Easy as πœ‹

I realise that my last few posts are a bit more advanced than my other posts. So let’s take it down a few notches. We will still be talking about angles however.

In my posts on trigonometry, I used angles measured in degrees. But that is a rather arbitrary measure. The origin of measuring angles based on 360Β° for a full circle is obscure, but one theory has it that it is based on ancient calendars that had 360 days in a year. In science and engineering, a different measure of angles is used, radians.

The number πœ‹ crops up a lot, especially when talking about circles. This is because πœ‹ is the ratio of a circle’s circumference (perimeter) to its diameter:

\[
\mathit{\pi}\hspace{0.33em}{=}\hspace{0.33em}\frac{C}{d}
\]

where C is circumference and d is diameter. This holds true no matter if the circle is as small as “o” or as large as the orbit of earth around the sun. The symbol “πœ‹” is a Greek letter with the name “pi”, and that is the symbol used to represent this ratio. In fact, it is the only way to write down this number exactly if we all agree what πœ‹ represents. This is because πœ‹ is an irrational number which means you cannot ever write it down exactly with numbers. Approximately, πœ‹ is 3.14159265359… where the decimal part goes on forever without ever a repeating pattern. Irrational or not, πœ‹ is a natural number to associate with circles.

Now if you replace the diameter d with twice the radius, that is 2r, the equation becomes

\[
\mathit{\pi}\hspace{0.33em}{=}\hspace{0.33em}\frac{C}{2r}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}{2}\mathit{\pi}\hspace{0.33em}{=}\hspace{0.33em}\frac{C}{r}
\]

which is again true for any size circle. Now notice that the 2πœ‹ on the left side is unitless. What do I mean by that? The right side of the equation is a circumference measured say, in meters, divided by a radius, again measured in meters. If you remember my post on units, you’ll remember that units can be cancelled in a fraction just like variables. So in this case, we have meters divided by meters which cancel each other and we are left with no units. So a full circle is associated with the unitless number 2πœ‹. What about a half circle? A half circle will have half the circumference. Half of 2πœ‹ is πœ‹. What about a quarter circle? A quarter of 2πœ‹ is πœ‹/2. As these numbers are all unitless, they were seen as a natural way to define angles.

So a half circle (also called a semi-circle), can be seen as an arc formed by the familiar angle 180Β°. The quarter circle is an arc formed by the angle 90Β°. But instead of degrees, we can call these angles πœ‹ and πœ‹/2 since these are the length of the arc formed by the angle divided by the radius. The term used to distinguish this measure of angles from degrees is radians.

This can be generalised for any angle. Please look at the below diagram:

So an angle in radians is the arclength formed by the angle divided by the radius. By the way, radians are abbreviated as rad and this is what you will probably see on your calculator when you use the radians mode.

So there are 2πœ‹ radians in a full circle, πœ‹ radians in a semi-circle, and πœ‹/2 for a quarter circle. We can use the fact that 180Β° = πœ‹ radians to convert between the two measurements. To change degrees to radians, multiply by πœ‹/180. To change radians to degrees, multiply by 180/πœ‹.

For example, 30Β° is 30 Γ— πœ‹/180 = πœ‹/6 rad. It is customary to keep πœ‹ as πœ‹ when working with radians.

One advantage of using radians becomes immediately apparent when rearranging the equation in the figure above. The arclength formed by an angle measured in radians is simply rπœƒ.

So now that you know what radians are, my future posts will use this measure almost exclusively.

A Springy Thingy, Part 2

So last time, I presented the equation that describes the position of a mass on a spring:

Animated portion is licensed under the Creative Commons Attribution-Share Alike 3.0 Unported license

The equation is

\[ {x}\hspace{0.33em}{=}\hspace{0.33em}{A}\hspace{0.33em}\sin\left({\frac{180t}{\mathit{\pi}}\sqrt{\frac{k}{m}}}\right) \]

where x is the position of the mass at a given time t in seconds, k is the spring constant, and m is the mass in kg. This was developed from the equation that describes the forces on the spring (gravity and the spring), and through calculus, out pops the equation above. This equation is the sine of stuff in brackets multiplied by a number A.

Even though the stuff in the brackets looks rather ominous, we are still just taking the sine of it and the sine only goes from -1 to 1. So the maximum extent of the mass is from –A to A. Now let’s look at the stuff in the brackets.

The 180 and πœ‹ are just there to change the rest of the expression so that you can press the sine button on your calculator in the “degrees” mode. Normally, when engineers model something like this, they use radians and not degrees. I have not explained what radians are yet so I’ve included an adjustment (the 180 and the πœ‹) so that you can continue to use degrees. I think I’ll explain radians in my next post.

The rest of the numbers, t, k, and m are the real meat of the model. For simplicity, I have started time at 0 seconds when the mass is at its rest position and is moving upwards in the postive direction. So you would expect the position of the mass to change with time and that is what the t in the expression does. The k and the m determine how fast or how slowly the mass oscillates. Let’s actually use some numbers instead of letters here for a specific mass and spring.

Now let’s assume the spring has a spring constant of 1 kg/sΒ² (I’ll discuss these units in my next post), and the mass connected to it is 1 kg. That means the stuff in the square root sign (called a radical) is just 1 and the square root of 1 is 1. And let’s further assume that I start the spring moving by stretching the spring 5 cm from its rest position. So now, the position equation above simplifies to

\[
{x}\hspace{0.33em}{=}\hspace{0.33em}{5}\sin\left({\frac{180t}{\mathit{\pi}}}\right)
\]

Starting at time as 0, you can choose various values of t, compute the stuff in the brackets, use the “SIN” button on your calculator which is in the “degrees” mode, and then multiply by 5. So for example, at t = 1 sec, 180/πœ‹ is 57.2958. Taking the sine of that gives 0.8415 and then multiplying that by 5 gives 4.2073 cm. So at 1 sec, the mass is 4.2073 cm above its resting position. You can plot this point and many others to graph this, or you can be lazy like me and use a graphing calculator. The graph of the position of the mass versus time for this scenario is

So no surprise, a sine wave. Now remember when I first talked about sine waves, I talked about the wavelength. Here I have indicated the wavelength as 6.28 sec. When dealing with time, the wavelength is called the period and usually represented with the symbol T. The period is the length of time it takes for one full cycle of motion. So it takes the mass 6.28 seconds to make one complete bounce. It turns out that you do not have to graph the curve to find this:

\[
{T}\hspace{0.33em}{=}\hspace{0.33em}\frac{{2}\mathit{\pi}}{\sqrt{\frac{k}{m}}}
\]

Since our k and m are each 1, T in this case is just 2πœ‹. Funny how πœ‹ keeps cropping up. Again, I’ll explain that in my next post on radians.

Associated with the period is something called frequency. The period is how long it takes for one complete cycle to occur, whereas the frequency is how many complete cycles occur in 1 second. Frequency is the reciprocal of the period and vice versa. That is f = 1/T. So for our mass, the frequency is 1/6.28 or 0.16 cycles per second. The term “cycles per second” is given a special unit called hertz which is abbreviated as hz. You may have heard this term before.

As you change the values of k and m, the values of T and f will change as well. If the spring gets stiffer (a higher k), you would expect the frequency to increase, that is it will bounce faster. You would expect a heavier mass to slow down the frequency and it does. I will leave it as an exercise for the student to check this using a graphing calculator or Excel.

A good simulation on the web that shows the effect of changing mass an spring constant is at https://www.physicsclassroom.com/Physics-Interactives/Waves-and-Sound/Mass-on-a-Spring/Mass-on-a-Spring-Interactive. This sets up the graph a bit differently than I do here, but the frequency changes are easy to see. Also, you can add damping to this which I did not include in this post to keep it simple, but you can play with that as well on this site.

A Springy Thingy, Part 1

From my last three posts, I think we are ready to model the motion of a mass on a spring:

Animated portion is licensed under theΒ Creative CommonsΒ Attribution-Share Alike 3.0 UnportedΒ license.

So, just like I did with the tennis ball, I will show you equations that describe this motion.

Now again, to develop these equations requires calculus, so I will just provide the final result. But before I do, just how does one begin modelling a physical process like this?

What is usually done, is to write down known equations that describe the forces acting on the mass. If you think about it, there are two: a force due to gravity and a force from the spring. There are other forces as well like resistance from the air, but as before, we will assume these to be zero to simplify the development.

Let’s first set up the picture. We have a weight on a spring. The weight has mass m. Now weight is different than mass, but on earth, the units are the same. So on earth, a 1 kg weight has a mass of 1 kg. But on the moon, the mass is still 1 kg but its weight is 0.165 kg because gravity is weaker there. We have a spring with a spring constant of k which is a measure of how stiff the spring is. The higher the value of k, the stiffer the spring.

To start the mass moving, we have stretched it A centimeters down from its resting position, then let go. We set up a one dimensional coordinate system where the rest position of the mass is 0 and up is positive.

The force due to gravity is –mg where m is the mass and g is the acceleration due to gravity. From my post on the tennis ball, remember that g is 9.8 m/sΒ². It’s negative because the force is acting in the down direction. This comes from Isaac Newton’s second law that says that force is equal to mass times acceleration. That is, F = ma. The force due to the spring comes from something called Hooke’s Law: F = kx where k is the spring constant and x is the amount that the spring is stretched (negative) or compressed (positive) from the resting position.

So the force equation for this setup is:

F = ma = kxmg

This is the equation engineers start with before they do calculus on it. So now in this post, this is the part where a miracle happens, and I’ll give you the final result.

So the equation that shows where the mass is at a certain time is below where t is time in seconds. It is assumed that time starts (that is t = 0) when the mass is travelling upwards and is at the 0 position:

\[
%Translator MathMagic Personal Edition Mac v9.41, LaTeX converter, 2019.1.28 09:06
{x}\hspace{0.33em}{=}\hspace{0.33em}{A}\hspace{0.33em}\sin\left({\frac{180t}{\mathit{\pi}}\sqrt{\frac{k}{m}}}\right)
\]

In my next post, I’ll dissect this a bit and put some actual numbers in it and plot the results.

Graphs of Trig Equations, Part 3

I know it’s taking a while before I use maths to model a mass on a spring, but that will only make sense by fully describing the graphs of sine equations. Hopefully, this development is interesting in its own right.

Now if you were to plot the daylight length at a certain latitude against days, and if you plotted for a full year, you would see a shape that looks amazingly like the sine graph I showed you in my last post. Except at the equator, the length of a day gets longer in the summer and shorter in the winter. Without actual taking a year to collect the data, I’ve plotted the daylight length in Melbourne Australia against days using the equation

\[
{L}\hspace{0.33em}{=}\hspace{0.33em}{2}{.}{63}\sin\left({\frac{360t}{365}}\right)\hspace{0.33em}{+}\hspace{0.33em}{12}{.}{165}
\]

where L is the length of daylight in hours and t is the number of days after 22 September of any year. Why I chose 22 September is an interesting topic which I may eventually discuss, but it has to do with what are called equinoxes. The plot is below and is almost exactly the plot created if I actually measured the day length each day and plotted these for a year:

Now the shape of this curve is a sine wave but you can see several differences from the standard sine wave explored in my last post:

  1. The amplitude is 2.63 instead of 1
  2. The wavelength is 365 days instead of 360 degrees
  3. The wave is centered at 12.165 instead of the x-axis
  4. We are evaluating the sine of time instead of degrees.

Let me explain these differences.

  1. Amplitude – The value of sin(x) , regardless of what form x is in, only has values from -1 to +1. So if I multiply the sine by any number, say 2.63, so that I now have y = 2.63 sin(x), then this results in values from 2.63 Γ— (-1) = -2.63 to 2.63 Γ— (+1) = 2.63. This make the amplitude of this new equation 2.63, that is, the number I multiply the sine by. So in general, the amplitude of y = AΒ sin(x) is A.
  2. Wavelength – Notice that the wavelength 365 is the denominator in
\[
\sin\left({\frac{360t}{365}}\right)
\]

The 360 in the numerator is the wavelength of the standard sine wave. The common symbol to represent wavelength is the Greek letter lambda, 𝝀, so in general, when you are taking the sine of something that looks like

\[
\sin\left({\frac{360t}{\mathit{\lambda}}}\right)
\]

the denominator, 𝝀, is the wavelength.

Now for those of you who have had exposure to this before, you may have expected to see 2πœ‹t in the numerator instead of 360t. This would be the case if we were taking the sine of numbers expressed as radians. But this series of posts is doing everything with calculators in the degree mode. I will explain radians later in a different post.

3. WaveΒ center – Notice that the center of the sine wave is at 12.165 which is the number added to the sine in the daylight length equation. The effect on a graph of adding a number to an equation is to raise or lower it – it does not change shape. So if you can graph and know the shape of y = something, then y = something + 10 will be the same shape, just shifted up 10 units. So adding 12.165 to the sine, doesn’t change its shape, it just changes where it is on the graph.

4. Time – The big change here is that we are no longer finding the sine of an angle. It may appear that we are now taking the sine of numbers in seconds, hours, or days – whatever the units of t are. However, the 360 in the numerator serves the purpose of making the number we are taking the sine of, unitless. That is, 360t/365 does not have any units – it is just a pure number.

Mathematicians/scientists long ago discovered that many periodic physical processes, have motions that follow a sine wave. In fact, when equations were formed that represented the forces on objects that were experiencing periodic motion, the sine of numbers involving time appeared when solving these equations.

And so it is with the length of the day throughout the year. The earth is rotating around the sun and this motion repeats, that is, is periodic. It is no surprise then that the graph of the day length is a sine wave.

I think we are now ready to model a mass on a spring. Let’s do that in my next post.

Graphs of Trig Equations, Part 2

In my last post, I showed that angles repeat every 360Β°. So an angle of 45Β° is the same as 45 + 360 = 405Β°. I also showed how angles can be negative if a reference line, like the positive x-axis is set up and angles created from that line going in the counter-clockwise direction are positive and going clockwise are negative. And I also showed that for angle 0Β°, sine 0Β° = 0 and sine 90Β° = 1. Please read my last post if needed.

Now without going through the development, it turns out that the sine has values that range from -1 to 1. Angles between 0Β° an 180Β° have positive sines and angles between 180Β° and 360Β° have negative sines. This repeats as one continues rotating around the x-axis.

Now we have already covered plotting equations so let’s plot the equation

y = sin x

where x is the angle:

So this is what a sine curve looks like. You can see that as you move along the x-axis, the curve moves up and down and repeats itself every 360Β°. The cosine curve is very similar but it is shifted to the left so that it begins at 1 when x = 0. So you see that the sine equation may prove useful when modelling something that repeats, like a mass on a spring bobbing up and down or a pendulum.

Now to prepare us for the modelling exercise which I will get to eventually, I want to define some characteristics of this sine curve.

First it has an amplitude. Amplitude is how high the curve goes above or below the center-line of the sine curve (or sine wave as it is frequently called). In this case, the center-line is the x-axis and the amplitude is 1 since the maximum extent of the curve is 1 unit above and below the center-line.

The sine wave has wave length. This is the distance between successive peaks (the highest points) or troughs (the lowest points). Lets look at the curve and measure the distance between any successive peaks. There is a peak at x = 90 and the next one is at x = 450. The distance between these two points on the x-axis is 450 – 90 = 360. This is what we expected as we know the sine curve repeats every 360Β° which is what wavelength means.

Associated with wavelength is something call frequency, but this will not make sense until I do a bit more development and include time in the mix. Stay tuned for the next post!

Graphs of Trig Equations, Part 1

Happy New Year and welcome to my first post of 2019!

My last post introduced the idea of modelling physical things with math equations. To do this from scratch, requires calculus but seeing the final result is very interesting. So in my last post, I modelled the simple physical event of a ball thrown into the air. Another common example when introducing modelling to students is a mass on a spring. But before I develop this, I want to show what the graphs of some trigonometric equations look like as they will be needed to describe any kind of motion that is cyclic, that is, repeats like a mass on a spring bobbing up and down.

So in a previous post, I defined what sin πœƒ and cos πœƒ are in terms of a right triangle. Given the below triangle

the sine and cosine of πœƒ are defined as\[ \sin\mathit{\theta}\hspace{0.33em}{=}\hspace{0.33em}\frac{\mathrm{opp}}{\mathrm{hyp}}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\cos\mathit{\theta}\hspace{0.33em}{=}\hspace{0.33em}\frac{\mathrm{adj}}{\mathrm{hyp}}\]

Let’s look at the sine for now. For very small angles, the opposite side will be small compared to the hypotenuse. Graphically, I think you can see that for an angle of 0Β°, there would be no opposite side so sin 0Β° = 0.

In the other extreme, as the angle gets close to 90Β°, the opposite side is close to the length of the hypotenuse, so the sine approaches 1. In fact,

sin 90Β° = 1.

Now angles are periodic in that they repeat every 360Β°. That is, an angle of 30Β° is also 30 + 360 = 390Β°. Another full circle of 360Β° can be added again to get an equivalent angle 390 + 360 = 750Β°. Angles can also be negative based on a convention of which direction you move to create the angle. Even with negative angles, multiple of 360Β° can be added or subtracted to get an equivalent angle whose sine will be the same. The below diagram shows these variations based on angles generated from the positive x-axis:

The angle in red is a positive angle, that is it is formed by going in the counter-clockwise direction from the x-axis. From that angle, you can go 1, 2, 3, etc complete circles to form the same angle. The angle in blue is a negative angle, that is it is formed by going clockwise from the x-axis. One can also go multiple complete circles around this angle to get the same angle. The point is that as you measure angles from 0, either in the positive or negative direction. you eventually repeat the same angles and these same angles will have the same sine value.

In my next post, I will plot the sine values against the angle values and show graphically what “periodic” means.

Latitude Aptitude

How did the early sailors determine their latitude position without GPS? That is the topic of today’s post.

Now first, a little background. The earth’s axis is tilted with respect to its orbit about the sun. The angle of this tilt is approximately 23.5Β°. This causes the northern and southern hemispheres to get more sun in summer and less in winter, which is the reason for seasons to exist. The tilted axis also causes our days to be shorter in the winter and longer in the summer. There are two times during the year when the days and nights are equal in length. The times are called the vernal and autumnal equinoxes. In the northern hemisphere, these  equinoxes occur on the first days of spring and autumn. Here in Australia in the southern hemisphere, we elected to call the start of spring on the 1st of September and the fall on the 1st of March, about 21 days short of the respective equinox. Perhaps this is because it is easier to remember. The main point here is that twice a year, at an equinox, the days and nights are equal.

At any time of the year other than an equinox, the highest height of the sun around noon is affected by the tilt of the earth’s axis. But at an equinox, the earth is in a neutral position where the axis tilt does not affect the highest sun height. At the equator (0Β° latitude), the sun would be directly overhead and a vertical stick in the ground would cast no shadow. As you go up or down in latitude, the highest sun height goes down and a vertical stick would cast the shortest shadow when the sun is at its highest. The below graphic shows the earth at an equinox with the sun at its maximum height. If a vertical stick is placed in the ground at your location, the sun’s rays would make an angle with it that is the same as your latitude angle.

Below is a blow-up of the vertical stick. You can see from the above picture that at the equator. The sun would be directly overhead at noon and there would be no shadow. At the poles, the sun would be at the horizon and the shadow would be very long (technically infinite). But in between, a measurable shadow would be made. 

Now you could measure the angle directly with a sextant, but I hardly know what a sextant is. let alone use one. But I am good at maths and I have a good calculator. The shadow, stick, and the line from the top of the stick to the shadow end forms a right triangle. If you remember the post on trig functions, the tangent of an angle is the length of the opposite side divided by the length of the adjacent side. We want to measure the angle 𝝀, so the adjacent side is the stick and the opposite side is the shadow: 

\[\tan\mathit{\lambda}\hspace{0.33em}{=}\hspace{0.33em}\frac{{\mathrm{length}}\hspace{0.33em}{\mathrm{of}}\hspace{0.33em}{\mathrm{stick}}}{{\mathrm{length}}\hspace{0.33em}{\mathrm{of}}\hspace{0.33em}{\mathrm{shadow}}}\]

On your calculator, if you have the trig functions, you would also have keys labelled “arctan” or “tan-1“. These keys mean “what is the angle that has what you entered as its tangent”. So if you enter the results of the division and then hit this key (making sure that your calculator is in “degrees” mode), you will get your latitude.

Now this method will not tell you if the latitude is positive (North) or negative (South). But if you are so lost that you don’t even know what hemisphere you are in, finding your latitude is probably the least of your troubles.

Also, waiting for noon to find your latitude is not too bad, but waiting for an equinox is fairly restrictive. Fortunately, our early sailors had tables to correct the angle found depending on the time of the year.

The Attitude of Latitude

There are many applications of trigonometry in everyday life. One of these is the position of anything on the earth. You have undoubtedly heard of latitude and longitude. These are coordinates defined on the earth’s surface that provide the position of a point, much like defining the position of a point on a cartesian coordinate system ( see my posts on Graphing). As of this writing, my position is -37.6836353 latitude and 144.746533 longitude. A more traditional way to express this location is 37Β° 41′ 1.086” S, 144Β° 44′ 47.5182” E. Notice that in the second way, the degree symbol is used. This is because latitude and longitude are angles. I’ll explain that in a minute.

Also notice that in the second method, that there are the symbols ‘ and ”. These mean “minutes” and “seconds” respectively. Just like hours, a degree is split into 60 minutes, and each minute is is divided into 60 seconds. So one difference between the using two methods is whether you want to use minutes and seconds for parts of a degree or decimals. This is the same as saying the time is now 9 hours, 4 minutes, 40 seconds or 9.0778 hours

Another difference between the two methods is that the first uses negative and implied positive signs while the second uses S and N for latitude and E and W for longitude. For latitude, S means South and N means North. For longitude, E means East and W means West. 

So how are these angles measured? The below graphic helps explain these angles:

Let’s first look at longitude. This angle is frequently represented with the greek letter phi, Ο•. Imagine circles around the earth that pass through both poles. These are called great circles as they are the largest circles that can be drawn on the earth’s surface and divide the earth into two hemispheres (assuming that the earth is a perfect sphere). One of these circles was designated as a reference circle called theΒ prime meridian. This reference circle was finally (after a history of development) agreed upon in 1884. It is designated as the 0Β° circle of longitude on the side that passes through Greenwich, England. Other great circles, like the one in red above, are an angle away from the prime meridian. One can go east (positive) from this reference circle or west (negative) until you get to the other side of the prime meridian. So the largest longitude angles are Β±180Β°. Let’s now look at latitude.

The latitude angle is often represented with the greek letter lambda, 𝝀. The reference for this angle is a much more natural geometric one, the equator. The equator isΒ 0Β° latitude. Along any great longitude circle, like the one in red above, one can go north (positive) or south (negative) to form an angle with the equator, measured at the earths centre like longitude. So the largest latitude angles are Β±90Β° which are the poles.

Now before GPS and other navigational aids, it was a real trick to know your position on earth. Longitude was particularly tricky, but measuring latitude was very possible. How that was done, will be the topic of my next post.

To the Stars

Maths can be a dry subject to learn when you do not know of its many applications.This post will be about one of the many applications of trigonometry. Trigonometry is used extensively in the GPS system to determine your position. But I’d like to go a bit farther out in space and show how distances to stars are determined.

Let me first set up the example and then define some new measurement units before I actually develop the solution. This is because of the large distances and small angles used in astronomy.

A common method to calculate the distance of nearby stars is to use parallax. You can see parallax in action by stretching out your hand and extend your index finger, then alternately close one eye and see how your finger changes position compared to objects further away. This effect is rather pronounced with your finger because the distance between your eyes is not much smaller than the distance to your finger.

What astronomers do is to see the relative position of of a nearby star at different places compared to farther stars that do not change their position much over time. But any distance on earth is very small compared to the distance to a star so the effect is too small to measure. What to do?

Well as we orbit our star (the sun), we could make a measurement on one side of the orbit, say in June, then on the other side in December. This would be a much larger distance than anything we could do on earth and the resulting parallax effect would be measurable. So the set up of the problem looks like this:

So the two readings are needed to determine the angle πœƒ. Once πœƒ s known, we can use it to calculate d. From the previous post on trig functions, you can see that for this problem:

\[\tan\hspace{0.33em}\mathit{\theta}\hspace{0.33em}{=}\hspace{0.33em}\frac{r}{d}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{d}\hspace{0.33em}{=}\hspace{0.33em}\frac{r}{\tan\hspace{0.33em}\mathit{\theta}}\]

But there are some practical problems here. Even though we are using the largest possible distance while restricted to the earth’s surface to measure the parallax angle, this angle is still very very small. So astronomers use arc seconds (arcsec). This is 1/3600 of a degree. Also, as distances to stars are very large, the unit of distance between the earth and the sun used is 1 AU (astronomical unit). 1 AU equals 149.6 million kilometers. Using these units results in the distance expressed in parsecs (yes, parsecs is a distance measurement, not a time measurement as used in the Star Wars movies). However, we will just use AU’s for distance and degrees for the angle in the example.

So how far is a star with a measured parallax angle of 0.0002 degrees?

\[{d}\hspace{0.33em}{=}\hspace{0.33em}\frac{{1}\hspace{0.33em}{AU}}{\tan\hspace{0.33em}{(}{0}{.}{0002}{)}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{0.00000349065}\hspace{0.33em}{=}\hspace{0.33em}{286479}{.}{6}\hspace{0.33em}{\mathrm{AU}}\]

You can see why astronomers use arcsec. Using the all knowing Google, converting 286479.6 AU gives 42,900,000,000,000 km or 4.53 light years or 1.39 parsecs.

What’s Your Sine?

Continuing with the trigonometry theme, there are other relationships between the angles and the length of the sides of a right triangle. You may have a calculator with buttons labeled as “sin” or “sin x” or “sin ΞΈ”. There would be similar buttons using the prefix “cos” and “tan”. These are the basic trig (trigonometric, hence the abbreviation) functions. What do these functions do?

As with the Pythagorus theorem covered in my last post, there are other relationships that apply to right triangles regardless of their size. But unlike the Pythagorus theorem which relates the lengths of just the sides, the trig functions relate the side lengths with the internal angles.

But before I show these, remember that there are several ways to measure angles, degrees and radians being the most common. When using the trig functions, your calculator needs to know what measurement you are using. As we have been and will continue to use degrees, you need to make sure that in your calculator settings, you have set the degrees mode. In most calculators, this will show up as an abbreviation “deg”. Radians will display as “rad”. As a full circle angle is 360Β° and approximately 6.28 radians, there is quite a bit of difference between the two types of measurements. So let’s begin.

So here is your basic, everyday right triangle with the angle of interest, πœƒ, and the sides labelled with respect to that angle. The side could be adjacent to it or opposite it. The hypotenuse, as seen before, is the side opposite the right angle. For any size right triangle, the basic trig functions are defined as follows:

\[
\sin\mathit{\theta}\hspace{0.33em}{=}\hspace{0.33em}\frac{\mathrm{opp}}{\mathrm{hyp}}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\cos\mathit{\theta}\hspace{0.33em}{=}\hspace{0.33em}\frac{\mathrm{adj}}{\mathrm{hyp}}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\tan\mathit{\theta}\hspace{0.33em}{=}\hspace{0.33em}\frac{\mathrm{opp}}{\mathrm{adj}}\hspace{0.33em}{=}\hspace{0.33em}\frac{\sin\mathit{\theta}}{\cos\mathit{\theta}}
\]

So the trig functions show the ratios of the various sides. In the above formulas, the following abbreviations are used: sin = sine, cos = cosine, tan = tangent, opp = opposite, adj = adjacent, hyp = hypotenuse. If you type 45 into your calculator and hit the “sin” key, you should see a decimal number 0.7071… . This means that for a right triangle with a 45Β° angle, the length of the opposite side divided by the length of the hypotenuse is always 0.7071…

The power of these functions is that you only need the length of one side and an angle (other than the right angle) of a right triangle to determine the lengths of the other two sides.

Example:

What are the lengths of the other two sides of the above triangle?

\[
\begin{array}{l}
{\sin{30}\hspace{0.33em}{=}\hspace{0.33em}\frac{\mathrm{opp}}{5}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{5}}\\
{{\mathrm{opp}}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{5}\hspace{0.33em}\times\hspace{0.33em}{5}\hspace{0.33em}{=}\hspace{0.33em}{2}{.}{5}}\\
{\cos{30}\hspace{0.33em}{=}\hspace{0.33em}\frac{\mathrm{adj}}{5}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{866}}\\
{{\mathrm{adj}}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{866}\hspace{0.33em}\times\hspace{0.33em}{5}\hspace{0.33em}{=}\hspace{0.33em}{4}{.}{33}}
\end{array}
\]

So the lengths are 2.5 and 4.33. The units (centimeters, inches, etc) are whatever the units of the given hypotenuse is.

Let’s try this on a practical problem:

You are 500 meters from a tall tower. Using a sextant (a device to measure angles), you measure the angle between the ground and the line between your feet and the top of the tower to be 15Β°. How high is the tower?

The side adjacent to the angle is 500 meters. We need to find the side opposite (the tower). Looking at the trig functions, the tangent looks like it is the best one to use since it is the ratio of the side opposite (unknown) and the side adjacent (known). Finding the tangent of 15Β° on my calculator gives 0.2679. So,

\[
\begin{array}{l}
{\tan\hspace{0.33em}{15}\hspace{0.33em}{=}\hspace{0.33em}\frac{\mathrm{opp}}{\mathrm{adj}}\hspace{0.33em}{=}\hspace{0.33em}\frac{{\mathrm{tower}}\hspace{0.33em}{\mathrm{height}}}{{\mathrm{distance}}\hspace{0.33em}{\mathrm{from}}\hspace{0.33em}{\mathrm{tower}}}\hspace{0.33em}}\\
{{=}\frac{{\mathrm{tower}}\hspace{0.33em}{\mathrm{height}}}{500}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{2679}}\\
{{\mathrm{tower}}\hspace{0.33em}{\mathrm{height}}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{2679}\hspace{0.33em}\times\hspace{0.33em}{500}\hspace{0.33em}{=}\hspace{0.33em}{133}{.}{97}\hspace{0.33em}{\mathrm{meters}}}
\end{array}
\]

Now that saves a lot of effort to physically measure the height, doesn’t it?