Week 42 was a bit weird. I feel like I made a lot of progress and got a lot of work done but paradoxically didn’t actually get through many videos or exercises. As I mentioned last week, I’ve been working through what are known as Trigonometric Identities which are essentially different types of formulas that let you add/subtract Sine, Cosine, and Tangent values or double/half them. I learned that there are other identities that can do more than just that but I haven’t gotten to them yet. Just like last week, I still felt thoroughly underprepared as I struggled to work through the videos and exercises this week BUT did manage to start figuring it all out by the end of the week (or at least I managed to get a somewhat fundamental grasp on the main concepts).

Covid Update – Nothing new to report. I still have a slightly irritated/scratchy throat and think it could be Covid but I’m not stressed about it as it seems to be getting better each day. I found out this week that the tests aren’t always 100% accurate (someone told me they’re only 80% accurate if they’re performed correctly and, from what he told me, it doesn’t sound like the nurse who gave me my test did it correctly) so I still think there’s a chance I might have it. To be honest, I hope I do since what I have in my throat isn’t that bad and it seems like I’m getting over it anyways.

Back to KA – I started the week by getting a better grasp on what’s known as the Sine Angle Addition Identity/Formula. The formula for this identity is:

• Sin(Theta + Beta) = Sin(Theta)Cos(Beta) + Sin(Beta)Cos(Theta)

Here is a photo of my notes that helps to explain how the formula is derived:

In the above photo you’ll see that I’ve drawn a diagram which can be viewed as two right angle triangles stacked on top of each other, ABC and ACD, which create a third triangle, AFD, which is layered overtop of the other two triangles. As you can also see from the photo, the Sine Angle Addition Identity more-or-less means, “you can find the Y-value of triangle AFD by stacking triangles ABC and ACD on top of each other and using this formula/identity”. It’s difficult to explain in words how to derive the identity through this diagram, but you start off by stating that the sum of Sine of x and y equal line segment DF, then stating that DF = DE + CB, then stating that line segment AD = 1 and then using SOH CAH TOA to label the other segments. You can then conclude that line segment DE = Cos(y)Sin(x) and that line segment CB = Sin(y)Cos(x) which leads you to the Sine Angle Addition Identity. It’s worth noting that this same formula can be used when subtracting two angles from each other as well, although I wasn’t shown a video on how/why that also works and haven’t taken the time to try and think it through.

The next Identity I learned was the Cosine Angle Addition Identity (#2 in the list below) which is exactly the same as the Sine Addition Identity in that you use it when stacking two right triangles on top of each other to find the Cosine value of the third triangle that’s created.

The following are a list of all the identities I learned this week which I was able to successfully use during the exercises:

1. Sin(x +/- y) = Sin(x)Cos(y) +/- Sin(y)Cos(x)
2. Cos(x + y) = Cos(x)Cos(y) – Sin(x)Sin(y)
3. Cos(x – y) = Cos(x)Cos(y) + Sin(x)Sin(y)
4. Sin(2x) = 2(Sin(x)Cos(x))
5. Cos 2x =
   • = Cos^2x – Sin^2x
   • = 2Cos^2x – 1
   • = 1 – 2Sin^2x
6. Sin^2x = ½(1 – Cos2x)
7. Cos^2x = ½(1 + Cos2x)
8. Cos^2x + Sin^2x = 1^2 = 1
9. Tan(x + y) = (Tan(x) + Tan(y))/(1 – Tan(x)Tan(y))
10. Tan(x – y) = (Tan(x) – Tan(y))/(1 + Tan(x)Tan(y))

I should note that in the Sine Angle Addition Identity (#1 on the list) it shows +/- meaning that you can use the same formula for either adding or subtracting the angles. When adding, you use (+) in both instances and when subtracting you of course use (–) in both instances. Numbers 3, 4, 5, 6 and 7 are all derived algebraically by manipulating #1, #2, and #8.   

There is a “Trig Identity Reference” page that lists all the trig identities, many of which I’m not very familiar with yet. I didn’t include the identities in the list above that I haven’t used before. The categorical names of the groups of identities on the reference page that I haven’t used before are Half Angle Identities, Symmetry and Periodicity Identities, and Confunction Identities. Each category of identities had between 5 and 6 individual identities.

Next I was then introduced to a new way of thinking about the reciprocal try ratios Cosecant, Secant, and Cotangent. Before I got to this section, I simply new each of the three to be the reciprocal of Sine, Cosine, and Tangent without really understanding how or why this was the case (i.e. I just knew that instead of SOH CAH TOA it was C(sc)HO S(ec)HA and C(ot)AO). Below is a diagram I drew based now what I was shown this week which helped give me a better understanding of how the Cosecant, Secant, and Cotangent lines are derived from a unit circle:

I copied this diagram directly form the Trig Identity Reference page but still have a difficult time understanding a few parts of it. One thing I find confusing is that the Tangent line runs outside the unit circle but I’ve always understood the Tangent line to run from somewhere on the radius of the circle to the centre point, (0, 0). I think I’m starting to understand how it works but don’t have nearly a good enough understanding to put it into words. That being the case, clearly I need to keep working through these types of questions/concepts to get better at them.

Below is a diagram of why Cosine/1 = 1/Secant:

It may be difficult to understand, but the way the above formula is derived is by determining that there are two similar triangles, triangles ABC and ACD, which makes their angles the same. Then you can conclude that the ratio between Cosine in triangle ABC (i.e. the adjacent side of angle BAC) and the hypotenuse (line segment AC, which equals 1) is congruent to the ratio between line segment AC and the Secant line segment AD in the triangle ACD .

Below is another diagram of why Sine/1 = 1/Cosecant which is derived in the same way as Cosine vs Secant:

The last thing I worked on this week was adding and subtracting values from ‘special’ triangles using the Sine and Cosine Angle Addition/Subtraction Identities. The ‘special’ triangles I’m referring to are 30-60-90 triangles and 45-45-90 triangles. As a reminder, here’s a table that gives the Sine, Cosine, and Tangent ratios of side lengths for these special angles:

	Cos(Theta)	Sin(Theta)	Tan(Theta)
Theta = 30 degrees	{3}/2	1/2	1/{3} = {3}/3
Theta = 45 degrees	½ = {2}/2	½ = {2}/2	1
Theta = = 60 degrees	½	{3}/2	{3}

Working through these questions, I would be asked to “find the Sine/Cosine of “x”. I would start by drawing out 30-60-90 and 45-45-90 triangles and labeling the ratios of the sides. I would also have to convert two angles to values that shared the same denominator (most often making the denominator equal 12) so I could add or subtract the numerators to equal “x”. I would then use the Sine and Cosine Addition/Subtraction Angle Identities and plug in the Sine and Cosine side ratio values that I’d labeled on the 30-60-90 and 45-45-90 triangles into the equations. An example of a question I was asked would be:

• Find Sin(Pi/12) exactly using an angle addition or subtraction formula.
  • Sin(Pi/12) = Sin(4Pi/12 – 3Pi/12)
    • Sin(x – y) = Sin(x)Cos(y) – Cos(x)Sin(y)
  • Sin(4Pi/12 – 3Pi/12) = Sin(4Pi/12)Cos(3Pi/12) – Sin(3Pi/12)Cos(4Pi/12)
    • = Sin(Pi/3)Cos(Pi/4) – Sin(Pi/4)Cos(Pi/3)
    • = Sin(60 degrees)Cos(45 degrees) – Sin(45 degrees)Cos(60 degrees)
    • = {3}/2 * {2}/2 – {2}/2 * ½
    • = {6}/4 – {2}/4
    • = ({6} – {2})/4

If it all seems confusing, that’s because it was…

Looking back on this past week, I would say it was one of the tougher weeks I’ve had in terms of learning new, difficult material. I’m pleased with what I accomplished but think it’s pretty obvious I don’t fully understand everything I got through. There are only 7 videos remaining in the unit (and course for that matter) which doesn’t sound too bad, however they’re all labeled as “Trig Challenge Problem: …. ” so I think they’ll all be fairly difficult. Hopefully going through them will help me wrap my head around some of the tougher things I learned this week. In any case, it seems like there’s a good chance that I’ll be able to finish the course in the next 9 days, before the end of the month. In fact, I might even be able to get it done by the end of this week! Here’s hoping.