ( 5, 53.1o)

2) Change ( 4, 150o) to rectangular coordinates

Solution:

Using property 1 from above x = 4 cos 150. Using your calculator you get x = -3.46 rounded to hundredths.

Using property 2 from above y = 4 sin 150. Using your calculator you get y = 2

Therefore, the point is ( -3.46, 2)

3) Change (-3, -7) to polar coordinates

Solution:

Using property 3 from above find r. Square -3 and add to the square of -7 you get 58 and taking the square root on your calculator means r = 7.6

To find theta we use property 4, tan 0
= 7/3. Notice I used positive values! I did this to find the reference angle in quad I. The reference angle is equal to 66.8 degrees. Since we are in quadrant III, (look at the signs of the original problem) we should add 180. Why? ( we are in quad III and we memorized we add 180 in that
quad, right?) So the angle is 246.8o.

Therefore, the point is (7.6, 246.8o)

Note

You can use the special functions on the TI-82 to find all the above answers. Make sure your calculator is set in degree mode(if you want the answer in degrees) and press 2nd function angle. This takes you to a menu that has choice 5 and 6 to change to polar form. Choice 5 gives you the radius and choice 6 gives you the theta value. Example (1,3) Get to choice 5 on your calculator: You should see this R arrow Pr ( you type 1, 3) and the calculator gives you 3.16227766. This is r. Round it and your all set. Then go back to the angle menu and choose 6. You should see R arrow P0 type 1,3) again and get 71.56505118 round this and the point would be ( 3.2, 71.6o)

Do basically the same thing to find x and y by using choices 7 and 8. If the problem is in degrees set calc to degrees and if in rads make sure calc is in rads.

Sketching graphs in polar form using the TI-82

Graphs are sketched at the bottom of the page!
To use your graphing calculator is extremely easy to graph in polar form. First, get to the mode screen. Make sure you highlight the radian button and drop down one line and highlight the pol button. Now hit the y= key. Instead of y= it's now r= ! When you hit the x, t, theta button, it now shows a theta!

Try a few problems

1) r = 3 sin 0 Go ahead, type it in and hit graph. Did you get something that sort of looked like a circle? Hit zoom then choice 5. Look better? It should be a circle tangent at the origin and translated up. The top point is (0, 3)

2) r = 4 cos 0 Try this one.
Hit y= and type it in. Hit graph. This time,
you should get a circle tangent again to the origin but moved to the right. The point farthest right is (2, 0). Wonder if this is a pattern? Sine up or down and cosine right or left?

3) r = - 5 sin 0
Circle moved down

4) r = -3 cos 0
Circle moved left.

5) r = 3 sin 40 Interesting
graph. Looks like 8 leaves on a rose. Hit zoom and take choice 1. Move the cursor to a corner of the picture and hit enter. Hit the down arrow and make the line bigger than
the picture. Hit enter when it is. Now stretch it across until it makes a box around the picture. Hit enter Now you see a better picture of it. The number of leaves depend on the number multiplying 0. If it is odd that's how many leaves you get. If it is even like this one, you get double the number
of leaves. Let's see, the number for this problem was 4, I double it and I should have 8 leaves. Count them on the screen.
How about that. To be a picture like this, the number multiplying 0 must be bigger than 1 and a whole number.

6) r = 4 cos 30
Same as above with 3 leaves.

7) r = 3 cos 20
Another leaved rose. This time 4 leaves. Remember after each problem to reset the calculator to standard zoom by hitting zoom then 6

8) r = 2 + 2 cos 0. A
cardiod. (heart- shaped)

9) r = 1 + 2 sin 0
A limicon

10) r = sec 0 What is this? A vertical line if you remember sec is 1/cos

Graphs

..........

..........

..........

..........

..........

## Tuesday, August 9, 2011

### Laws of Logarithms

5-6: Laws of Logarithms

1) Log
b MN = Logb M + LogbN

2) Logb M/N = Logb M - LogbN

3) Logb M = Logb N if and only if M = N

4) Logb Mk = k LogbM

5) Logb b = 1

6) Logb 1 = 0

7) Logb bk = k
8) bLogbx = x

Sample problems

1) Log5 xy2 =

Solution:

Log5 x + Log5 y2
= Log
5 x + 2 Log5 y

2) Log7(xy/z2) =

Solution:

Log7 x + Log7 y - 2 Log7 z

3)

Express each as a single log.

1) Log x + Log y - Log z =

Solution:

Log (xy)/z

2) 2 Ln x + 3 Ln y =

Solution:

Ln x2y3

3) (1/2) Ln x - (1/3) Ln y =

Solution:

Writing logs as single logs can be helpful in solving many log equations.

1) Log2(x + 1) + Log2 3 = 4

Solution:

First combine the logs as a single log.

Log2 3(x + 1) = 4

Now rewrite as an exponential equation.
3(x + 1) = 24

Now solve for x.

3x + 3 = 16

3x = 13

x = 13/3 Since this doesn't make the number inside the log zero or negative, the answer is acceptable.

2) Log (x + 3) + Log x = 1

Solution:

Again, combine the logs as a single log.

Log x(x + 3) = 1

Rewrite as an exponential.

x(x + 3) = 10

Solve for x.

x2 + 3x = 10

x2 + 3x - 10 = 0

(x + 5)(x - 2) = 0

x = -5 or x = 2 We have to throw out 5. Why? Because it makes (x + 3) negative and we can't take the log of a negative number. So the only answer is x = 2.

3) Ln (x - 4) + Ln x = Ln 21

Solution:

Notice, this time we have a log on both sides. If we write the left side as a single log, we can use the rule that if the logs are equal, the quantity inside must be equal.

Ln x(x - 4) = Ln 21

Since the logs are equal, what is inside must be equal.

x(x - 4) = 21

Solve for x.

x2 - 4x = 21

x2 - 4x - 21 = 0

(x - 7)(x + 3) = 0

x = 7 or x = -3 Again, we need to throw out one of the answers because it makes both quantities negative. Throw out -3 and keep 7. Thus, the answer is x = 7.

Simplify each log

1) ln e5

Solution:

This is rule number 7. The answer is 5!

2) Log 10-3

Solution:

This is again rule #7. The answer: -3 (This answers the question: what power do you raise 10 to get 10 to the third?

3) eln 7

Solution:

This is rule #8. The answer is 7.

4) e2ln 5

Solution:

We can use rule #8 as soon as we simplify the problem.
Rewrite as: e
ln 25
= 25 The 25 came from 5
2.

5) 10Log 6

Solution:

6) 102 + log 5

Solution:

We need to simplify before we can apply one of the rules.
Rewrite as: (10
2)(10log5) Adding exponents means you are multiplying the bases.

= 100(5) Use rule #8 again.

= 500