Saturday, March 29, 2014

Conclusion to Half-Tau Month

Though pi day passed a few weeks ago, my brother made the interesting observation that this month is "Pi Month." It is March of 2014, or 3/14. Since I spent the month focused on trigonometry, I never had a chance to honor this joyous occasion until today.

Interestingly, pi does play a huge role in trigonometry. From wrapping functions to sine curves to angle measurements, pi is always popping up. Though this is kind of interesting, trigonometry is also one of the areas where tau really shines. Having just finished topics such as trigonometry, polar coordinates, and wrapping functions, I have found that it is a real struggle to use pi. I found myself converting most of my problems to tau before solving them just because pi made it too confusion.

No video describes these sorts of issues better than Vi Hart's "Pi is (still) Wrong" video, which gets into some of the issues involved with using pi, one of these being trigonometry.

I would also like to make you all aware that next month is Math Awareness Month. The theme this year is Mathematics, Magic, and Mystery, in part to honor the centennial of Martin Gardner's birth. At, there is a poster with 30 squares on it to represent the 30 days of April. On each day of the month, the next square becomes active. I will try to keep an eye on these webpages, as I will probably use April to comment on the topics posted there. Also, April 1st is a page on magic squares, and I am extremely honored to be hosting that day.

Saturday, March 22, 2014

Rediscovering Trigonometry Part 4: More Useful Formulas

Click here to see part one of this four week series.
Click here to see part two of this four week series.

Click here to see part three of this four week series.

Now that we have discovered some useful trigonometric identities, we can continue to build on them and create many more. There are an infinite number of trigonometric identities out there (not all of them have been created of course), but we will stick to two in this post: the product-to-sum formulas and the sum-to-product formulas.

Take the four angle addition/subtraction formulas we discovered in our first week. I will use A and B as our letters rather than alpha and beta.

1. sin(A + B) = sinAcosB + cosAsinB
2. sin(A – B) = sinAcosB – cosAsinB
3. cos(A + B) = cosAcosB – sinAsinB
4. cos(A – B) = cosAcosB + sinAsinB

These can be messed with very easily to create some new formulas. For instance, adding together the first two formulas gives:

sin(A + B) + sin(A – B) = sinAcosB + cosAsinB + sinAcosB – cosAsinB
2sinAcosB = sin(A + B) + sin(A – B)
sinAcosB = [sin(A + B) + sin(A – B)]/2

We now have a new identity. This can be now be used to solve a whole new range of problems and generate a whole new range of identities. The same steps can be done by subtracting the second equation from the first, adding the third and fourth together, and subtracting the fourth from the third. This creates the four product-to-sum identities.

These four formulas can be rewritten in a way that converts the sum into a product. Let's rewrite the variables as the following:

a + b = A
a – b = B

Making this change, we can then perform some operations to get a whole new set of formulas. These are called the sum-to-product identities.

These can then be built upon to generate whole new sets of formulas as well. Though the actual mathematics here might be a bit complicated, the idea is simple. Mathematics is always continuing to be developed, and this can be done through building upon previous ideas to form new ideas that help solve new problems. Trigonometry is a great place to see this sort of thing happen.

Saturday, March 15, 2014

Rediscovering Trigonometry Part 3: Half Angle Formulas

Click here to see part one of this four week series.
Click here to see part two of this four week series.

Last week, we figured out a way to figure out trigonometric functions for twice a given angle. This week, we will do the same, but for determining the trigonometric functions for half a given angle. First, let's discuss what half of an angle means.

Remember that there are 360° in a circle, or 360° in a full revolution. This means that a number like 370° can also be expressed as 10°. Though they are different measurements, plugging 370° into a trigonometric function will yield the same answer as 10°. It is also equivalent in most other situations in mathematics.

For double angle formulas, we did not need to discuss this. This is because when doing double angle formula calculations with these measurements, there would be no issue.

sin(2 • 10°) = sin(20°)
sin(2 • 370°) = sin(740°) = sin(740° - 720°) = sin(20°)

Note that 720° is two full revolutions around a circle, and thus, it can be subtracted off when performing a trigonometric operation.

But performing a half angle calculation will create more of an issue. Let's use 10° and 370° again.

sin(1/2 • 10°) = sin(5°)
sin(1/2 • 370°) = sin(185°)

These answers are not the same. Since they are both in the 0° - 360° interval, we cannot make any assumptions. We do know that the sine and cosine of 185° are the negative sine and negative cosine of 5° respectively, but this proves that they are not equal. If one were go up to 730°, they would be back to normal, however.

sin(1/2 • 730°) = sin(365°) = sin(365° - 360°) = sin(5°)

This means that for every angle, the half sine and half cosine function should yield two answers. The two answers should have the same absolute value, but different signs (they are the same number, but one is negative and one is positive). You may already have a function in your head that can create this type of situation, but we will be able to derive it as well.

Take a variation of the cosine double angle formula that we derived last week:

cos(2α) = 1 – 2sin2α

Let's try to isolate sinα. We would first add that to the left hand side and subtract the cosine of 2α over to the right hand side.

2sin2α = 1 – cos(2α)

Divide through by 2 to get:

sin2α = (1 – cos(2α))/2

And square root both sides to get:

sinα = ±√((1 - cos(2α))/2)

Notice how there is a ± sign in front of the square root. This is because when one squares a positive or negative value, it becomes positive. For instance, the equation x2 = 25 would be solved as x = ±5 because (5)(5) = 25 and (–5)(–5) = 25. The same thing happened here. But also remember what we found before. We proved through logic that the half sine and half cosine of an angle has two answers, one negative and one positive. With that in mind, we can see that this is the accurate way to write the square root (some derivations call for just a positive answer such as the Distance Formula).

Let's replace angle α with α/2 to keep the half angle definition. This gives a formula of:

sin(α/2) = ±√((1 - cosα)/2)

Let's derive a cosine half angle formula. We can take another variation on the cosine double angle formula and go forward.

cos(2α) = 2cos2α – 1

This time, we will only need to add one to both sides to isolate the cosα term. Let's also flip the equation around to make it simpler.

2cos2α = 1 + cos(2α)

Divide through by 2 to get:

cos2α = (1 + cos(2α))/2

And square rooting both sides yields:

cosα = ±√((1 + cos(2α))/2)

Again, we end up with a ± sign in the formula. This means that we probably did everything correctly, as logic shows we will need this sort of sign to create two answers. Rewriting α as α/2 gives a final formula of:

cos(α/2) = ±√((1 + cosα)/2)

It is tough to see what these formulas actually look like in this formatting, so I have written them out in LaTeX so you can see what is going on more conveniently.

These formulas themselves are pretty cool, but the logic involved in finding them is also very interesting. The fact that we could predict the nature of the function before we even found it is really helpful. This can be huge in trying to figure out the right way to go about solving a problem.

Saturday, March 8, 2014

Rediscovering Trigonometry Part 2: Double Angle Formulas

Click here to see part one of this four week series.

Last week, we developed some formulas that could calculate the sine, cosine, and tangent of the sum and difference of two angles. This proves useful when trying to calculate the exact sines/cosines/tangents of angles that are not in special right triangles (45-45-90, 30-60-90, 18-72-90).

What if we wanted to find the exact values for double a certain angle. Let's say we know the following:

sin18 = (√(5) - 1)/4
cos18 = (√(10 + 2√(5)))/4

How could we calculate the sine of 36°? We know the sine angle addition formula from last week, but it would be much easier to have a generalized version of this. Let's take a look at it.

sin(α + β) = sinα cosβ + cosα sinβ

What if we were finding the sine of 2α? Let's see what the formula would tell us:

sin(2α) = sin(α + α) = sinα cosα + cosα sinα = 2sinα cosα

So with some pretty simple computations, we get the formula:

sin(2α) = 2sinα cosα

That's a pretty simple formula. To find the sine of 36°, we could just do some easy manipulation with the values from up top.

sin36 = sin(2 • 18) = 2 • ((√(5) - 1)/4) • ((√(10 + 2√(5)))/4) = (√(10 - 2√(5)))/4

This sort of computation is clearly very useful when you are studying trigonometry. It wouldn't be practical in the sense that we use it in our day-to-day lives, but I think it is clear how useful this is to mathematicians and astronomers. I also find it really cool that these seemingly random irrational values can be derived exactly using just some basic mathematics.

Let's create a cosine formula. We know from last week that:

cos(α + β) = cosα cosβ – sinα sinβ

Substituting α in for β gives:

cos(2α) = cos(α + α) = cosα cosα – sinα sinα = cos2α – sin2α

This is the formula that naturally comes out:

cos(2α) = cos2α – sin2α

Knowing that sin2α + cos2α = 1, this can be rewritten in a few different ways:

cos(2α) = cos2α – sin2α
cos(2α) = 2cos2α – 1
cos(2α) = 1 – 2sin2α

This is extremely convenient, as the problem can be made much easier depending on what information you have. If you only know the sine of the angle, the third formula will work. If you only know the cosine, the second formula will work. There are also times where the first formula might be most convenient.

As you can see, it is pretty simple to do the work to come up with the double angle formulas, probably even easier than applying them in many cases. I encourage you to do the same process as we have done for sines and cosines to generate one for tangents, using the tangent formula we found last week. You will get a very pretty result.

Also, it can be fun to play around with these and find the exact sines and cosines of various angles. You will also see that there are many ways to write these different values. This is because they are irrational numbers. You could find the sine of 105° by doing sin(45 + 60), sin(90 + 15), sin(180 - 75), or many other variations. These could very well give different looking answers, but if the math was correct, the actual results will be equal.

Saturday, March 1, 2014

Rediscovering Trigonometry Part 1: Addition and Subtraction Formulas

In January, I did a series with some heavier mathematics, involving lots of number theory, combinatorics, and some fairly difficult manipulation. Last month, we took a little rest from this by talking a bit about casino games and optimal strategies. The mathematics we used was more probability and game theory. This month, I thought we could explore something completely different that is also extremely interesting. It is taught in school (much of the material in this post actually comes straight from my math class), but maybe not in a fun, thought-provoking way. This month, we will dive into trigonometry.

If you want to brush up on your basic foundational trigonometry, please click here. It is my post on the Law of Cosines, but also explains at the beginning what the sine, cosine, and tangent functions are. I will make sure to prove any other trigonometric identities we use in this series (or link to a proof), so all you need to know is what a sine, cosine, and tangent is.

Now we are ready to derive our first identity. We will try to figure out a simpler way of writing sin(α+β). There are many proofs of this, but I am using one that I found at Let's start with a diagram to help explain this idea.

Create the line AB.
Rotate it to point C, creating the angle α.
Rotate it to point D, creating the angle β.
Draw DE perpendicular to AB.
Draw DF perpendicular to AC.
Draw FG perpendicular to AB.
Draw FH perpendicular to ED.

We are trying to figure out the sine of α+β, which can be written as (ED/DA). First, let's determine the value of the angle HDF (which has already been written in for us).

Note that the lines HF and AG are parallel, with a transversal (a line that intersects both lines) of AF. In geometry, there is a theorem called the Alternate Interior Angle Theorem, that essentially tells us that in this circumstance, the measure of angle GAF is equal to the measure of angle AFH. In other words, angle AFH has a measure of α. Noting that angle AFD has a measure of 90 and AFH has a measure of α, that means angle HFD would have a measure of 90–α. Since a triangle has 180 degrees, having two angle measures in a triangle is enough to get the third. Triangle HFD has an angle of 90, an angle of 90–α, and one more angle that creates a sum of 180 degrees. Do the math and you will find that the angle must be α, and thus, angle HDF has a measure of α as depicted in the diagram.

Now we can try to find the formula. Let's start with the following equality:

ED = GF + HD

Note that we are trying to find what (ED/DA) equals. Because of this, let's divide both sides by DA.

(ED/DA) = (GF/DA) + (HD/DA)

Let's complicate some terms on the right hand side. Multiply the fraction on the left by (AF/AF) and the one on the right by (FD/FD)

(ED/DA) = (GF/AF)(AF/DA) + (HD/FD)(FD/DA)

(ED/DA) is simply the sine of α+β. Looking back at the diagram, (GF/AF) would be the sine of α. (AF/DA) would be the cosine of β. (HD/FD) would be the cosine of α. (FD/DA) would be the sine of β. This gives us the following formula:

sin(α + β) = sinα cosβ + cosα sinβ

This is a beautiful result. The sine of the sum of two quantities can just be pulled apart so easily. This is really cool on its own, and furthermore, it ends up being the foundation of tons more identities. Let's say we wanted to find the sine of α–β. We could just plug (-β) in for β and see what we get.

sin(α – β) = sinα cos(-β) + cosα sin(-β)

These negative terms can be analyzed in a number of ways, but the easiest way to do it is to look at the graphs of the sine and cosine functions.

Graph of y = sin(x)

Graph of y = cos(x)

Notice how the sine graph is symmetric over the origin. Any sin(x) will equal -sin(-x) and any sin(-x) will equal -sin(x). On the other hand, the cosine graph is symmetric over the y-axis. Any cos(x) will equal cos(-x) and -cos(x) will equal -cos(-x). This enables us to simplify the formula much more.

sin(α – β) = sinα cos(-β) + cosα sin(-β)
sin(α – β) = sinα cosβ + cosα (-sinβ)
sin(α – β) = sinα cosβ – cosα sinβ

Fantastic. What if we wanted to find the cosine of the sum of two angles? Going back to our original diagram, we can start with the following expression:

EA = GA - FH

Dividing through by AD creates the desired left hand side.

(EA/AD) = (GA/AD) - (FH/AD)

Multiply the term on the left of the right hand side by (AF/AF) and the right term by (DF/DF) to get:

(EA/AD) = (GA/AF)(AF/AD) - (FH/DF)(DF/AD)

And turning this into the appropriate functions gives:

cos(α + β) = cosα cosβ – sinα sinβ

Doing the same thing as before gives a subtraction formula of:

cos(α – β) = cosα cosβ + sinα sinβ

Finally, let's try to create a tangent formula. Knowing that dividing the sine by the cosine will give the tangent, let's try dividing the two formulas by each other.

sinα cosβ + cosα sinβ
cosα cosβ – sinα sinβ

Divide each term through by cos cosβ.

(sinα cosβ)/(cosα cosβ) + (cosα sinβ)/(cosα cosβ)
(cosα cosβ)/(cosα cosβ) – (sinα sinβ)/(cosα cosβ)

Many terms will now cancel.

(sinα)/(cosα) + (sinβ)/(cosβ)
1 – ((sinα)/(cosα))((sinβ)/(cosβ))

Turning the sine and cosine ratios into tangents gives a formula of:

tanα + tanβ
1 – tanα tanβ

And there's the formula! If you wanted to create a subtraction formula, you would just note that the tangent function is symmetric over the origin (or it is an odd function), and plug in the necessary values to get:

tanα – tanβ
1 + tanα tanβ

I think that these are really cool formulas. But they are also very practical. For instance, let's say you want to find the exact sine of 75°. We know that sin(45) = (√2)/2, sin(30) = 1/2, cos(45) = (√2)/2, and cos(30) = (√3)/2. So, this can be determined with the sine angle addition formula.

sin(75) = sin(30 + 45) = sin30cos45 + cos30sin45 = (1/2)((√2)/2) + ((√3)/2)((√2)/2) = (√(2) + √(6))/4

Knowing that the Greeks did some manipulation with the regular pentagon to find an exact sine of 72°, we can plug all of this information into the sine angle subtraction formula to find the exact sine of 3°. We can then eventually make our way down to 1°, which gives us the ability to generate the sines of all of the angles on the trig tables that mathematicians now use today.

Next week, we will use these formulas to create more generalizations that will make many of these trig table computations much much simpler. And give us some really pretty formulas in the process.