Math Concepts Explained

Introduction to Statistics - Mode

2009-07-26T17:19:00.000-07:00

The third concept of introductory statistics that I will explain here is the mode. (Refer to my previous posts about mean and median.)

The mode is also a measure of average, but again, different from how you have likely thought of an average before.

The mode is simply the value in a data set that is represented the most times.

Again, let's refer to my ongoing test grade example:

Example:
The grades received for a test in a math class, composed of 13 students, were:
65, 98, 92, 43, 76, 64, 69, 72, 75, 85, 96, 90, 90

When you rearrange them from lowest to highest, you can quickly identify which value appears the most times:
43, 64, 65, 69, 72, 75, 76, 85, 90, 90, 92, 96, 98

So, for this data set of math scores, the mode was 90... that is, more students scored 90 than any other single grade achieved.

Hopefully this series of posts has helped to explain the calculation of median, mean, and mode. As you will see in your studies, each has its own applications and are useful in their own ways. It is important to realize though that they are all related as forms of average, and they all describe the centeredness of a data set.

As always, leave a comment if you need clarification or more information on the topics I've posted. As well, as I have done with this series of posts, I will try to address any requests students may have as well in future posts. :)

Introduction to Statistics - Median

2009-07-26T17:04:00.000-07:00

The next introductory statistical concept I will discuss is the median. It is similar to the mean in that it is a form of an average, though most likely not in the terms that you have thought of an average before.

While the mean is the sum of a group of values divided by the number of values, the median is the point at which half of the data points of the set are below it, and half of the data points are above it. In other words, the median is the midpoint of the data set, with 50% of the data points on either side of it. As you will see, while this number CAN be equal to the mean, it does not have to be.

Let's continue with the example given in the post about means.

Example:
The grades received for a test in a math class, composed of 12 students, were:
65, 98, 92, 43, 76, 64, 69, 72, 75, 85, 96, 90
What is the median of this set of data?

The first thing to do is to rearrange the data points from lowest to highest.
43, 64, 65, 69, 72, 75, 76, 85, 90, 92, 96, 98

To determine the median, you simply have to pick out the MIDDLE value of this data set. For data sets with an odd number of values, this is easy. This data set, however, has 12 values, so the median is actually represented by the AVERAGE of the center TWO values. In this case, the middle 2 values are the 6th and 7th values, 75 and 76. Therefore, the median of this data set is the average of 75 and 76, which is 75.5

Let us pretend that one student was sick on test day, and when he took it later on, he scored a 90 on the exam. If we now factor this in, we have a data set of 13 values (an odd number), and so as you can see, the middle point is the 7th point, which is 76.

Also, convince yourself that that addition of this student's score increases the mean to 78%.

The final concept to discuss is the mode, which I will explain next.

Introduction to Statistics - Mean

2009-07-26T16:47:00.000-07:00

As I have received several requests to do so, I am going to put up a couple of posts that explain some basic statistical concepts. The first few will be the mean, median, and mode. Several students find these concepts difficult to grasp at first, but you will see that they are really quite simple. This post will explain the mean.

The mean may sound like a bad thing, but the mean is actually just another word for a concept that you have UNDOUBTEDLY used SEVERAL times up to this point... the AVERAGE! That's right, the mean is what you have always known as the average. (In fact, the average is not the most precise word to use to describe this function... mean is the correct name.) It is remarkable how many times this connection is not immediately presented to the students, and so they feel they are struggling with a new concept, and one that is probably not being taught well! I will briefly go over the calculation of the mean, but please leave a comment if you wish to have any additional details about it.

The arithmetic mean is the sum of a group of values, divided by the number of values used to determine that sum. A familiar example would go something like this:

Example:
The grades received for a test in a math class, composed of 12 students, were:
65, 98, 92, 43, 76, 64, 69, 72, 75, 85, 96, 90
What is the mean grade received on this test?

To compute the mean of this set of data, first determine the sum of the grades... this is 925.
Next, you determine the number of values, which was stated as 12.
Divide the sum by the number of values to give.... 77.083333
So the mean grade received on this test was 77%.

It's that simple. The sum of the values, divided by the number of values. :) Let me know if you want more. Otherwise, I will now move on to median and mode.

Check out this website! www.homeschoolbytes.com

2009-06-27T16:40:00.000-07:00

Just a quick post here to mention a great site worth a visit. Homeschool Bytes has put out a recent post called Math At Play Blog Carnival #7 - Onomatopoeia. They even mention my site! It's a great resource for math, homeschooling, and other educational topics. Pay them a visit if you get a chance and explore their site. It's really informative and well done. :)

Square Roots - Part III (in progress)

2009-02-03T23:10:00.001-08:00

Reducing an irrational number!

What is the most simplified way of writing the square root of 24?

To do this, you want to know the numbers that can multiply up to 24 (factors).

Square Roots - Part II

2009-02-03T22:49:00.001-08:00

Picking up where I left off in my previous post (Square Roots - Part I), I'm going to briefly explain IRRATIONAL NUMBERS. (Also be sure to check out Square Roots - Part III!)

A RATIONAL number is one that can be expressed as a quotient of two integers (ie. as a fraction). So, all fractions are rational. And all whole number integers are rational (since they can be expressed as "something over one").

Conversely, IRRATIONAL NUMBERS can NOT be expressed as a fraction. The number pi is commonly given as an example of an irrational number, since it cannot be written as a fraction, and the numbers after the decimal point keep going and going. Going back to my last post, the square root of 24 is also irrational. If you punch it into a calculator, you will see that it is 4.8989794855663....... On this note, you can say that the square roots of non-perfect squares are irrational.

Now, to find the square root of a non-perfect square by hand, you just do the same trial and error method we learned in the last post, only this time, as you narrow your guesses down, you can use numbers with more and more decimal points. Observe:

Find the square root of 24:
4 x 4 gives 16....
5 x 5 gives 25...
4.5 x 4.5 gives 20.25
4.9 x 4.9 gives 24.01.... needs to be a bit smaller
4.89 x 4.89 gives 23.912.... needs to be a bit bigger, but less than 4.9
4.898 x 4.898 gives 23.9904.... bit bigger still... but still less than 4.9
4.8989 x 4.8989 gives 23.999221.... even a bit bigger, but still less than 4.9

As you can see, this can go on and on, until you have as many decimal places as you want.

Alternately, and more commonly, you would leave your answer as an irrational number, rather than recording decimal places (since technically, writing to so many decimal places, unless you write FOREVER, can be expressed as a fraction.... tenths, hundredths, thousandths, ten thousandths, etc...)

However, you don't want to leave your irrational number in a form that hasn't been reduced yet, do you? Check back to my next post to find out more.

Square Roots - Part I

2009-02-03T10:36:00.001-08:00

Make sure you check out my Square Roots - Part II and Part III posts!

The concept of square roots often gives students trouble. (In fact, it is the topic most searched for on my site!) However, the initial uncertainty and hesitation with this topic is quite unnecessary. Square roots sound daunting, but they're really a simple concept.

The SQUARE ROOT of a number "x" is some number "y", such that when "y"is multiplied by itself, its product is "x". Sounds confusing... but you will see that it's not. The square root sign looks like this: √25 (usually with a line over the top of the number.)

Example:
Find the square root of 25.

So, going along with the definition I gave above, let's say x = 25. So then, we want to know y... that is, what number, when it is multiplied by itself, will equal 25. In this example, most people will be able to say immediately "5 times 5 equals 25!" And they will be right. The square root of 25 is 5, because when 5 is multiplied by itself, it gives 25.

Now, I'm sure that most of you will be saying something like "That's easy! But what about when the numbers are big... or weird... like 529?" While most calculators can tell you the square root by the touch of a single button, figuring it out by hand can take a little more guesswork. To find the square root of, say, 529 (by hand), you have to just keep trying to multiply numbers by themselves to reach it. Watch:

10 x 10 = 100..... not big enough
15 x 15 = 225..... still not big enough
20 x 20 = 400..... getting closer
25 x 25 = 625..... too big! so we've narrowed it down to between 20 and 25
22 x 22 = 484
24 x 24 = 576
23 x 23 = 529 BINGO!

It takes a bit of work, but you see how it can be done. Now, having seen this, there are a few things to note.

1) While it can be said that the SQUARE ROOT of a number "x" is some number "y" that, when multiplied by itself, gives "x", the SQUARE of a number is the result you get when you multiply a number by itself. So, the square root of 25 is 5, whereas the square of 5 is 25. It is important to understand these definitions and not to mix them up. Pay attention to what the question is asking!

2) The examples and method I described use PERFECT SQUARES. A perfect square is a number whose square root is an integer (whole number). So, the square root of 25 is 5, but the square root of 24 is.... less than 5.... but more than 4 (since the square of 4 is 16). Therefore, 25 is a perfect square, but 24 is not. We would call the square root of 24 an IRRATIONAL NUMBER.

I'll have a bit more to say about squares and square roots in following posts. Stay tuned!

Converting Point-Slope Form to Standard Form

2009-01-16T20:31:00.001-08:00

I previously described how to obtain the equation of a line, and how to express that in both point-slope form and standard form. While both equations describe the exact same line, sometimes you may be asked to express the line in a specific way, and you need to be able to manipulate and rearrange the provided equation to make it look like the other form. I will show an example of how this can be done.

Reminders (refer to the posts linked above for more details)

Point-slope looks like this:
(y-y1) = m(x-x1), which is the general way of saying y=mx+b

Standard form looks like this:
Ax + By = C

Example: Express the equation y=5x-10 in standard form. State the values for A, B, and C.

Basically, what you want to do is move all the x and y terms over to one side, and move the constants (terms with no variables) over to the other. Combine and simplify where possible. That's all there is to it. "A" will be the term left over in front of x, "B" will be with y, and C will be the value not attached to a variable.

y=5x-10

10=5x-y

So:
5x-y=10
A=5, B=(-1), C=10
(remember the standard form has a "+", so a "-" in your answer implies a coefficient of (-1).

Let's try another one:

Example:
Express the equation y=(4/3)x+2 in standard form. State the values for A, B, and C.

This one works the same way, but there is something else that can be done, as I will demonstrate.

y=(4/3)x+2
(-2)=(4/3)x-y
So:
(4/3)x-y=(-2)
A=(4/3), B=(-1), C=(-2)
There is nothing wrong with this answer. It is properly rearranged, and the coefficients have been stated. However, usually it is a good idea to not have fractions (ie. have nothing in the denominator). So, to do this, you work our final answer a bit further, so that all the values are in the numerators.

(4/3)x-y=(-2)
Multiply all terms by 3, to remove it from the denominator of the first term. This gives:
4x-3y=(-6)
A=4, B=(-3), C=(-6)

Again, this answer describes the exact same line as the initial answer without the extra moves, so technically, they are both right. It is just a common convention to keep things in the numerator wherever possible.

Converting from the Standard Form to the Point-slope form is basically just the reverse. Try it for yourself with these examples!

Inverse Trigonometry Functions Part 1 - arcsine, arccosine, arctangent

2008-10-15T22:40:00.000-07:00

Coming soon...

Arcsin(x), sin-1(x)
Arccos(x), cos-1(x)
Arctan(x), tan-1(x)

Special Angles in Trigonometry - Part 2 (Similar Triangles)

2008-06-10T18:02:00.000-07:00

This post will hopefully clarify how to work with special triangles when the sides are not the standard lengths (as I described in the original post for Special Angles).

Maybe this would be a good time to describe SIMILAR triangles. Similar triangles are triangles that have different side lengths, but have the same angles. Don't let the fancy name fool you... just think of them as smaller or larger versions of the same triangle. Examine this picture:

If you were to cut out all the triangles, and shrink or enlarge them, you would see that they all would fit on top of each other. This is possible because each triangle has the same angles, despite having different side lengths. When triangles are like this, they are said to be SIMILAR.

Now, to apply this to my previous post explaining the special angles in trigonometry. I explained how to derive the trig functions using the simplest triangles. However, in all likelihood, you will find triangles of different sizes, rather than these same simple triangles. You need only remember the rules of SOHCAHTOA to be able to evaulate the trig functions.

For example:

Take a right angle triangle with an angle of 30 degrees, and you know that the short side is a length of, say, 5. (Try to sketch this out... being able to draw a triangle from a description is important to learn as well!) There are a few strategies you could use here to solve the other 2 sides and last angle.

1) You know that all the angles in a triangle sum up to 180 degrees. So, 180-90-30 = 60!

2) You can apply the rules of SOHCAHTOA to determine either one of both of the unknown sides. In this case, you have the 30 degree angle known, and the short side is opposite this angle. So, you can use the Sine function to determine the length of the hypotenuse (try it! how do you solve Sin(30). Once you have the hypotenuse figured out you can then turn around and use it to to solve the Cosine function... OR you could just use the Theorem of Pythagoras, since it is a right-angle triangle.

3) Remember that you could also use either the Sine Law or Cosine Law in there as well (especially on triangles that are NOT right-angle triangles)... Sine Law works whenever you know an angle and it's opposite side, and then either of an angle or a side to complete the identity. Or the Cosine Law will work when you know 2 sides and the angle between those 2 sides.

Hopefully this quick post addresses some concerns that some of you may have had. Sorry about any confusion or lack of clarity in the first post. As always, please continue to drop me a line if you have any questions, concerns, or topics you would like me to cover!

**EDIT** - My apologies... as pointed out in the comment section, triangles as I have discussed in this post are in fact called SIMILAR, and not CONGRUENT (I have edited the post to reflect this). Similar triangles are triangles with the same angles but can have differing side lengths. Congruent triangles, on the other hand, have the same angles AND sides. This means they look either the exact same, or are a mirror image of the original. Sorry for the confusion, and thanks to the astute reader who picked out my error. :)

Word Problems - Hints and Tips to Solving Them

2008-03-11T21:08:00.000-07:00

Word problems are one kind of math question that ALWAYS gives students troubles, probably because making a connection between the math and the 'real world' situation isn't necessarily the easiest thing to do. Despite that, there are a few pointers that you can try to which will hopefully lead you to the correct solution.

The first tip is to MAKE A LIST OF ALL OF THE INFORMATION THAT YOU ARE GIVEN.

Example:

Joe wants to save money to buy a new Playstation 3. It costs $400 to buy from the store. He has a job which pays him $15 per hour, and he works 6 hours per day. However, to get to and from his job, he has to take the bus, which costs him $2 each way, and he buys lunch each day he works for $6. He also owes his dad $500 from when he borrowed it to buy a new stereo. Taking all of these costs into account, how many days will it take Joe to save up enough money to buy his new Playstation? How much more money does Joe need to earn if he wants to buy it in 5 days?

LIST WHAT YOU KNOW:
PS3 cost = $400
Wages = $15 / hr
Work day = 6 hr / day
Bus fare = 2 x $2 / trip
Lunch = $6 / day
Owes Dad = $500

Next, it is important to IDENTIFY WHAT YOU WANT TO FIND OUT. In this case, we want to find out how many days it will take Joe to save enough to buy his Playstation. Specifically, we want to know how long it will take him to save $400.

The next pointer is to try to CLASSIFY WHAT YOU ARE GIVEN INTO GROUPS. In this question, you can classify things into "Things Related to Making Money" and "Expenses."

Money:

Wages = $15 / hr

Work day = 6 hr / day

Expenses:

Bus fare = 2 x $2 / trip

Lunch = $6 / day

Owes Dad = $500

From here, it might start becoming more apparent what need to be done. In this case, you would find out how much money Joe makes in a day, and also how much he spends in a day.

Earnings = $15 / hr x 6 hr / day = $90 / day

Expenses = 2 x ($2 / trip) + $6 / day = $10 / day

Overall = 90 - 10 = $80

So, Joe saves $80 per day.

Now, he owes his Dad $500, and wants to buy his Playstation for $400... totalling $900. At $80/day, Joes can pay off his Dad and then be able to afford his Playstation in:

$900 / ($80/day) = 11.25 days.... = 12 days (since he can't stop working at 11.25 days). :)

Of course, now that you have an answer, make sure that you answer the question completely.... there's still another part!

If Joe wants to make his purchase, after paying back his dad, in 5 days.....

KNOW:

total cost = $900

time = 5 days

NEED:

savings per day = ?

When you separate what you know from what you don't know, it is very helpful in seeing how to get to the answer. In this case, Joe needs $900 / 5 days = $180. Compared to what he makes right now, he needs to save another $100 per day!

So to summarize, they keys to word problems are to:

IDENTIFY WHAT YOU KNOW (classifying into groups helps here),

DECIDE WHAT YOU NEED TO KNOW,

and MAKE SURE YOU READ AND ANSWER THE ENTIRE QUESTION.

Simple rules, but important nonetheless. :)

Word Problems and Unit Conversions

2008-02-27T21:05:00.000-08:00

For some people, working with variables that have different units is very easy. They can conceptualize the relationship between the different variables in a way that seems natural to them, yet remains very difficult to explain to others. There are even more people who cannot do this type of math so automatically, and often struggle to solve problems... myself included. However, there is a trick to solving these questions, and the solution lies, quite simply, in keeping track of the units. I will try to demonstrate.

I will start simply, with multiplication and division. You know that if one number is on the top and on the bottom of a division line, you can cancel them out to reduce things:

You also know that you can do this when you are working with variables, like x and y:

So then, it's only natural that you can do this with units as well!

Now, here is the trick. If you are struggling with the concept of a word or other problem that deals with units, change the way that you are looking at it, to focus on the units involved. They will guide you towards the final solution.

Let's try a basic physics problem: How far will an object travel in 10 seconds if it is moving at a constant speed of 25 meters/second? This one might not be too difficult to conceptualize, but it demonstrates my point. Watch what I do with the units at the start, if you are not familiar with this part.

As you can see, the units cancel out to leave just the 'meters.'

Let's try one that is a bit more complicated: If a force of 10 N pushes a 10 kg block along a flat, frictionless surface for 5 seconds, what is the block's average velocity? This one is undoubtedly more difficult to do in your head, but if you keep track of the units (what you have, and what you need), then you should be able to do it. For those who aren't in physics, it is easier to know that the unit Newton, N, is equal to (kg) x (m) / (s)^2... blogger is still terrible with superscripts and equations. :(

As you can see, we wanted velocity (m/s) in the end, and so I just worked with the units until I got what I wanted. From a physics point of view, I used Force = Mass x Acceleration, and Velocity = Acceleration x Time.

Currency conversions are a common everyday example of where you can use this trick. If I say that the exchange rate is 0.985 $US / $CAN, and I want to know how much $US I can exchange for $232 CAN:

Naturally, it's best to try to understand the concepts of the questions, and the more complex questions won't be quite so easy to muddle through just by tracking the units. However, hopefully you can see the benefit of looking at these types of problems from this angle, and it will help you to solve them. I learned this technique in high school, and I still benefit from it. As always, feel free to drop me a line if you would like me to clear anything up. :)

Polynomials

2008-02-12T22:19:00.000-08:00

I have a request to go over polynomials, and to specifically explain what FOIL means.

"Polynomials" refer to mathematical expressions that contain multiple terms (poly = multiple, nomials = terms/numbers (roughly)). "Monomial" refers to a polynomial that has only a single (mono) term. Furthermore, the product of multiplying monomials together also results in a monomial. Basically, think of it along these lines:

- 1 term (bunch of things multiplied/divided/raised to a power) = monomial

- more than 1 term (monomials) that are added or substracted = polynomial

Some examples hopefully clear this up:

1) Monomials:
x,
2x,
x^3,
xy,
5xy,
(10xy^4)/3

2) Polynomials:
x+5,
2x-5,
(5xyz^3)/4 + 7x

Doing calculations with polynomials is fairly straight-forward, but you do have to keep a few things in mind.

For addition or subtraction of polynomials, you must remember that you can only combine 'like terms.' 'Like terms' contain the sample variables and differ by the numeric coefficient in the front. 3x and 5x are like terms, and can be added to get 8x. However, 3x and x^2 are not like terms and CANNOT be added like in the first example.

x + 2x = 3x
4x + 4x = 8x
15x + 100x = 115x
50x - 10x = 40x

You cannot add terms in this way that have different variables:
x^2 + 5x
3x^4 - 2x^3
x + 7

**Technically, some expressions such as these can be SIMPLIFIED by pulling out a common factor, but they still cannot be added up, as with like terms:
eg. x^2 + 5x = x(x + 5)

Multiplying polynomials is a little more complicated, but still straight-forward, provided that you keep track of what you are doing. When multiplying two polynomials together, all that you need to do is add up the products of multiplying each term in the first polynomial by each term in the second polynomial. You have probably heard the expression "FOIL" when talking about multiplying polynomials. FOIL is short for "First, Outside, Inside, Last" and refers to which terms you multiply together and add up when multiplying two polynomials, each composed of two monomials (specifically, this is the product of two BINOMIALS).

Here is an example of this:

(2x + 3)(x + 5) = ?

First: 2x * x
Outside: 2x * 5
Inside: 3 * x
Last: 3 * 5

Adding these up gives:
2x^2 + 10x + 3x + 15

Furthermore, as I showed earlier, about adding polynomials with like terms, you can simplify this expression:
2x^2 + 13x + 15

And that's the final, reduced answer. You can apply the FOIL principle to any two binomials to arrive at their product. When you have more complicated polynomials, such as those composed of 3, 4, 5, or more monomials, you do the same type of thing... what I find easiest is to take the first term of the first polynomial, and multiply it with every term of the second polynomial. Then do the same for the second term in the first one, multiplying with every term in the second one, and so on.

Like this:
(x^2 + x + 5)(x^3 + x^2 + 1)

First group (multiply x^2 with all in the second polynomial)
x^2 * x^3
x^2 * x^2
x^2 * 1

Second group (x)
x * x^3
x * x^2
x * 1

Third group (5)
5 * x^3
5 * x^2
5 * 1

Now, you just add up all these terms, and simplify where you can:
(x^2 * x^3) + (x^2 * x^2) + (x^2 * 1) + (x * x^3) + (x * x^2) + (x * 1) + (5 * x^3) + (5 * x^2) + (5 * 1)

(x^5) + (x^4) + (x^2) + (x^4) + (x^3) + (x) + (5x^3) + (5x^2) + (5)

(x^5) + 2(x^4) + 6(x^3) + 6(x^2) + x + 5

And that's it. A little more complicated, but as long as you keep track of what you're doing and work your way through it, you will arrive at the answer!

Graphing Sine and Cosine Curves

2007-10-24T19:17:00.000-07:00

As I usually include in most of my posts, please feel free to ask questions or leave comments on any material that I post, and I will try my best to address them. I recently received a request to go over how to graph sine and cosine curves.

Sine and cosine functions are PERIODIC FUNCTIONS, meaning that the function will evaluate to a repeated number after a defined length (period). You've probably seen these graphs several times before and not realized what they were. Sine and cosine graphs are virtually identical, EXCEPT that they are slightly shifted relative to one another... or they are "out of phase." This is an important distinction to make, as confusing which is which, of course, will get you the wrong answer.

The EASIEST way to remember, or to quickly plot out the graph, is to create a table of values. Let's work on the Sine curve first. The cosine one works exactly the same way. Creating this table of values will be a good review of how these trig functions work. On that note, we will evaluate for values for x that represent the angles of our special triangles, and then continue with more values at the same intervals. (I'm going to represent the square root function by "sq.rt.") If you want, you can continue to make the list longer if you wish, but I think this is sufficient to demonstrate the graph.

y = sin(x)

x |||||y
-----------
0 |||||0
30 |||||0.5
60 |||||(sq.rt.3)/2
90 ||||| 1
120||||| (sq.rt.3)/2
150||||| 0.5
180||||| 0
210||||| -0.5
240||||| -(sq.rt.3)/2
270||||| -1
300||||| -(sq.rt.3)/2
330||||| -0.5
360||||| 0

As you can probably see from looking at the list of values, the numbers seem to have a repeating pattern. If you continue to add values, they will begin to repeat themselves again. Essentially, we have gone in a full circle (360 degrees), and so that is why they repeat.

So, taking these values that we have and plotting them on a graph, we can see what a sine curve looks like:

The cosine curve will look very similar, and you can deduce it exactly the same way. As I said before, they look the same but are out of phase. The important difference to remember is that the sine curve passes through (0,0), and the cosine curve passes through (0,1), but the 'periodicity' is the same... that is, they repeat at the same regular intervals. You can extend the curve into the negative x direction, or further in the positive direction.

I hope that explains what the original poster was asking me to go over. These functions, like any other function, can also be modified by adding coefficients etc, in the same way as other functions that I've reviewed before. I will explain these concepts in a future post. Or for fun, add a coefficient to the front of the sine function (eg. A*sin(x)), or to the x value itself (sin(2x)) and create tables of values and compare the resulting graphs!

Trigonometric Identities - Pythagorean Identities

2007-09-13T20:56:00.000-07:00

In addition to the basic trig identities that I've already covered, a little bit of derivation leads to a whole bunch of new identities. In addition to the Pythagorean Trigonometric Identity I covered in my last post, there are a few other identities that can be derived from that (also sometimes called Pythagorean Identities).

If we start with the first identity:

we can divide each side by the Sine term to give something new.

[Sin^2(theta) + Cos^2(theta)] / Sin^2(theta) = 1 / Sin^2(theta).

The first term reduces to 1:
1 + Cos^2(theta) / Sin^2(theta) = 1 / Sin^2(theta)

And the remaining terms can be simplified and rewritten in terms of inverse functions:
1 + [Cos(theta) / Sin(theta)]^2 = [1 / Sin(theta)]^2
1 + Cot^2(theta) = Csc^2(theta).

More likely, it will look something like this:

Similarly, using the same steps as above (which I will leave for you to play with if you want), you can start with the first identity, and divide by the Cosine term (rather than the Sine term), to come up with the third Pythagorean Trigonometric Identity:
1 + tan^2(theta) = sec^2(theta)

There are still several other Trig identities which I will show you how to derive in coming posts. Stay tuned!

Trigonometric Identities - Basic Identities

2007-06-26T00:09:00.000-07:00

Trigonometric identities are specific equalities that express one trig function in terms of other trig functions. They are fairly straightforward, but they take some work to derive them. If you are comfortable with simple derivations, you shouldn't have any problems though. Personally, I find it easier to remember the basic set of identities, and derive the more complex ones from those, rather than trying to memorize all of them... although some people are more comfortable just to memorize them.

The basic identities are traditionally visualized with a triangle formed by a radius r, length x, and height y:

The basic trig definitions can easily be seen:
Sin(theta) = y/r..... opposite/hypotenuse
Cos(theta) = x/r..... adjacent/hypotenuse
Tan(theta) = y/x..... opposite/adjacent

If we now apply the Theorem of Pythagoras, we can see:

r^2 = x^2 + y^2

Dividing everything by r^2 gives:

1 = (x^2)/(r^2) + (y^2)/(r^2)
1 = (x/r)^2 + (y/r)^2

And then, subbing in the basic definitions, we get:

1 = [Cos(theta)]^2 + [Sin(theta)]^2

And that is the first basic identity. Nothing to it. It's proper name is the Pythagorean Trigonometric Identity. I'll rewrite it in proper notation to clean it up a bit... (Blogger is a pain with superscripts and fonts)

Another basic relationship starts with:

Tan(theta) = y/x

But, then sub in the Sine and Cosine definitions (isolated for x and y, respectively) to give

Tan(theta) = (r*Sin(theta)) / (r*Cos(theta))
Tan(theta) = Sin(theta) / Cos(theta)

And that's it again. This is called the Ratio Identity:

Those are now two of the simplest trig identities from which most of the others can be derived.

Stretching and compressing graphs

2007-06-03T19:59:00.000-07:00

If you understand how to shift a curve horizontally or vertically, stretching or compressing isn't much different. Once again, it's only a small modification to the equation that causes the stretch or compression.

Stretching or compressing a graph is determined by the coefficient in front of the x (or more specifically, in front of the other direct modifications to x).

Let's look at a basic example.... f(x) = x^2, a standard parabola.

Now, to compress this curve, you put a 'fraction coefficient' in front of the x component of the graph. i.e. f(x)= (1/2)*x^2. This squashes the graph down by a factor of 2. Or, another way to look at it, every y value in this curve is 1/2 of the value in the starting curve. Plot your own points to convince yourself of this. Note that the curve crosses (-2,4) and (2,4) in the original curve, and the new one crosses at (-2,2) and (2,2).

Now, naturally, if you put a whole number coefficient in front of the x term, you will be stretching the graph. For example: f(x) = 2x^2

You can see that this has caused the parabola to stretch upwards. Note that it now crosses (1,2), not (1,1). Or once again, to look at it from a different angle, every y value is now twice the value as in the original graph.

The only other thing that you should keep in mind is that the coefficient to stretch or compress the graph MUST be in front of any brackets that might be surrounding x, and the coefficient will act on any horizontal translation component and the exponent. Convince yourself of this by looking at graphs such as:

f(x) = (x-3)^2....... and f(x)=2(x-3)^2
f(x) = (x+1)^3...... and f(x)=1/2(x+1)^3
f(x) = x + 5...........and f(x) = 4x + 5
f(x) = (x-1)^5 + 7...... and f(x) = 4(x-1)^5 + 7

As you can see, stretching and compressing graphs really aren't that difficult. They are just an extension of what you already know, building on your knowledge of horizontal or vertical shifts. Keep practicing, and you'll get it in no time. :)

Shifting graphs horizontally

2007-05-30T22:19:00.000-07:00

Shifting graphs horizontally is slightly different, but still pretty straight-forward. Perhaps it would be helpful to review my posting on vertical shifts of graphs. Recall from that section: "Picture all the complex stuff that is happening to x as being one "chunk" of the height component, and then when you add the "+ 5" to the equation, you are really just adding an additional "height chunk" to the total height for a given x."

To shift a graph horizontally, you include the shift amount WITH x and perform whatever is being done otherwise just to x, to x and the shift amount.

Let's look at an example:

f(x) = x^2 + 4 ................. and f(x) = (x-3)^2 + 4.

You can see what I mean by including the shift amount WITH x. The 'square' function acts on the entire (x-3) term. This will cause the graph to shift 3 units to the RIGHT. This may seem somewhat counter-intuitive, but it is correct.

In this example, x-3 shifts 3 units right... if it were x+3, it would shift 3 units left. It may be easier to remember this by analyzing the x and shift amount, letting that small term equal to 0, and then solving for x. Like this:

x - 3 = 0
x = 3

OR

x + 3 = 0
x = (-3)

That shows you how far over, and in what direction, the new x values are!

I hope these postings on graph manipulations are helpful. Please leave a comment if you are unsure of something, as there are likely many other students who might be wondering the exact same thing!

Shifting graphs vertically

2007-05-29T20:55:00.001-07:00

The first graphing manipulation I will demonstrate is shifting a given graph vertically, and how you can identify this transformation by looking at the equation. This is also called vertical translation.

Let's look at a simple equation first.... y = x. This equation represents a line that goes through (0,0), (1,1), (2,2), etc. Now, what if I asked you to draw the graph represented by the new equation y = x + 2. Some students may suggest to make a table of values and then plot the points on the graph... never a bad way to do things, but not necessarily the easiest.

Let's think about things SLIGHTLY differently. In the first equation, let's think about y specifically as being the height (and not just a different variable), and so the height for a given x value will always be equal to that x value. Not a huge change in our thinking, but it may help some to see graphical changes easier. Now, let's apply this thinking to the second equation, and we see that the height is always going to be 2 greater than the x value.

Now, let's spice things up a bit. Let's look at a function (which, recall, is still the same as saying "y").

f(x) = x^2

You may recognize this equations as being the most basic equation that describes a parabola that opens up. The lowest point on the graph is (0,0).

Now, if I said to draw the graph of f(x) = x^2 + 1, apply the same type of logic, but keep in mind one VERY IMPORTANT THING. The change in height is dictated by the single number that is not associated with the x variable... this may be clearer after another example, but let us focus on this one for the moment. You should hopefully be able to see that this change in our graph will result in a shift up of 1 unit.

One final example, hopefully to clear things up for good. Let's take a complicated function:

f(x) = (x-4)^3 and f(x) = (x-4)^3 + 5

Picture all the complex stuff that is happening to x as being one "chunk" of the height component, and then when you add the "+ 5" to the equation, you are really just adding an additional "height chunk" to the total height for a given x. So in the end, the second equation above looks exactly like this one, only shifted 5 units up:

Remember to combine terms if necessary, so that you are left with a single number to add to the x term (and whatever operator is acting on it). You will quickly find that vertical translations of graphs are far simpler than they may sound at first!

Manipulating graphs

2007-05-19T19:33:00.000-07:00

In the next few days, I'm going to begin posting on graph manipulations. For any graphical expression, small changes to the expression can result in a very predictable change in the position of the graph. For instance, the graph can be shifted along the x-axis (horizontally), or along the y-axis (vertically). Similarly, the graph can be stretched or compressed along either axis as well. This may sound confusing, but the alterations to the equations are simple and easily recognizable. I will provide examples and go into detail in the coming days.

Trigonometry - Secant, Cosecant, Cotangent

2007-05-11T11:56:00.000-07:00

In addition to the three basic trig functions we've already looked at (Sine, Cosine, Tangent), there are three other related functions. These are Secant, Cosecant, and Cotangent. These functions have similar meanings as the first three, in that they represent the ratios of various side lengths of a right angle triangle, and can be used to find angles or unknown side lengths. I will not go into extensive detail on these functions, as they are less commonly required, but I will show you what they mean.

So far, with the help of SOHCAHTOA, we have seen that:

Sine = opposite / hypotenuse
Cosine = adjacent / hypotenuse
Tangent = opposite / adjacent

These new functions are related to the originals because they represent the inverse ratios.

Cosecant = hypotenuse / opposite... (compare to Sine)
Secant = hypotenuse / adjacent....... (compare to Cosine)
Cotangent = adjacent / opposite...... (compare to Tangent)

Also, these functions can be abbreviated: Cosecant = Csc, Secant = Sec, Cotangent = Cot.

At the middle school and high school math level, you will rarely have a need to use these functions, but it is good for you to know what they are. However, most trig problems at this stage can easily be solved with the original three functions.

Area of a Triangle

2007-05-07T19:05:00.000-07:00

With all this talk about triangles and their side lengths and angles, we shouldn't forget to discuss how to find the AREA of a triangle.

You are probably familiar with one formula for finding the area of a triangle:

Area = 1/2 (base)(height)

Compare this to finding the area of a rectangle:

The area of the rectangle is equal to the product of (base) x (height)..... (or length x width). However, by drawing a diagonal within the rectangle which joins two opposite corners, you can see that each newly-formed triangle is equal to half of the area of the original rectangle. Therefore, the area of a triangle is one-half the area of the rectangle, as shown by this triangle area formula. Even if you are looking at a triangle that doesn't immediately look like it is half of a rectangle, this formula still applies.

To prove it, you can draw a line in to represent the height, as I have shown here, thus creating two smaller triangles, and you can rearrange them to see that they indeed are equal to the area of half a rectangle:

That is one way to find the area of a triangle. However, if instead of base and height measurements, you are given lengths of sides or angles, this method won't work for you. In this case, you need to use a trig equation to solve for the area of a triangle.

Let's start with the first equation we had above, and modify it. By the standard trig identities, we can show that:

height = (a)(SinC)

So substituting that into our formula:

Area = 1/2(base)(height)
Area = 1/2(b)(a)(SinC)

And this is the trig formula for solving the area of a triangle!

Area = (1/2)abSinC

You can use this to find the area of a triangle where you know any two sides and the angle between them! It's that easy!

Special Angles in Trigonometry

2007-05-01T21:18:00.000-07:00

The values of trig functions of specific angles can be represented by known ratios, and are good to commit to memory to help you work through problems faster. These 'special' angles can be remembered by examining two different triangles. Specifically, the trig functions for angles of 0, 30, 45, 60, 90 degrees are the special ones.

Let's look at the first triangle, and this will hopefully become clear. Take a right angle triangle with two 45 degree angles, and with sides of 1 unit length. By the Theorem of Pythagoras, the hypotenuse of this triangle is of length √2. This is what this triangle looks like:

So then, from these values and SOHCAHTOA, you can obtain the trig values for this special angle of 45 degrees. You can see that:

Sin(45) = 1/√2
Cos(45) = 1/√2
Tan(45) = 1

Don't worry if you can't remember these exact ratios... the simplest thing to remember is how to construct the triangle... which is as easy as remembering a right angle triangle with a 45 degree angle and 2 sides of length 1... you can easily fill in the rest, and then work out the trig ratios yourself!

The second triangle to remember is slightly more complex, but still straightforward. Take a right angle triangle with angles of 30 and 60 degrees. The simplest lengths of the sides of this triangle are 1, 2, √3 (with 2 being the longest side, the hypotenuse). This triangle looks like this:

So then, using the these values, you can obtain the trig values for these special angles as well:

Sin(30) = 1/2
Cos(30) = √3/2
Tan(30) = 1/√3

Sin(60) = √3/2
Cos(60) = 1/2
Tan(60) = √3/1 = √3

Again, just remember how to construct the triangle, and the ratios are easy to come up with!

For 0 and 90 degrees, there isn't a simple triangle scheme to remember the values (although please feel free to correct me if I am wrong!). However, these aren't scary square root numbers or weird fractions:

Sin(0) = 0
Cos(0) = 1
Tan(0) = 0

Sin(90) = 1
Cos(90) = 0
Tan(90) = undefined

If you can familiarize yourself with all of these special angles, or at least understand how to derive them, then you will have a much easier time working on trigonometry questions.

The Theorem of Pythagoras

2007-04-30T14:15:00.002-07:00

The Theorem of Pythagoras is a specific case of the Cosine Law that applies specifically to right angle triangles. With it, and given any 2 sides of a right angle triangle, you can find the third side. Then having solved all the sides of the triangle, you can use the standard trig identities (Sine, Cosine, Tangent) to evaluate the size of all the angles.

Specifically, if we have a right angle triangle with sides a, b, and c, with c being the hypotenuse, we can write:

c^2 = a^2 + b^2

Or: "The square of the hypotenuse is the sum of the squares of the other two sides."

This is one of the most basic trig concepts, and is probably one of the first concepts that were taught in trig... very important and very useful!

Trigonometry - Cosine Law, Theorem of Pythagoras

2007-04-27T14:10:00.003-07:00

The Cosine Law works similarly to the Sine Law that I have already discussed. The Cosine Law is the general form of the Pythagorean Theorem, which itself applies strictly to right angle triangles. Therefore, this law allows us to work with any triangle. It's a bit more of an equation to remember than the Sine Law unfortunately, but it is extremely useful. Here is the equation:

c(squared) = a(squared) + b(squared) -2abCosC

(I don't see a way to type superscript text on blogger, unfortunately.)

So, as you can see, the Cosine Law is useful for finding the third side of a triangle when any 2 sides and the angle between them is known. Let's try an example:

So by the Cosine Law:
c^2 = 6^2 + 8^2 - 2(6)(8)Cos(60)
c^2 = 36 + 64 - 96Cos(60)
c^2 = 36 + 64 - 48
c^2 = 52
c = 7.2

Now that you have done that, you have obtained a complete ratio to use with the Sine Law! And now, you can find the rest of the angles to fully describe the triangle!

Sin(60)/7.2 = SinB/8
and
Sin(60)/7.2 = SinA/6

I'll leave those for you to solve. But that's it! Using a combination of the Cosine Law and Sine Law, you can completely solve any triangle that you are given. They are extremely powerful and useful equations!

Also, on a side note... as I mentioned at the start of this post, the Cosine Law is a generalization of the Pythagorean Theorem, which specifically applies to right angle triangles. You can see that if you are working with a right angle triangle and substitute in 90 degrees to the Cosine Law, it reduces down to the Pythagorean theorem:

c^2 = a^2 + b^2 -2abCosC
but
Cos(90) = 0... so the 2abCos(90) term reduces to 0
and so
c^2 = a^2 + b^2... The Pythagorean Theroem! :)