Prior to Calculus: Trigonometric Identities

Hey, guys. Trigonometric identities never go away. Ever. If you need to do anything with math, this is sure to come up at some point. They are so important to learn and use properly! There are many different ones, so I couldn’t possibly list and teach all of them to you. There is a whole course devoted to trig, that many people take in highschool. I’ll show you the ones that I use most often, and all the ones that I can remember off the top of my head. I’m only going to show you the ones I can remember without looking them up, because that is obviously the ones that are most useful to me. If I need to know it, I would have remembered it by now.

So, trigonometry isn’t just about triangles. It’s about the relationship between the angles and lengths of the sides. We describe these relationships using trig functions. There are six of them, and you should become comfortable with them all. You can do some really cool things with them. They come up in derivatives, integrals, and can be used to simplify answers in many ways.

Reciprocal Identities

These ones are fairly easy to remember, as there is a symmetry about them. Here’s my trick: List out the 6 trig functions. The two on the outside go together, the second one and the second to last one goes together, and the two middle ones go together. Let me show you how I picture it. Write them out vertically, and draw lines to match them up. This will tell you that sin goes with csc, cos goes with sec, and tan goes with cot.

See? If you are ever in doubt, just write this out very quickly. You’ll never get it wrong!

Tangent and Cotangent

These two identities are extremely important, and not very hard to remember. Keep them in mind, because you will definitely need them.

Other Important Identites


These three are also very important to know. The first one comes up most often, and the third one the least often. However, I’ve seen them all come up often! Remember that these last three use the squares of trigonometric functions. Try not to mess that up.

How are they used?

I’m going to teach you the technique I use to simplify complex trigonometric functions. You see these often throughout calculus. I’m not going to show you specific problems, just big expressions that I’ll simplify. You need to be on the lookout. If you have an expression with many different trig functions inside it, try to think if it can be simplified. Follow my method, and see what you can do. Sometimes you’ll be surprised, and things can simplify dramatically.

Step 1. Look for obvious simplifications. That should be your first step all the time. If you can blatantly see one of the identities, simplify it right away. Don’t skip over them just to follow my process. I tend to have a gut feeling about simplifying expressions, because I have a lot of experience with it. If you feel like something might work, go with that gut feeling. It’s good to explore these identites and try out things on your own.

Step 2. Convert everything to sin’s and cos’s. Do this using reciprocal identities and tangent and cotangent identities. Now, you can see if you can combine like terms or eliminate fractions.

Step 3. Now, do all the cancelling that you can! Use factoring and the last three identities to simplify the expression.

Step 4. The last step is to convert back to the most proper expressions. For example, if you have a sin in the denominator, move it to the numerator, making it a csc. Convert back to tangents and cotangents when necessary, and other things like that. Get rid of as many fractions as you can. That’s how you make it look clean. That’s what the final answer should look like.

Here are two examples of complex expressions, that I’ll simplify as much as possible. There are many other things that happen, but I can’t show every type of example. There are too many. Following this process with help you simplify just about any trigonometric expression, which is necessary in very many situations!

Immediately, I don’t see any easy simplifications. So, I convert everything to sin’s and cos’s. Now, I see that the numerator can be turned into just a sin squared, because of one of the identities. Also, the cos is in the denominator of the denominator, and we can fix that. When you divide by a fraction, it is the same as multiplying by the reciprocal. So, if you flip the denominator, and multiply by it, the cos moves to the numerator, and the sin squared stays in the denominator. Then, there is a very easy cancellation of the sin squareds, and you are left with just a cos!

In this expression, something sticks out to me. I’m always on the lookout for seeing a sin squared and a cos squared near each other. If they are being added together, you can cancel them out using a trig identity. They become just a 1, and it is so nice to simplify a large expression into something so small! So, the first thing I did was factor the cosecant, to get the sin squared plus cos squared on its own. Then, I cancel them, using a trig identity. The next thing I did was make everything a sin or cos. After that, I see one more cancellation. Then, I converted the cos in the denominator to a sec, just to make it a little cleaner looking. I hope you could follow me!

I hope this has helped you! As always, leave a comment below or email me at with any questions or comments.

Prior to Calculus: Polynomial Division

If you’re having trouble with this subject, you’ve come to the right place. If you don’t know what this is, and you’re curious, this is also the right place for you! Polynomial division is fairly important, and is used throughout calculus.

What is polynomial division?

Polynomial division is when you divide one polynomial by another. A polynomial just means an expression where there are multiple powers of x. Here are a couple of examples that require polynomial division.

Long Division

Long division is really important to know, because it can be used to divide any two polynomials. There are no exceptions or rules to this. Synthetic division is a nice trick to know, but it is only for a single type of polynomial division. Long division covers them all. Let me set up a problem to show you the technique of long division. I’ll explain it all, step by step. For some reason, you would like to do the division in this expression:

First, we need to write this expression in long division form. You write it the same way you do normal long division, as if they are only simple numbers. I’ll show you what I mean very soon. There is one catch to this. You need to use place holders if you have powers of x that are missing. The dividend, the thing on the inside being divided, needs to have all powers of x represented in decreasing order. The first term must be of the highest power of x that is in the expression. The second term must be one lower power of x. If there is none of that power, you put a zero. In this first example, there is no x². I’ll show you how it’s done.

This is the correct format that needs to be used when doing long division. So, what is the first step? We need to look at the divisor, the thing outside of the box, on the right side. In this case, x – 2. In the divisor, you look only at the highest power of x. In this example, we have only an x. The coefficient is just 1. So, we look at the x. What times ‘x’ is equal to the first term of the dividend? In other words, what times x equals x³? This is pretty simple, and the answer is obviously x². This is the first term of the quotient (our answer). Write this on top of the line, exactly above the first term of the dividend.

Now, like in normal long division, we multiply this first part of our answer by the divisor, the expression on the right side. Then, you subtract that from the dividend! After this subtraction, you bring down the next term in the dividend, and repeat the process. I know this seems complicated when you are reading it for the first time, but I’ll show you what I mean.

See how I did that? Now we repeat the process until the first term of the divisor can no longer “go into” the term we are looking at in the dividend. After that, I’ll explain what we do. For now, look at the first term of the divisor, the x. Of this new expression, 2x² – 5x, look at the first term. What times x equals 2x²? Then we just continue doing what we just did. I’ll complete the rest of the process.

Now, I’ve run out of terms in the dividend. There is nothing more to “bring down”, therefore there is nothing more that x can “go into”. I must stop dividing, and reveal my answer. The solution to the division is everything that is on the top line. That is your answer. However, there is one more part that we cannot forget. This is exactly the same as regular long division. When you run out of things to bring down, you must look and see what your remainder is. You remainder is whatever is left over after you do the last subtraction in the long division. Here, we have 0 left over. This means we do not have a remainder. If you have a remainder, you must adjust your answer. You need to add one more term to the expression on the top line. This term is the remainder divided by the divisor. If we had a remainder of 3, for example, we would add .

The final answer can be written like this.

I’ll do one more example for you.

There you go! This technique can be used to divide any two polynomials. Try it out, as it can be useful in several situations in calculus!

Synthetic Division

Alright. You are proficient in long division now, right? You’ve got that down? Now you want a trick? Is that what I’m supposed to do now? Okay, well I have one for you. This is called synthetic division. Now, it only works in a specific situation, but it can be extremely useful sometimes. If you don’t believe me, check out this post where I needed it to solve the problem. I actually give the full explanation of synthetic division in that post, but I’ll do it again here just so you guys can see it together with long division. The post was a question submitted by a reader, asking about how to take the limit of something fairly difficult. Click here to go right to the post and see how I used synthetic division.

So, when can we use synthetic division? A condition must be met in order to use it. We need to look at the divisor. This is the expression we are dividing by. This needs to be a single x plus or minus a constant number. This could be x – 3, or it could be x + 10. It doesn’t matter if it is a plus or if it is a minus. It just needs to be one of these simple forms. Let me show you how it is done through an example.

Now, there are a few similarities to long division. First, you need to use place holders in the exact same way. If you don’t have any x² term like in the example above, you need to put a 0. Also, the last number on the bottom is the remainder, and is treated the same way as it is in long division. The remainder over the divisor is added to the solution, in order for it to be correct. If the remainder is zero, you can ignore this step. Okay, so below I will show you the way to set up synthetic division. First, you must think of the divisor, the x – 3, always in this for. x – a. a can be anything, even a negative number. If the divisor were x + 5, a would be -5, right? Place -3, the a, in the upper right hand corner. Then, in a straight row, right the coefficients of each term. If there is zero of a term, make sure to write 0. Below is the proper way to set up the division. Leave room for a second row under the first, and draw a line under that space.

To begin, bring the first term in the top row straight down.

Next, multiply the number in the top right corner by this number you just brought straight down. In this example, multiply the 3 by the 1. Write this number in the second row underneath the next number of the first row.

Add the two numbers that are now right above/below each other, and write this number underneath them.

Now, repeat the process again. Multiply the upper right hand number by this new bottom number. Add that to the next number of the top row. And repeat until there are no more numbers in the top row. The final number in the bottom row is the remainder. Place this in a box, and leave it off to the side where it belongs.

Not including the remainder, these are your new coefficients! The new polynomial has the highest power of x that is one less than the highest power that was in the dividend. In this problem the original highest power of x was x³, so the new highest power will be x². Just place on the coefficients, and add the remainder if necessary. This is the solution!

I believe that I explained this better in the solution I posted to a reader’s question. I encourage you to check out that page if you have any remaining questions. Click here if you’d like to view that post. I’ll show you one more example.

Okay! Now you know how to divide polynomials. But, do you know how to check if your answer is correct? Let me teach you! To find out if the answer is correct, multiply the answer by the divisor, and it should equal the dividend. If it does not, you have messed up. If it does, then you have done it correctly! It is as simple as that. If you have any questions or comments, please post them below or send me an email at Thanks and good luck!

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