4.7 Inverse Trigonometric Functions

We have learned previously that in order for a function to have an inverse that is also a function, it must pass the horizontal line test. In other words, that function must be one-to-one, meaning that each point in the domain of the function must correspond to a unique point in that function's range.

When examining the graph of , or any other trigonometric function, we can see that the function is periodic, and therefore not one-to-one.

The inverse of would appear like so:

This is not a function. However, by constraining the domain of our function to , The graph of becomes becomes one-to-one, and therefore its inverse, shown below, is also a function.

This is known as the inverse sine function, and is denoted by either or .

In the same vein, an inverse cosine function stems from the parent function , but unlike the sine function, the domain of the cosine function is restrained to in order to maintain the function as one-to-one. The result is displayed below.

By constraining the function by the interval , an inverse tangent function may also be defined.

2.5 Fundamental Theorem of Algebra

This theorem is easy to understand: the highest degree in a polynomial function equates to the total number of zeroes that that function has. Before reading this, most likely you were told to find all the real or rational zeroes of a function. Sometimes the only zeroes a function has are real, sometimes not. Sometimes the zeroes are imaginary.

Here is an example:

The given zeroes of x for f(x) are:

A casual glance at this might make one think that this is, at the very least, a cubic function (a polynomial whose highest degree is 3). But that would be a wrong assumption to make. The presence of the 3-i automatically implies the presence of its conjugate, 3+i. So this function is at least a quartic function, and the zeroes actually are:

The easiest function that could be formed from these zeroes would be:

The easiest thing to do would be to then to multiply the lone x and the (x-2) together, and then to multiply the two imaginary zeroes together:

To deal with the imaginary zeroes, we have two options. We could simply distribute all of the variables in the first imaginary zero to the other, but that is tedious and has a high potential for error. Instead, we will ignore the presence of the x's in the imaginary zeroes:

Then, we merely add up these numbers to get a middle term, and then multiply/foil them to get the last term. To demonstrate:

So now our equation looks like this:

All done! One more thing: a problem asking for all of the x-intercepts and a problem asking for all of the zeroes are not asking for quite the same thing, although they're very, very close. Imaginary zeroes do not appear on a normal x-y graph, so when a problem asks for all of the x-intercepts, it is another way of asking for all of the real zeroes. In other words: you don't need to bother finding the imaginary zeroes. But if the problem wants all of the zeroes, then you have to find all of them, both real and imaginary.

2.4 Complex Numbers

You may have learned in previous math classes that there is no taking the square root of a negative number. But your other teachers kind of lied to you about that. In fact, you can take the square root of a negative number. But your answer is going to be some quantity of the imaginary unit i.

In this same vein, the square root of -36 would be 6i.

The standard form of i is a + bi. a is some real number, and bi is just a quantity of i. This is a type of complex number. But be aware that any real number can be a complex number if you assume that b = 0!

This brings us to our next order of business - working with i. In adding and subtracting, you can really just treat i as any old variable. i can be negative.

(5 + 2i) - (6 - 7i) = ?
5 + 2i - 6 + 7i
5 - 6 +2i + 7i
=
-1 + 9i

Multiplication and division is slightly different. Use the same steps until you get to the end. Then look at your i term.

Now, this will be true not only for these four exponents, but for any exponent of i. But because there are only four options, you can figure out which applies to your exponents.
Take your exponent and divide it by 4 (the number of options we have). Go until you have a remainder. This remainder should be either 1, 2, 3, or 4; that number represents which of the original four (illustrated above) applies.

One more thing. Sometimes when you're dividing or just when you're being asked to put a number into standard form, you'll have to deal with an ugly fraction. That's where this magical thing called a conjugate comes in. It's pretty easy to use. Take your denominator and get that into standard a + bi form if it's not already. Then swap the plus sign for a minus sign (or, if you've got a minus sign, swap it for a plus sign). Do NOT make changes to the positivity or negativity of anything else. Put your new standard form in both the numerator and the denominator of a new fraction and multiply across with your original ugly fraction. When your fraction reduces, you'll get your standard form.