Trigonometry: Evaluating Base Angles - The Hand Method

Notice: On 12/6/2012, I uploaded a YouTube video that corresponds to this blog post. You can find it by following this link.

Most Trigonometry students are encouraged (and sometimes required) to memorize the "base angle" evaluations for the three fundamental trigonometric functions. These five core angles (in degrees) are 0, 30, 45, 60, and 90. The three fundamental trig functions are sine, cosine, and tangent. So students spend hours attempting to memorize 15 values... or they don't memorize them and do poorly on the test.

Of course, a calculator will give you these values quite easily. Memorizing them is useful though because (1) these angles show up frequently in introductory trigonometric and calculus courses and (2) having the answers memorized saves time on exams and homework. But memorization is a chore and is difficult for most students. That's why I really love this "hand method" of remembering (or rather, determining) these values. I did not create this--I learned it while taking Pre-Calculus my junior year of high school, and I've been using it ever since!

Each finger on the left hand with the palm facing towards you (thumb up, pinkie down) represents one of the five base or core angles. Visualize your hand as the first quadrant--the pinkie is in the 0 degree position and the thumb is in the 90 degree position. Then it is easy to see that the ringer finger is the 30 degree finger, the middle is the 45 degree finger, and the pointer represents the 60 degree finger. Here is a visual with angles in degrees and radians. Knowing the fingers in both measurements will save you the time it would take to convert all radian measures into degrees!

Once you have the fingers identified, we can begin evaluating the base angles, those found in the first quadrant. We will learn later in the post how to evaluate angles that are constant multiples of these base angles.

First, let's look at evaluating the cosine or sine of one of these core angles. In my visual below, I've chosen to evaluate at 60 degrees. Step 1 is to identify the finger that corresponds to the angle. Here, the pointer finger represents 60 degrees. Step 2 is to lower that finger, just fold it down, as in the image below. To find the cosine of this angle, simply count the number of fingers still standing above the lowered finger. Then, take the square root of this number and divide by 2. This is the cosine of the angle. To find the sine of this angle, count the number of fingers still standing below the lowered finger, which is 3 in this case. Take the square root of this number and divide by 2.

Easy enough, right? Evaluating tangent is simple too. You only have to do one extra thing. After lowering the required finger, flip your hand over so that the pinkie is on top and the thumb is on bottom. The fingers still represent the same angle values. Avoid confusion by lowering your finger first, then flipping your hand. After flipping your hand over, to find the tangent of the angle, count the number of fingers above (which is 3 here) and the number of fingers below (which is 1). Take the square root of the number above and divide by the square root of the number below.

See, I told you it was just as easy! Only one extra step, flipping the hand over. You can quickly evaluate any of these base angles at any of the three fundamental trig functions using this method. I've compiled a few for you to try. Evaluate these trig functions at the given angles using the hand method. Don't convert between radians and degrees--practice learning the fingers in both units. After you've evaluated all of them, use a calculator to check your answers (your calculator will give decimal approximations; enter your answer to see if the decimals match). Make sure your calculator is in the right mode for the type of angle you are evaluating!

You can also see the answers quickly by clicking here.

It is also possible to evaluate the trig functions at any constant multiple of the five core angles using this method. There is just an extra step involved, and that is to determine the sign (positive or negative) the answer will have. You know that all angles can be found on the unit circle. The unit circle contains 360 degrees (or 2pi radians). Angles between 0 and 90 degrees are in quadrant I of the standard x,y-plane; angles between 90 and 180 are in quadrant II; angles between 180 and 270 are in quadrant III; and angles between 270 and 360 are in quadrant IV.

But there are angles greater than 360 and angles less than 0. We discussed coterminal angles in another post. Review How to Find a Coterminal Angle in that post.

The first step to evaluating a trig function at a multiple of one of the five core angles is to determine which core angle is multiplied. To do this, find the angle that is coterminal with the given angle. For example, given the angle -405, the angle coterminal with this is 315 degrees. This angle is a multiple of 45 degrees: 315 = 7 * 45. Find this by trial and error--divide the angle by each of the five base angles. Whichever base angle gives a whole quotient, that is the base angle you are looking for.

The second step is to evaluate the trig function at the base angle you found.

The third step is to determine the quadrant of the original angle. Since -405 is coterminal with 315, and since 315 lies in the fourth quadrant, I know that -405 lies in quadrant IV.

Now, you can determine the sign of your final answer. Refer to the diagram below:

This little phrase will help you remember which trig functions are positive in which quadrants: All Students Take Calculus. "All" stands for, well, all. That is, in quadrant I, all of the trig functions are positive. "Students" stands for sine; so only the sine function is positive in the second quadrant. "Take" stands for tangent; so only the tangent function is positive in the third quadrant. And "Calculus" stands for cosine; so only cosine is positive in the fourth quadrant.

In my example, -405 degrees lies in the fourth quadrant where only cosine is positive. If I'm evaluating cosine at this particular angle, the answer is positive; if I'm evaluating sine or tangent at this angle, the answer is negative.

Here are a few examples:

In each instance, the steps are the same. First, identify the base angle by finding the angle that is coterminal to the given angle. Divide the given angle by each of the core angles to determine which angle to evaluate. Then determine which quadrant the angle is in and use it to determine the sign of the final answer.

Practice this method of angle evaluation enough, and it will be very helpful to you in future classes. After trigonometry, you will probably use it in calculus and maybe other subjects as well. You can impress your teachers or classmates with the speed at which you can answer evaluation questions. Of course, it only works for multiples of the five core angles, but you will see that these angles appear frequently in homework and test questions because trig functions evaluated at these angles yield nice numbers.

Edit: This post was edited on 8/16/2012 to correct a mistake in the "Examples" photo.

Edit #2: This post was edited on 12/6/2012 to add the notice at the top.

Calculus: Rates of Change in Other Sciences

There are many uses for derivatives in the natural and social sciences, including physics, chemistry, biology, and economics, just to name a few. You will encounter a lot of word problems dealing with these subjects and how they use derivatives. Most people become weary right after reading the problem because the physics/chemistry/biology/economics wording can make the question and information difficult to understand. Don't let that scare you off, though! You don't need to know these other subjects in order to work these word problems, and I'll show you why.

First, let's talk about the typical format. You will (almost always) be given an equation with variables that represent a numerical value in that field of study. Then, you will be asked a specific rate of change question, either to find a formula for the rate of change, or to find the rate of change at a particular variable value. Sometimes you will be asked to explain what the rate of change means in the particular situation.

That's it! Doesn't sound too hard, does it? Let's look at some examples. First, a physics problem.

I know this is a loaded problem, so we'll take it one step at a time! First, notice the format. You don't need to know anything about physics to work this problem, but you do need to understand derivatives. We are given an equation that represents the particle's position to the time. We know this because the variable x is given in meters, which is a position unit.

In the first question, we are asked for the velocity at time t. This means we need to find a formula for the rate of change of position with respect to time (see your textbook for the definition of velocity). We simply need to find the derivative of the position function with respect to time.

I'm not going to show you part (b) because it is simple. To find the velocity at a particular time (any time, but in this case, 3 seconds), simply plug 3 in place of t in the velocity equation you just found. That's simple algebra.

Part (c) might be a challenge if you've never been faced with something like it before. The term "at rest" means just what it sounds like--"not moving." We are asked to find when the particle is not moving. If a particle is at rest, then its position is not changing. Therefore, the change in s is 0, but time is still passing, so the change in t is a value other than 0. So then, it's velocity is 0 divided by some number--which is 0. Thus, when a particle is at rest, its velocity is zero. All we need to do here, then, is find when the velocity of this particle is zero.

(See image above.) Anytime you are solving for a variable that appears in a quadratic equation, you have to use the quadratic formula to find the value. In some cases, you can factor the quadratic equation, but more often than not, it's faster to use the formula than to try and guess the factors. Also remember that when using the formula, you will obtain two answers--make sure both answers make sense! In this case, both times are positive, so we should report both answers, but if one were negative, we could ignore that answer.

Part (d) might also be a challenge for you. Again, you don't need to know physics to solve this problem, you just need to use your brain! Let's think of an example situation. Envision a particle on top of a ruler at the 0 meters mark. That particle moves forward 1 meter in 1 second. The particle's average velocity is the (second position minus the first position) divided by the total time,which comes out to (1-0)/1, or 1 m/s, and the value is positive. Now the particle moves backward 1 meter in 1 second. The average velocity now is (0-1)/1, or -1 m/s. It is negative because the particle moved from the 1 meter mark to the 0 meter mark, instead of the other way.

I hope you can see now that when a particle moves forward, it has a positive velocity, and when it moves backward, it has a negative velocity. In this way, velocity is different than speed.

For this question then, we need to find at what time interval the particle's velocity is positive. To do so, we can look at the graph of the velocity function, or we can find the roots of the velocity function and then determine where the velocity is positive. Notice that we already found the roots of the velocity function in part (c).

(See imagine above). I like to use a chart/table for this. In the first row, divide your time into segments using the roots you found in part (c). Then, in the next row, input the sign that the velocity has in those time intervals. For example, to find the sign for the first interval (t < 2), I input t = 0 into the velocity equation and get v = 36, which is positive. For the second interval, I chose t = 3 and got v = -9. You can choose any number greater than 6 to find the last sign. Then, since we want to know when the particle is moving forward, we answer with the intervals in which the velocity is positive.

In part (e), we are asked when the particle is speeding up and slowing down. This question is asking about the acceleration of the particle, which is the rate of change of the velocity of the particle with respect to time. Similarly to velocity, speeding up occurs when the acceleration is positive, and slowing down when the acceleration is negative. So, these answers can be found in a similar way to part (d), with the only difference being the equation we use. We first have to find the acceleration function (which is the derivative of the velocity function), and then find the roots of the acceleration function before finally using a chart to find the intervals desired.
That wasn't too hard, right? Now let's look at a problem from economics.

First, a few words about the cost function. The cost function describes how much it will cost to manufacture a number of goods. Notice that as x increases, that is, if the company produces a large number of jeans, the cost increases less and less. We know this to be true--we know that if a company produces 250 light bulbs, it will be cheaper per bulb than if they only produced 50. This is the relationship that the cost function describes. In this little scenario (using the function from the problem), it costs $2200 to produce 50 lightbulbs, which comes out to $44/bulb (I know, it's not realistic). It costs $6,500 to produce 650 bulbs, which comes out to $26/bulb, which is less than the cost per bulb to produce only 50 bulbs.

Now onto this problem. We are first asked to find the marginal cost function, which is simply the derivative of the cost function given to us.

In part (b), we are told to find the marginal cost at x = 100 and to explain its meaning. We get a number of 11 for this number, but what are its units? The derivative dC/dx is said "the rate of change of cost with respect to x". In this case, x is the number of pairs of jeans. So, we have cost ($) over jeans (or pairs). This number then represents the cost per pair of jeans, but it is an instantaneous rate of change that is occurring at the exact moment the 100th pair of jeans is produced. That is, once the 101st pair of jeans is produced, the cost will change yet again, and before the 100th pair of jeans was produced, the marginal cost was not $11/pair! Therefore, we can conclude that this number predicts the cost of producing the 101st (next) pair of jeans.

In part (c), we are to compare the marginal cost at the 100th pair of jeans with the cost of producing the 101st pair of jeans. We found the marginal cost at x = 100 to be $11/pair. What about the cost of producing the 101st pair of jeans? Well, C(100) is the cost to produce 100 pairs of jeans and C(101) is the cost to produce 101 pairs of jeans--thus, the cost of just that last pair of jeans is simply the difference C(101) - C(100)!

That's all we have time for now. Other problems may come from the fields of chemistry or biology and might deal with reaction rates or growth rates. These problems, in all honesty, are similar to the ones presented here, only in a different context. If you understand the concept of a derivative and use formulas given in your textbook, you should be able to answer questions.

If you have a word problem type not covered here that you need help with, leave a comment below. If you have any other questions or comments or suggestions for future topics, let me know!

Calculus: Continuity

A definition for "continuous" from Dictionary.com is "being in immediate connection or spatial relationship." Mathematical continuity is the same, roughly speaking. A function f(x) is continuous at a value a if the limit of the function as x approaches a is equal to the functional value at a. In other words: 

This simple definition actually contains three implications.

Many functions are continuous over their entire domains, such as polynomials and rational functions. The domain of a polynomial function is all real numbers and the domain of rational functions are everywhere except at vertical asymptotes. Thus, to find the limit of a polynomial or rational functions (that are not restricted in any way) at a particular value, one must simply find the functional value at the desired x-value. Easy enough, right?

A function can be discontinuous at a value a if the above requirements are not met. There are multiple types of discontinuity.

A function can be continuous from only one side and we call that left-hand or right-hand continuity. A function is continuous over an interval if the function is continuous at every value in the interval. You can use the definition of continuity (using limits) to determine if a function is continuous over an interval or at a value.

You can also use the properties of continuity to determine if a function is continuous. Let f and g be continuous functions at a and let c be a constant. Then the following functions are also continuous at a:

1. f + g
2. f – g
3. cf
4. fg
5. f/g, provided g(a) does not equal zero

If you are asked to determine if a function is continuous at a value a, first see if you can break it into two functions (f and g) that you know are continuous, probably some combination of rational and polynomial functions. Then the above properties tell you that these combinations of two continuous functions are also continuous.

Many functions are continuous over their entire domains, not just polynomials and rationals. Here is a list: polynomials, rational, root, trigonometric, inverse trigonometric, exponential, and logarithmic functions are all continuous at every number in their domains. Thus, any sum or difference (properties 1 and 2) of these, any multiple of these (3), any product of these (4), and any quotient of these (5; provided the denominator is not zero) is also continuous at a, provided that a is in the domain of both continuous functions.

If g is continuous at a and f is continuous at g(a), then the composite function f(g(x)) is also continuous at a. Note that here, f(x) does not need to be continuous at a, but at g(a) because g(a) is what is being "fed" into the f(x) function in the case of composite functions.

Intermediate Value Theorem: Suppose f is continuous on the closed interval [a, b] and let N be any number between f(a) and f(b), where f(a) does not equal f(b). Then there must exist a number c in (a, b) such that f(c) = N.

This sounds confusing, but it makes perfect sense. If a function is continuous over an interval, then it is continuous at every number in that interval. If there is a y-value, N, between two other y-values, f(a) and f(b), which are all in this continuous interval, then there must be some x-value whose functional value is N, and we call that value c. This value occurs in the open interval (a, b). Since the function is continuous over this interval, we can think of it as not being "broken" between these x-values a and b. Since it's not broken, every x-value in this interval has a y-value and vice-versa: every y-value in the interval (f(a), f(b)) must have an x-value in the interval (a, b).

We will use continuity in further Calculus topics and the Intermediate Value Theorem (IVT) is used in many Calculus proofs, so you should become familiar with it and work some problems involving it before you move to the next topic of interest.

Calculus: Finding Limits with Limit Laws

The limit laws are sort of like the properties of logarithms or the trigonometric identities you've used in the past. They are laws (that have been proven to be true) that make complicated calculations easier. I've typed up the basic limit laws into a Google Document that you can view by following this link. You must have the linke to view this document. If you save the link/bookmark it and then find it is not working, come back to this page and get the link again. It's possible I may have to refresh the link from time to time, so if you can't get it to work, leave a comment on this post and I will try to fix it!

The image is too large to put in this post for reference, so open the link above and have it in a separate window as you go through this topic.

Along with the limit laws I've given you, you'll need a couple of other theorems.

The first is simple. If a function f(x) is less than a function g(x) when x is near a, then the limit of f(x) as x approaches a is less than the limit of g(x) as x approaches a. This makes sense. If one function is less than another at a particular value, then its limit will be less than the limit of the other at that value.

The "Squeeze Theorem" is similar, but uses three functions.

This also makes sense. If a function (here, g(x)) is "squeezed" between two other functions near x = a, and if the two outside functions have limits of L, then the function squeezed between (g(x)) must have the same limit.

Let's practice using the limit laws and the squeeze theorem.

Here, I've detailed each step of the process. The first step, I used property 3. In the second, I used the property of polynomials and property 3. Next, I used property 6, then property 10, then the polynomial property again to arrive at my final answer.

And here is a problem that uses the squeeze theorem.

Here, we are told that f(x) is between these other two functions whenever x is greater than or equal to 0. We are asked to find the limit of f(x). So, by the squeeze theorem, we know that the limit of f(x) is equal to the limit of the other two functions. I found the limit of each of the other two functions (you really only need one, but it is good to check that these two limits are equal), and then I know that the limit of f(x) is the same.

Sometimes, you will be asked to find limits regarding piecewise functions, like below. Questions are in bold, answers in regular font.

Notice that for handed-limits as in part (i), (iv), and (v), in a piecewise function, you need to know which piece of the function you are considering. In part (i), we are approaching 1 from the positive side, so we use the piece of the function where x > 1, which is the third portion. In part (iv), we use the first portion because we want to approach -1 from the left side, that is, where x < -1. Also notice that the limit in part (vi) doesn't exist because the right- and left-handed limits (parts (iv) and (v)) are not equal.

I hope this post has helped you. Limits are not difficult--you just need to know the limit laws. Practice limits. If you are struggling to find an answer using the limit laws, graph the function and see if you can get an estimate. This isn't a sufficient answer on homework or a test, but maybe can help you see where you are going.

Also, if a limit does not exist, you cannot use the limit laws for it! You will need to graph the function and estimate a limit or use the squeeze theorem. Find two functions, one that is less than the questioned function and one that is greater at the point of interest, and then find the limits of those. Note that polynomials are the easiest limits to find, so try to find polynomials when using the squeeze theorem.

Calculus: Introduction to Limits

Follow this link to YouTube to see the video that corresponds to this post.

Limits are usually the first topic covered in a Calculus course. They are the foundation of most higher mathematics and especially for derivatives, which will come next, so it's important to have a firm grasp on the topic.

The limit of a function is a value, L, if we can make the values of f(x) arbitrarily close to L (as close as we wish) by taking x to be sufficiently close to a (on either side of a), but not equal to a.

I know, this is not a very firm definition, but it is the best I've found. We write a limit as follows:

And we say "the limit of f(x) as x approaches a equals L." What this means, then, is that as x gets very close to a (but does not necessarily equal a), f(x) gets very close to L (but again, does not necessarily equal to L). It is important to note that while it may happen that this limit is actually equal to the functional value of f(x) at a (that is, L = f(a)), it is not always the case. In fact, when finding a limit, we are not actually looking at what happens at the point x = a. We are only considering what happens as x approaches or gets closer and closer to a.

You've seen something like this before, probably, in a class like College Algebra. You have probably heard of asymptotes. We will cover this in more detail later, but an asymptote is a limit--as x approaches a vertical asymptote, f(x) approaches either positive or negative infinity. You know that the function never actually crosses this vertical asymptote, so we are only looking at x-values sufficiently close to the asymptote value.

When dealing with limits, how close is "sufficiently" or "very" close? Well, it's as close as you want. Mathematically, we want to consider values infinitely close to a.

In the beginning, you will probably use a table of values to estimate a limit. This is done by choosing x-values on either side of the value in question, and choosing them closer and closer to the value (but never actually choosing the value in question), and finding the f(x) values at those points. You can then see what number y or f(x) is approaching as x approaches the value in question. This is a great way to explore limits in the beginning, but you will soon find it cumbersome and not very accurate.

Somtimes, a limit may be "one-sided," that is, the limit coming from one side of a is different than the limit coming from the other. We write one-sided limits as follows: 

 The first is the right-hand limit of f(x). We say "the limit of f(x) as x approaches a from the right." The second is the left-land limit. In the first case, you are approaching the value a from the right--i.e. from the positive x-value side, and in the second, you are approaching from the left. This seems strange, but there are cases in which these limits may not be the same.

In fact, if the right- and left-hand limits at a particular value are the same, then we say that the limit as x approaches a exists, and is equal to L. Therefore, if the right- and left-hand limits at a particular value are NOT the same, then the limit at that value does not exist, and we write "DNE".

You can use the graph of a function to guesstimate a limit value. For example, estimate the following limit:

Begin by graphing the function inside the limit. On your calculator, you will use the variable x in place of the variable t. It is fine to replace variables in that way. Use the zoom feature to zoom in once. Press ZOOM, then press 2: ZoomIn, then press enter to zoom in at the origin. Here is what you should see.

Now, we are looking for what y-value is approached as x approaches 0. Well, we know that L is between 0 and 1 from looking at the graph. Go over to the Table screen (2ND, then GRAPH) and find x = 0. You'll notice that the y-value given says "ERROR." Press "+", and then type in 0.01 to see the table in increments of 0.01. You may need to arrow up or down to find x = 0 again. Still, it says "ERROR", but notice the values around 0.

Notice that f(-0.02) = 0.16667, as does f(-0.01), f(0.01), and f(0.02). Since these x-values are very close to 0, and since the y-values are changing very little, we can guess that the limit in question (as x approaches 0) is probably equal to 0.16667, which is a rounded decimal approximation of 1/6.

While using the calculator is helpful for finding limits (I use it on more complicated limits if I am struggling), it is not the best way and is not always correct. The next thing we will do is look at limit laws and how to use them to calculate exact values of limits.