Thread: Help me with trigonometry please.

Trig can be a beast but it's pretty once you get the hang of it. I have a few questions.

Have you studied complex numbers? Those make it a lot simpler.

Do you understand the geometric interpretations of the functions?

If you're working mostly on a computer, I suggest you always have pen and paper ready, so you can make quick sketches of whatever you're working with. It is infinitely easier than any solution you have on your PC, and it helps you a ton to have a drawing to look at, if something doesn't make sense inside your head.

The reason you suck at math is that you want to memorize the functions. Understand how it works from the ground up, and you will never have to memorize a thing. If you can draw a right triangle with angles A, B, C and edge lengths a, b, c, and if you can express "sin(A)" or "cos(B)" in terms of a, b or c, then you can easily solve just about any problem, and derive just about any trig inequality.

It takes practice at first, but trust me: Memorization is the dumbest way to do math. I know, I've seen people do it.

Thanks everyone. I don't know complex numbers yet. I'll do what you said Replicon.

With math in general I find the following important:

The most important thing for math and most subjects: don't fall behind. In math, a lot builds up from what you've been learning in the last few classes. If you make sure you understand each lesson right after learning it, you'll find there actually isn't much to learn or memorize, and you'll probably find that each lesson, taken individually, isn't that difficult. It's only when you wait two weeks without grasping the material and then decide to study it all in one day that it will seem very hard and stressful.

Do a lot of practice problems. First, try to understand the material by reading your notes or the relevant textbook sections, until you think you understand it. You don't have to understand it perfectly on the first run, just try to get the general idea. Then, start doing practice problems, referring back to the notes or textbook sections if you get stuck. And only do the questions for which the answers are given in the back of the textbook, the odd-numbered questions for example. Getting a lot of practice is the key, I think, in math. Even if you think you understand the basics, chances are you'll run into situations on tests that you'll never have considered before unless you've gotten a lot of practice.

wolframalpha.com is a great site. You can type in mathematical formulas, algebra, etc, and it will solve the problem for you and often give you the graph. It recognizes a lot of English words and formats. For example, I just typed in "cos(x)+y = 5, solve for x" and it gave me the answer and a graph. It can be a great help. For example, if you've just solved an algebra problem and got the wrong answer but don't know where you went wrong, you can type it into wolframalpha will often show you how to solve it step-by-step so you can see exactly where you went wrong. It can speed up your practice a lot.

You will have to memorize a few things, but not much. If you refuse to memorize anything you won't succeed. But I find it's mostly memorizing maybe 2 or 3 facts or rules per section. I find understanding math is really just understanding a few very basic concepts and rules. Once you understand them, it's just problem solving which, when you understand the rules, can even be a bit fun.

Thanks so much Dianeva wolframalpha makes my life way easier.

Replicon is totally right. The way to learn maths is to understand it, which is also the case with most of science. This has many advantages: firstly, you can pretty much never forget something if you understand why it is true. Secondly, you will be a lot more confident with applying your knowledge, especially to novel situations. Thirdly, it's a lot easier to learn more stuff which builds on this knowledge, and the thing about maths is that new stuff always builds on what came before.

The only things you need to learn are definitions, and the number of these is tiny. In the case of trig, there are three main definitions;



cos(A) is the length of the adjacent (the side touching A) divided by the hypotenuse, sin(A) is similar but is the opposite side divided by the hypotenuse, and finally tan(A) is the opposite side of A divided by the adjacent side.

Already you can deduce one formula: tan(A) = sin(A)/cos(A)

This is because by definition sin(A)/cos(A) is

(opposite/hypotenuse) / (adjacent/hypotenuse),

and then by the basic laws of fractions you just cancel the /hypotenuse on the top and the bottom to find that it

= opposite/adjacent,

which is tan(A) by definition.

This is a perfect example; by understanding the above formula, that it is a simple consequence of cancelling the hypotenuses, you will never forget it; you understand that it is obviously true. However, somebody who doesn't put in the half a minute required to understand it instead has to memorise the entire formula, which ultimately will take much longer, much more effort, and they will have no means of knowing if what they've put is actually right.

Xei, do you happen to know a good way to derive the formulas for e.g., cos(a + b), that doesn't use e^{i(a + b)} or rotation matrices? Those are the only two I know and it would be good to know one that doesn't require complex multiplication or matrix multiplication.

Thank you all so much for helping me

sin(A+B)

Originally Posted by PhilosopherStoned

Xei, do you happen to know a good way to derive the formulas for e.g., cos(a + b), that doesn't use e^{i(a + b)} or rotation matrices? Those are the only two I know and it would be good to know one that doesn't require complex multiplication or matrix multiplication.

Yes; I know a geometrical argument for sin(A+B) which is the simplest one. All variations you can then get using cos(x) = sin(pi/2 - x), sin(-x) = -sin(x), and cos(-x) = cos(x).

All the other stuff like trig(A)trig(B) and trig(A) + trig(B) also follows from that one geometrical argument.

Basically you have two right angled triangles one on top of each other; angles A and B at the origin, the bottom triangle's adjacent horizontal, and the top triangle's adjacent the same as the bottom triangle's hypotenuse. You give the top triangle's hypotenuse a length of 1 and the rest follows.

The form of a rotation matrix is usually deduced using the addition formulae so I don't think you can do it that way round. The same can apply to the Euler's identity way round, but that can also be deduced from Taylor series so that's fine. You could also do it straight from the Taylor series though it'd be a bit ugly.

I don't understand your construction. How can the bottom triangle's adjacent horizontal be the same as its hypotenuse? I'm also not sure which leg is the adjacent horizontal and which is the adjacent. I've seen them referred to as the adjacent and the opposite with respect to an angle. In this case a or b.

With respect to the rotation matrices, you don't need the addition formulae. Just note that a rotation should be linear because it should preserve parallelograms and scaling a parallelogram before rotation should be the same as scaling it after a rotation. But a linear transformation is uniquely determined by its effect on a basis. So (1, 0) should get mapped to (cos a, sin a) and (0, 1) should get mapped to (-sin a, cos a). Transpose everything in site and put it in a matrix. We can then take this as the definition of a rotation.

It's the top triangle's adjacent which is the same as the bottom triangle's hypotenuse.

You're right about the matrix thing. I was going over my matrices notes recently and it initially deduces the rotation matrix by explicitly working out the image of a general vector specified by its size and angle as a sort of introduction and that was what was in my mind, but of course you can do it without addition formulae by considering the effect on the Cartesian basis.

I suck at trig but one thing you should know is soh cah toa

PM me your trig questions... I'll destroy them

Get some programs for your calculator to do it for you. (Google it)

Do it in a lucid dream and ask for help.(Yea, im serious)

Originally Posted by Codename

Get some programs for your calculator to do it for you. (Google it)

What an excellent way to learn.

Originally Posted by Xei

What an excellent way to learn.

Yep!