Inverse Trigonometric Functions - Arcsine, Arccosine, Arctangent

Inverse trig functions are a core concept in trigonometry.  Specifically, they are named arcsine, arccosine, and arctangent.  We have already discussed how to find things such as the sine of 25 degrees, or cosine of 71 degrees.  However, we need to introduce a new math concept to figure out such problems as "what angle gives a sine value of 0.2?"  This is where inverse trig functions come in. Think of their relationship to the standard basic trig functions as being comparable to that between multiplication and division.  You do one operation, and then the opposite to go the other way and undo the first operation.

You may be taught these in school slightly differently.  Instead of using the names arcsine, arccosine, and arctangent (arcfunctions!), it is also very common to see these represented as the basic trig notation with a -1 exponent, like these:

arcsine = sin^-1

arccosine = cos^-1

arctangent = tan^-1

These are likely the symbols that you will see on your calculator.  Typically, they are the shifted function of the regular sin, cos, and tan buttons.  It is important to note the distinction between the basic trigonometric functions and these new inverse trig functions.  When using the basic trig functions, the value you are obtaining is the ratio of the two relevant sides of the triangle for the given angle.  When using the inverse trig functions, what you are solving for is the actual angle that produces the given ratio of sides.  So, make sure you push the right button on the calculator!

The most basic way of finding an inverse function, in general, is to take your given function and switch the x and y, and then rearrange.

f(x) = x² + 5
y = x² + 5.... now switch x and y
x = y² +5.... and rearrange
y² = x - 5
y = sqrt(x-5)

And there you have determined the inverse function.  This is the same strategy that is being applied when we are talking about inverse trig functions.  However, having the inverse trig buttons on our calculators really take all of this extensive and possibly difficult rearranging and calculating out of the picture.

Here is a basic example of one of these inverse trigonometric functions.  Hopefully you will see that they are extremely easy to work with.

Find the angle for the given trig ratio:
sin(θ) = 1 /√2
θ = sin^-1(1 /√2)
θ = 45°

Hopefully this introduction to the inverse trig functions has been useful for you.  There is a lot more information about this math concept that is probably quite beyond the scope of what is necessary to actually solve most of the basic trigonometry questions you will find in your maths homework.  I have found several good resources, if you would like to learn more about inverse trigonometric functions.  They can be found on the ubc.ca math website, or on The Math Page (among many other great math resources out there).  Please leave a comment below if you know of any other great sites that you would like to share with other readers. Also, once again, please hit the +1 if you've enjoyed this post!

Piecewise Functions Explained

The last post introduced you to functions and how they work, but there are few other things worth discussing - first I'll talk a bit more about function notations, and then after that, I will go over piecewise functions.

When I introduced functions, I referred to the standard notation as f(x). However, you will quickly come across problems that use different letters, but they mean the same things. You will probably see things like g(x) or h(x), but they are just used to differentiate between different functions. An example will probably clear this up:

f(x) = 2x + 3
g(x) = x - 8
h(x) = 4x + 1

That shows that function f (or g, or h) is equal to the given expression. So f(x), g(x), and h(x) all represent different functions. It's the equivalent of saying 3x = 12, or 5a = 10... the math is the same, just with different letters.

So, with multiple expressions, we can do slightly more complicated things. If we say solve f(1), we substitute 1 in for x, and solve it to get 5:

f(x) = 2x + 3
f(1) = 2(1) + 3
f(1) = 5

However, we can now ask for something like f(g(x))... read as "f of 'g of x'." Looks complicated, but keeping in mind what the function notation is saying, we can easily solve this by substituting the expression for g(x) in for x (instead of just subbing in a single number like before):

f(x) = 2x + 3, g(x) = x - 8
f(g(x)) = 2(x - 8) + 3
f(g(x)) = 2x - 16 + 3
f(g(x)) = 2x - 13

Furthermore, if we wanted to solve f(g(5)), we can do one of two things. We can evaluate the expression as we just did, then solve when x= 5... or we can solve g(5) first, then substitute that into f(g(x)):

Method 1:
f(g(x)) = 2x - 13
f(g(5)) = 2(5) - 13
f(g(5)) = (-3)

Method 2:
g(x) = x - 8
g(5) = 5 - 8
g(5) = (-3)...
then...
f(x) = 2x + 3
f(g(5)) = f(-3) = 2(-3) + 3
f(g(5)) = (-3)

Both methods get the same answer, because they are doing the same things. It's just a matter of what you are more comfortable with. Function notation can get very confusing like this, but for problems like this, you just have to basically work from the inside out, just as in any other math expression you've seen! (eg. (2x + 7(3x -4) + 2).... solve the inside bracket first, then the outside). They may seem tricky, but you will be surprised at how easy they are when you see through the notation and know what they are asking!

Another point to discuss is functions that are not described by a single equation, but remain continuous. Remember back to what we said a function is... a way of relating two variables, where each number in the domain corresponds to ONLY ONE number in the range. Also remember the vertical line test. But we never said anything about it being entirely expressed by one equation! So then, we can say that the following is also a function:

Obviously, it is composed of two straight lines, each having their own equation. This is called a piecewise function. Since it is just one line with two distinct 'pieces,' it is a single function, not two, and it is written as:

That says that f(x) is described by two separate equations, over the domains of x stated after each equation.  Given this graph or equation, if you were then asked to solve f(2), you must look to see within which domain x = 2 falls, and evaluate that expression. For this example, f(2) is in the top equation, so f(2) = 3. You can check by looking at the graph to see that when x = 2, the function evaluates to (y equals) 3. Alternately, f(4) is within the bottom expression, so f(4) =3/2.

Try to remember that as difficult and tricky as some functions look, they aren't that difficult if you relate to them what you know already about solving math problems and you pay attention to what is being asked! Remember, if you need something clarified further, don't hesitate to post a comment!

Math Carnival Submissions

The Math and Multimedia Carnival is coming up again soon, and the next one will be posted here on Math Concepts Explained!  It will be the 21st edition of the carnival, so hopefully it will have a great turnout by the carnival contributors.  The number 21 obviously has a lot of significance in all sorts of ways, so hopefully the 21st Math and Multimedia Carnival will have a wide variety of interesting articles to mirror the many different meanings for this special number.  I'm looking forward to hosting it and can't wait to see what great articles are going to be submitted for it!  Before that, I just want to briefly explain the math carnival submission process.

For those of my readers who are interested in submitting an article to the 21st Math and Multimedia Carnival that will be on my website, you can either submit your contribution to the carnival submission form site, or email it directly to me at mathconcepts101[at]gmail[dot]com.  Feel free to comment or ask me questions at this email address as well.

The carnival will go live on Monday, March 19, 2012.  I would appreciate if any submissions that you want considered for the carnival could please be submitted by Friday, March 16 (either through the blog carnival submission website linked above or emailed to me).

Personally, this is the first time I've ever hosted the Math and Multimedia Carnival.  Blog carnivals are new to me, so hopefully I'll be able to do a good job at it!  On that same note, I have also recently submitted my first article to a blog carnival, which can be found on the latest (20th) edition of this same math carnival.  

For those who are like me and are new to blog carnivals, Guillermo (who maintains this carnival, and is also the owner of the website Math and Multimedia) has provided several good resources.  Some of these are:

• "What is a Blog Carnival" is a very good overview that describes what they are and what can be expected.
• Index of all the previous carnivals.

So, here's hoping for another successful Math and Multimedia Carnival!  Remember to get your submissions in on time, and remember to come back to Math Concepts Explained on March 19 to see all the new articles!

Square Roots - Part II (The Irrational Number)

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 Part III - Factoring Square Roots!)

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 when I explained square roots in my 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 how to factor square roots and express them in simplified radical form.

Square Roots - Part III (Factoring Square Roots)

Where I left off in my previous post about irrational numbers, we were trying to solve for the square root of 24 by trial and error.  Reducing an irrational number can be a tedious job if you do it this way!  Luckily, unless specified otherwise, you are allowed to leave your answer in the "simplified radical form," which you can get by factoring square roots.

"Simplified radical form" is exactly what it sounds like: you simplify your expression and leave it expressed as some radical.  But, you have a little bit of work to do to reduce it.

Factoring square roots is quite simple.  It has to do with factoring perfect squares.  But first, you have to determine the factors of the number under the radical sign, and then if any of those factors are perfect squares, you can pull it (the square root of the perfect square factor) out from underneath the radical sign and put it in front to multiply by it.  That sounds awfully wordy and probably isn't the most concise definition, but I think an example will go a long way to helping you understand factoring square roots, and then leaving them in simplified radical form.

Let's continue with the example from my last post, looking for the square root of 24.  In this case, let's just reduce it to its simplified radical form, and not bother wasting time trying to find the exact decimal answer.

So, the first step is to ask yourself what the factors of 24 are.  (I'm going to ignore the √ for a minute)  For this, you can determine this quite easily by trial and error.  You can find that the factors of 24 are 1,2,3,4,6,8,12,24.  From these now, you want to see if any are perfect squares.  Again, you should be able to easily say that 4 is the perfect square.  So, 24 can be expressed as 4 x 6. That's what we want.  Now, let's look at what we have done so far:

√24 = √(4x6)

Now, as I said before, we can take the perfect square that is under the radical sign, and bring it outside the sign.  This is because of the properties of square roots (property #2), which in this case allows us to write:

√(4x6) = √4 x √6

With that property in mind, it should be a bit easier to see why we are interested in the perfect squares, because now, by factoring perfect squares, we can just rewrite the square root of 4 to 2!  So, we can finally write our expression in simplified radical form as:

√24 = 2√6

I hope this demonstration has explained to you the basics of factoring square roots, and leaving you answers in the simplified radical form.  Let me know in the comments if you'd like another example, and I'll do one for you!  Remember to +1 me if this helped you!  :)