How to find sines/cosines without memorization?

[QuantumCurt]

Hey everyone, I've got a question about finding the basic sines/cosines of the unit circle, without relying solely on pure memorization. I've got them all memorized personally, so it's not really an issue for myself, but I'm currently tutoring someone in trig, and they're having a lot of trouble memorizing the first quadrant of the unit circle. I vaguely know about this method, where if we write the basic angles from the first quadrant in this form, matched up respectively with the corresponding fraction in the series below the given angle-

$$0 \ \ 30 \ \ 45 \ \ 60 \ \ 90$$
$$\frac{0}{4} \ \frac{1}{4} \ \frac{2}{4} \ \frac{3}{4} \ \frac{4}{4}$$

Then we can take the corresponding values, and find that sine/cosine.

For example, for $\sin45$, we'll use $\frac{2}{4}$, which then simplifies to $\frac{1}{2}$, then we'll simply take the square root of that value, and we have$\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$, which is the sine of 45 degrees. The same process will work for any of the angles. To find $\sin60$, we'll take $\frac{3}{4}$, then take the square root to get $\frac{\sqrt{3}}{2}$ which is the sine of 60 degrees.

To find the cosines of the respective angles, you simply reverse the order of the fractions into descending order, and use the same process.

Are there any problems with this method? Obviously it is only going to work for the base values from the unit circle, but for those values it does seem to be quite effective. I've searched, and haven't managed to find anything in the way of reading about it. I'd like to try explaining this method to the person that I'm tutoring, in hopes that this will help them find these values, in lieu of having a unit circle. I want to make sure I fully understand it before explaining it to them though. It's really pretty intuitively simple, so I'm hoping it will be easier for them than simply trying to memorize it.

Is there a way to apply a similar method to the values of the other quadrants, or does that still rely on being able to reflect the angles into the different quadrants? I can see it being possible by selectively using negative signs, but that seems like it would be harder to memorize than memorizing how to reflect the angles into different quadrants.

 

[Mentallic]

If the student you're teaching is still memorizing the acute angle trigs then your best approach is to teach him/her with the method you have posted, while ignoring the larger angles for now. When you eventually get to those, the safest method would be to teach them about angle reflections, and of course you'll end up simplifying a sum of fractions which will naturally lead into memorizing the larger angles.

You should also present the student with the sine and cosine functions on cartesian coordinates and relate each angle you calculate on the unit circle to its location found on the graph.

 

[QuantumCurt]

Yes, she's still doing the basic right triangle trig. The semester just started a couple weeks ago, so they're really just getting going. They haven't gotten to graphs of trig functions yet, but when they do, I'm definitely going to go over how the unit circle corresponds to different points on the graph. That's something that my professor never really discussed when I took trig. I had to kind of intuitively figure it out on my own.

 

[symbolipoint]

QuantumCurt,

Is the standard unit circle not helpful enough for your student? We typically use only five reference angles so they should not be difficult to memorize, and they are easy enough to draw.

 

[LCKurtz]

Hey everyone, I've got a question about finding the basic sines/cosines of the unit circle, without relying solely on pure memorization. I've got them all memorized personally, so it's not really an issue for myself, but I'm currently tutoring someone in trig, and they're having a lot of trouble memorizing the first quadrant of the unit circle. I vaguely know about this method, where if we write the basic angles from the first quadrant in this form, matched up respectively with the corresponding fraction in the series below the given angle-

$$0 \ \ 30 \ \ 45 \ \ 60 \ \ 90$$
$$\frac{0}{4} \ \frac{1}{4} \ \frac{2}{4} \ \frac{3}{4} \ \frac{4}{4}$$

Then we can take the corresponding values, and find that sine/cosine.

For example, for $\sin45$, we'll use $\frac{2}{4}$, which then simplifies to $\frac{1}{2}$, then we'll simply take the square root of that value, and we have$\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$, which is the sine of 45 degrees. The same process will work for any of the angles. To find $\sin60$, we'll take $\frac{3}{4}$, then take the square root to get $\frac{\sqrt{3}}{2}$ which is the sine of 60 degrees.

To find the cosines of the respective angles, you simply reverse the order of the fractions into descending order, and use the same process.

Are there any problems with this method?

Yes, I do think there are problems with this method. You say your student has a problem with pure memorization yet that is what your "method" requires. And it gives zero insight or understanding of what is really being calculated.

Your mileage may vary, but my thinking is that for basic understanding of how to calculate the trig values for the "standard" angles, all that is required is to understand equilateral triangles and isosceles right triangles. Given an equilateral triangle with side 2 and an altitude drawn making a 30-60-90 triangle, hopefully your student could figure out the sides of that triangle are $2,~1,~\sqrt 3$. And given a right triangle with legs of $1$ that the hypotenuse is $\sqrt 2$. These two triangles in various orientations will fit the standard angles.

Sure, you have to memorize the Pythagorean theorem and the triangle definitions of sine, cosine, etc., but these two triangles themselves give all the ratios.

 

[symbolipoint]

Yes, I do think there are problems with this method. You say your student has a problem with pure memorization yet that is what your "method" requires. And it gives zero insight or understanding of what is really being calculated.

Your mileage may vary, but my thinking is that for basic understanding of how to calculate the trig values for the "standard" angles, all that is required is to understand equilateral triangles and isosceles right triangles. Given an equilateral triangle with side 2 and an altitude drawn making a 30-60-90 triangle, hopefully your student could figure out the sides of that triangle are $2,~1,~\sqrt 3$. And given a right triangle with legs of $1$ that the hypotenuse is $\sqrt 2$. These two triangles in various orientations will fit the standard angles.

Sure, you have to memorize the Pythagorean theorem and the triangle definitions of sine, cosine, etc., but these two triangles themselves give all the ratios.

That's great. QuantumCurt, maybe your student just needs to be retaught and have a great deal of extra practice.

 

[QuantumCurt]

You guys are probably right. I don't think this method would really help the problem out at all. The student in question has a mild learning disability, particularly where memorization is concerned. He also has a distinct aversion to learning math. He simply doesn't see the point, which is half of the problem. I tutored him last semester for college algebra, and he ended up doing pretty well, with a low B. That was honestly better than I was expecting.

We're still pretty early in this semester though, so we'll see. Hopefully after some repetition he'll catch on to it. He has a good understanding of the Pythagorean Theorem, so I think I'm just going to have to relate as much of the course to right triangle concepts as I can.

 

[Mentallic]

Yes, I do think there are problems with this method. You say your student has a problem with pure memorization yet that is what your "method" requires. And it gives zero insight or understanding of what is really being calculated.

Your mileage may vary, but my thinking is that for basic understanding of how to calculate the trig values for the "standard" angles, all that is required is to understand equilateral triangles and isosceles right triangles. Given an equilateral triangle with side 2 and an altitude drawn making a 30-60-90 triangle, hopefully your student could figure out the sides of that triangle are $2,~1,~\sqrt 3$. And given a right triangle with legs of $1$ that the hypotenuse is $\sqrt 2$. These two triangles in various orientations will fit the standard angles.

Sure, you have to memorize the Pythagorean theorem and the triangle definitions of sine, cosine, etc., but these two triangles themselves give all the ratios.

That's how I still differentiate the trigs of 30 and 60 in my head, yet I didn't think to comment on it!