In Part 0, Section 0 I gave an overview of two basic algebraic skills needed to learn physics effectively. In this post I will explain several trigonometric (hereafter abbreviated to 'trig') ideas that are indispensable when physics problems are in more than one dimension. Since the world we live in is three dimensional, trig is really quite useful. \(\require{cancel}\)
Now, there is a lot more to trig than I am going to discuss here, mainly because we only need this one small portion to be able to tackle physics at this level. But trig is a much deeper subject with really interesting insights, if your fancy is tickled by beautiful interconnected math.
So to begin, we are going to need a triangle. This should be clear just from the name, which comes from the Greek for 'triangle' and 'measure'. More specifically, we need a right triangle, which is a triangle that possesses a \(90\) degree angle (a 'right' angle). You may have run into them before if you ever had to do anything with the Pythagorean Theorem. Right triangles are very nice to work with because you automatically know one angle, so all you need is is either two side lengths or a side and one of the other two angles and you can deduce everything else. So let's begin with a basic right triangle:
This is just a run of the mill right triangle. Each of the sides is labeled with a term that both labels it and can be used to indicate or find its length. The little square in the bottom righthand corner of the triangle tells you (as it says in the picture) that that angle is \(90^{\circ}\) (degrees). One of the other angles, in this case the one in the lower left hand corner of the image, is specified using the Greek letter \(\theta \) (pronounced 'theta') to stand in for whatever the actual measurement is.
Before we go any further with triangles, I want to say a few words about the use of Greek letters in math and physics. When I was younger, I thought that weird looking symbols were the sign of TRUE MATH and required REAL SMARTS to use. An equation could look like pure gibberish to me but so long as it included some Greek letter it must contain a great truth about the universe. Then I started taking physics and math classes and realized NOPE. They are, basically, there because we ran out of useful letters in the Carolinian alphabet we use in western European languages. We set aside a bunch of Greek letters to stand for quantities that pop up a lot (e.g. \(\phi, \theta \) are generally reserved for angles) and free up \(a\), \(b\), \(c\), to be used as needed. In other words, Greek letters do not imply fancy-shmancy calculations. They say "we needed more letters and these were easy to co-opt".
So, getting back to our representative right triangle, which I am going to post again so you don't have to scroll back up to see what I'm talking about.
For this triangle, we are given all the information we could possibly want or need. Yes, even that unlabeled corner. First of all, all triangles* have angles which must sum to \(180^{\circ}\). The right angle eats up half of that (\(90 ^{\circ}\) ) automatically. The remaining \(90^{\circ}\) are divided between the other two angles. So if we know \(\theta\) we know the other angle is going to be \(90^{\circ}-\theta\). But secondly, and most importantly, we don't need that third angle. Trigonometry lets us get away with just one angle, using the power of
You may have seen or heard this mnemonic before. Maybe carved into a desk in a high school math classroom, maybe muttered by some flustered-looking college student. This acronym is a way to remember the names and definitions of the three most important trig functions: sine, cosine, tangent. Each of these functions describes a ratio, or a particular relationship between two sides of a right triangle and one of its angles. So in this picture,
our chosen angle is \(\theta\), so all of our ratios, all of our functions will be found with respect to this angle. The side of the triangle that connects our angle and the right angled corner is known as the 'adjacent' side. In our triangle, that would be side '\(a\)'. The other side (or leg) of the triangle that is connected to the right angled corner is called the 'opposite' side, which in our case would be side '\(b\)'. The last side that does not connect to the right angle at all is called the 'hypotenuse', and is the longest side of any right triangle. In this case, side '\(c\)' is our hypotenuse. The reason that the hypotenuse has to be the longest side is because of the Pythagorean Theorem which states that for a right triangle, the square of the first leg plus the square of the side of the second leg will equal the square of the side of the hypotenuse. Mathematically, $$a^2 + b^2 = c^2,$$ and so the hypotenuse ('\(c\)') will always be bigger than either leg. You can prove this to yourself with a bit of graphing paper. Although you do not have to do so, I can tell you it is easiest to demonstrate this with a 3-4-5 triangle (a triangle whose legs measure 3 units and 4 units and whose hypotenuse is therefore 5 units).
Now, let's get down to the nitty-gritty definitions. This is where most people start to have problems, because while you can express the trig functions as simple ratios, what they *are* is a little tricky. Lets start by expressing them as simple ratios, and work backwards to what it all means.
The function \( \sin{(\theta)} \) (said 'sine of theta') gives us the ratio between the 'opposite' side ('\(b\)') and the hypotenuse ('\(c\)')**. Explicitly, $$\sin{(\theta)} = \frac{b}{c}, $$ and thus the mnemonic for the sine function being "SOH": Sine [is] Opposite [over] Hypotenuse.
The function \( \cos{(\theta)} \) (said 'cosine of theta') gives us the ratio between the 'adjacent' side ('\(a\)') and the hypotenuse ('\(c\)'). Explicitly, $$\cos{(\theta)} = \frac{a}{c}, $$ and thus the mnemonic for the sine function being "CAH": Cosine [is] Adjacent [over] Hypotenuse.
And lastly, the function \( \tan{(\theta)} \) (said 'tangent of theta') gives us the ratio between the 'adjacent' side ('\(a\)') and the 'opposite' side ('\(b\)'). Explicitly, $$\tan{(\theta)} = \frac{b}{a}, $$ and thus the mnemonic for the sine function being "TOA": Tangent [is] Opposite [over] Adjacent.
I should note that there are functions for the opposite ratios (hypotenuse over adjacent, etc) but they are rarely, if ever, used in physics at this level and I honestly don't see them pop up in physics at any level.
That's all well and good, you might say to me through your computer screen, but what is 'the sine of theta'? What if I just have an angle? How do I find the sine of it then?
Simple answer? Calculator or an online tool like Wolfram Alpha. But let's face it, that's a cop out, not an explanation.
The way to find the sine (or cosine or tangent) of any angle is using a thing called a unit circle, illustrated so very nicely below.
A unit circle is simple a circle of radius \(1\). Right triangles are created using the radius of the circle as the hypotenuse of the triangle and drawing a vertical line connecting the x-axis and where the radius meets the circle. If you move the location of the radius, 'sweeping' though the full \(360^{\circ}\) of the circle, the triangle changes accordingly.
Now I want you to imagine that at every increase of \(1^{\circ}\), we measure '\(b\)', the vertical leg or the 'opposite' side. We then plot, or map out, that leg height against the number of degrees in our angle. We do the same for '\(a\)', the horizontal leg or 'adjacent' side. Since our hypotenuse has a value of \(1\), by plotting out the height and length of the two legs we are plotting the sine and cosine functions, as shown in this nifty graphic***.
This goes on forever. One of the cool, and occasionally frustrating to scientists, thing about the trig functions is that they go on forever. They never slow down, approach any sort of limit. If you could keep tracing the unit circle forever, you would keep producing sine and cosine waves.
So, how does this help us with our triangles back there? Remember that at their hearts, trig functions are ratios. And ratios scale, as any cook can tell you (though in cooking you do reach some practical limitations on the ability to scale). If I want to make a double batch of spaghetti sauce, I don't have to reinvent my recipe. If my one-batch recipe calls for one garlic clove for a pint of pureed tomatoes, I know that for my double batch I need two garlic cloves for \(2\) pints of pureed tomatoes, and I will still get the same tomato sauce. Similarly, if I know the sine of an angle, say \( \sin{(30^{\circ})} = 0.5 \), I am saying that on the unit circle, the vertical leg is half a unit long. So for any right triangle that has a \(30^{\circ}\) angle, the side opposite that angle will be half the length of the hypotenuse!
Let's work through an example of this. Here we have right triangle, with one angle given and one side length.
![]()
.jpg)
It's considered good practice, and it becomes much easier to do this when pictures become more complex, to label the picture with symbols and give the value to those symbols elsewhere. It makes the image cleaner, and leads to fewer errors.
So, what can we find out about this triangle? We have chosen our angle (or it was chosen for us) and we have the 'adjacent' side length. We could start by finding the 'opposite' side length using the tangent, which gives us the relationship between the angle and the two legs. Let's call the 'opposite' side '\(y\)'. Then we can say $$\tan{(\theta)} = \frac{y}{l}$$ and we can solve for \(y\) using just a touch of algebra $$l\times \tan{(\theta)} = \frac{y}{\cancel{l}} \times \cancel{l} $$ $$l \times \tan{(\theta)} = y$$ and then to keep with convention we flip the sides of the equation to the thing we are solving for is on the left hand side (frequently abbreviated LHS): $$y =l \times \tan{(\theta)}. $$ Now we can plug in our values if we like to find the value of '\(y\)'. Since \( \theta= 30^{\circ} \) , \( l = 5\, \text{cm} = .05\, \text{m} \) and $$y=\left(0.05\,\mathrm{cm}\right)\tan\left(30^{\circ}\right)= 0.03\, \mathrm{m} =3 \,\mathrm{cm} $$ You may be wondering why I converted from centimeters (cm) to meters(m) and then back again. This is again just a 'best practice' thing. It does not make a big difference in situations, like this, were there is only one unit (length), but when we start calculating things like forces, we will be combining different units into other units and it becomes very important to make sure all your units are base units (e.g. meters, seconds) or you can easily end up with an answer a thousand times bigger or smaller than it should be.
So now we have the 'opposite' side of the triangle. We now have several options on how to find the hypotenuse. We could use the sine or the cosine to find the hypotenuse (call it '\(h\)') in much the same way that we used the tangent to find '\(y\)'. Or we can use the Pythagorean Theorem to find it, which is my preferred method. Plugging in our particular terms to the formula we get $$l^2 + y^2 = h^2. $$ To extract the '\(h\)' value we need to take the square root of both sides: $$ \sqrt{l^2 + y^2} = \sqrt{h^2}$$This is the same as raising both sides to the power of 1/2, so \( (h^2)^{1/2} \) becomes just \( (h) \). So we find that $$ h = \sqrt{l^2 + y^2} = \sqrt{(.05\, \mathrm{m})^2 + (.03\, \mathrm{m})^2 } = .05773627... $$ $$ h \approx 6\, \text{cm} $$ And now we know everything about this triangle!
So, why do we need this? What arcane bit of knowledge am I trying to put in your head? What good does it do to know anything about this *one* triangle? How will this help me in physics?
Good questions. First of all, I want to emphasize that, while math is beautiful and worth knowing for it's own sake, for a student of physics I find it useful to think of math concepts less as "yet another thing I have to learn" and more like a tool in your tool kit. Just like you need multiple, different tools to attack house repair (a hammer to drive nails, a screwdriver for screws) you need different tools to break down a problem and then build up your answer. Having these basic trig functions in your back pocket gives is like having a drill and driver set for a power drill. You can use it to both break down and rebuild the problem.
Secondly, by knowing about this one triangle, you are now able to attack any right triangle, no matter its shape or orientation in space.
And by knowing about one triangle, you allow yourself to know about triangles which can be drawn as extensions of that main triangle. Why exactly you would want to do that will be made more clear in a few weeks when we get into physics, so you will have to take me for my word on that for now.
Lastly, by being able to translate between angles and the sides of a right triangle we have set ourselves up to deal aptly with another very important concept for physics--vectors!
As always, please leave any questions in the comments section!
*In Euclidean, or flat, space. If you drew a triangle on a sphere, it would not have angles totaling \(180^{\circ}\)
**You may notice that I kept 'opposite' in quotation marks, but not hypotenuse. This is because while the hypotenuse is fixed, which side is called the 'opposite' depends on which angle you chose for your reference.
***Here's another graphic that shows a little better the cosine function, but it's not as smooth and it's labels are in German
Now, there is a lot more to trig than I am going to discuss here, mainly because we only need this one small portion to be able to tackle physics at this level. But trig is a much deeper subject with really interesting insights, if your fancy is tickled by beautiful interconnected math.
So to begin, we are going to need a triangle. This should be clear just from the name, which comes from the Greek for 'triangle' and 'measure'. More specifically, we need a right triangle, which is a triangle that possesses a \(90\) degree angle (a 'right' angle). You may have run into them before if you ever had to do anything with the Pythagorean Theorem. Right triangles are very nice to work with because you automatically know one angle, so all you need is is either two side lengths or a side and one of the other two angles and you can deduce everything else. So let's begin with a basic right triangle:
.jpg) |
| You will be subject to the same drawings and handwritten labels as my students are/will be |
Before we go any further with triangles, I want to say a few words about the use of Greek letters in math and physics. When I was younger, I thought that weird looking symbols were the sign of TRUE MATH and required REAL SMARTS to use. An equation could look like pure gibberish to me but so long as it included some Greek letter it must contain a great truth about the universe. Then I started taking physics and math classes and realized NOPE. They are, basically, there because we ran out of useful letters in the Carolinian alphabet we use in western European languages. We set aside a bunch of Greek letters to stand for quantities that pop up a lot (e.g. \(\phi, \theta \) are generally reserved for angles) and free up \(a\), \(b\), \(c\), to be used as needed. In other words, Greek letters do not imply fancy-shmancy calculations. They say "we needed more letters and these were easy to co-opt".
So, getting back to our representative right triangle, which I am going to post again so you don't have to scroll back up to see what I'm talking about.
|
 |
| Repeated like a prayer at every physics 1 exam |
|
Now, let's get down to the nitty-gritty definitions. This is where most people start to have problems, because while you can express the trig functions as simple ratios, what they *are* is a little tricky. Lets start by expressing them as simple ratios, and work backwards to what it all means.
The function \( \sin{(\theta)} \) (said 'sine of theta') gives us the ratio between the 'opposite' side ('\(b\)') and the hypotenuse ('\(c\)')**. Explicitly, $$\sin{(\theta)} = \frac{b}{c}, $$ and thus the mnemonic for the sine function being "SOH": Sine [is] Opposite [over] Hypotenuse.
The function \( \cos{(\theta)} \) (said 'cosine of theta') gives us the ratio between the 'adjacent' side ('\(a\)') and the hypotenuse ('\(c\)'). Explicitly, $$\cos{(\theta)} = \frac{a}{c}, $$ and thus the mnemonic for the sine function being "CAH": Cosine [is] Adjacent [over] Hypotenuse.
And lastly, the function \( \tan{(\theta)} \) (said 'tangent of theta') gives us the ratio between the 'adjacent' side ('\(a\)') and the 'opposite' side ('\(b\)'). Explicitly, $$\tan{(\theta)} = \frac{b}{a}, $$ and thus the mnemonic for the sine function being "TOA": Tangent [is] Opposite [over] Adjacent.
I should note that there are functions for the opposite ratios (hypotenuse over adjacent, etc) but they are rarely, if ever, used in physics at this level and I honestly don't see them pop up in physics at any level.
That's all well and good, you might say to me through your computer screen, but what is 'the sine of theta'? What if I just have an angle? How do I find the sine of it then?
Simple answer? Calculator or an online tool like Wolfram Alpha. But let's face it, that's a cop out, not an explanation.
The way to find the sine (or cosine or tangent) of any angle is using a thing called a unit circle, illustrated so very nicely below.
 |
| Via WikiCommons, Public Domain |
Now I want you to imagine that at every increase of \(1^{\circ}\), we measure '\(b\)', the vertical leg or the 'opposite' side. We then plot, or map out, that leg height against the number of degrees in our angle. We do the same for '\(a\)', the horizontal leg or 'adjacent' side. Since our hypotenuse has a value of \(1\), by plotting out the height and length of the two legs we are plotting the sine and cosine functions, as shown in this nifty graphic***.
 |
| Via WikiCommons, Public Domain |
So, how does this help us with our triangles back there? Remember that at their hearts, trig functions are ratios. And ratios scale, as any cook can tell you (though in cooking you do reach some practical limitations on the ability to scale). If I want to make a double batch of spaghetti sauce, I don't have to reinvent my recipe. If my one-batch recipe calls for one garlic clove for a pint of pureed tomatoes, I know that for my double batch I need two garlic cloves for \(2\) pints of pureed tomatoes, and I will still get the same tomato sauce. Similarly, if I know the sine of an angle, say \( \sin{(30^{\circ})} = 0.5 \), I am saying that on the unit circle, the vertical leg is half a unit long. So for any right triangle that has a \(30^{\circ}\) angle, the side opposite that angle will be half the length of the hypotenuse!
Let's work through an example of this. Here we have right triangle, with one angle given and one side length.
It's considered good practice, and it becomes much easier to do this when pictures become more complex, to label the picture with symbols and give the value to those symbols elsewhere. It makes the image cleaner, and leads to fewer errors.
So, what can we find out about this triangle? We have chosen our angle (or it was chosen for us) and we have the 'adjacent' side length. We could start by finding the 'opposite' side length using the tangent, which gives us the relationship between the angle and the two legs. Let's call the 'opposite' side '\(y\)'. Then we can say $$\tan{(\theta)} = \frac{y}{l}$$ and we can solve for \(y\) using just a touch of algebra $$l\times \tan{(\theta)} = \frac{y}{\cancel{l}} \times \cancel{l} $$ $$l \times \tan{(\theta)} = y$$ and then to keep with convention we flip the sides of the equation to the thing we are solving for is on the left hand side (frequently abbreviated LHS): $$y =l \times \tan{(\theta)}. $$ Now we can plug in our values if we like to find the value of '\(y\)'. Since \( \theta= 30^{\circ} \) , \( l = 5\, \text{cm} = .05\, \text{m} \) and $$y=\left(0.05\,\mathrm{cm}\right)\tan\left(30^{\circ}\right)= 0.03\, \mathrm{m} =3 \,\mathrm{cm} $$ You may be wondering why I converted from centimeters (cm) to meters(m) and then back again. This is again just a 'best practice' thing. It does not make a big difference in situations, like this, were there is only one unit (length), but when we start calculating things like forces, we will be combining different units into other units and it becomes very important to make sure all your units are base units (e.g. meters, seconds) or you can easily end up with an answer a thousand times bigger or smaller than it should be.
So now we have the 'opposite' side of the triangle. We now have several options on how to find the hypotenuse. We could use the sine or the cosine to find the hypotenuse (call it '\(h\)') in much the same way that we used the tangent to find '\(y\)'. Or we can use the Pythagorean Theorem to find it, which is my preferred method. Plugging in our particular terms to the formula we get $$l^2 + y^2 = h^2. $$ To extract the '\(h\)' value we need to take the square root of both sides: $$ \sqrt{l^2 + y^2} = \sqrt{h^2}$$This is the same as raising both sides to the power of 1/2, so \( (h^2)^{1/2} \) becomes just \( (h) \). So we find that $$ h = \sqrt{l^2 + y^2} = \sqrt{(.05\, \mathrm{m})^2 + (.03\, \mathrm{m})^2 } = .05773627... $$ $$ h \approx 6\, \text{cm} $$ And now we know everything about this triangle!
So, why do we need this? What arcane bit of knowledge am I trying to put in your head? What good does it do to know anything about this *one* triangle? How will this help me in physics?
Good questions. First of all, I want to emphasize that, while math is beautiful and worth knowing for it's own sake, for a student of physics I find it useful to think of math concepts less as "yet another thing I have to learn" and more like a tool in your tool kit. Just like you need multiple, different tools to attack house repair (a hammer to drive nails, a screwdriver for screws) you need different tools to break down a problem and then build up your answer. Having these basic trig functions in your back pocket gives is like having a drill and driver set for a power drill. You can use it to both break down and rebuild the problem.
Secondly, by knowing about this one triangle, you are now able to attack any right triangle, no matter its shape or orientation in space.
 |
| All equally susceptible to attack by trig |
Lastly, by being able to translate between angles and the sides of a right triangle we have set ourselves up to deal aptly with another very important concept for physics--vectors!
As always, please leave any questions in the comments section!
*In Euclidean, or flat, space. If you drew a triangle on a sphere, it would not have angles totaling \(180^{\circ}\)
**You may notice that I kept 'opposite' in quotation marks, but not hypotenuse. This is because while the hypotenuse is fixed, which side is called the 'opposite' depends on which angle you chose for your reference.
***Here's another graphic that shows a little better the cosine function, but it's not as smooth and it's labels are in German
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