# Best Way to Learn Trigonometry

What are the best resources to learn trigometry? I recently decided to pursue a BS in mathematics at uni. I used to fail all my math classes with D's or F's until I started teaching myself, and so far I've done well enough teaching myself geometry, algebra, etc., to get consistent A's. However, within my first couple days of studying trigonometry I've noticed its already been a frustrating experience (I've tried about 3 books so far and still struggle understanding the first chapters). Of course, in order to do well in calculus and beyond I need to overcome this.

I personally would prefer a textbook recommendation I can download or pick up that is [preferably] not old and does not make trigonometry intimidating to approach (especially one that emphasizes understanding proofs behind properties/theorems). I'm fine with being recommended online lectures or series of videos if you insist, but I just think I learn best relying on textbooks based on my experience.

I would also like a platform one recommends to test my trigonometry skills (I don't want to rely on constantly flipping pages to find the answers to odd or even problems on the back of textbooks in order to test myself), preferably one that is cheap or free. I currently use Khan Academy's website, but for a variety of reasons they aren't ideal for me.

• Trigonometry is such a tiny part of what you need for calculus! I'd rather suggest concentrating on functions. – Peter Saveliev Aug 6 '19 at 11:25
• Is it really? That kind of relieves me because I'm already great with my Algebra, especially functions. Super comfortable with rational decomposition, function composition, logarithms, et cetera. I've only been worried about trig for awhile. I finally found a good trig textbook though and things are making sense now! – Lex_i Aug 6 '19 at 11:36
• In that case you should learn SOME trigonometry, the very basics things and not the 500 formulas every book seems to contain. – Peter Saveliev Aug 6 '19 at 11:47
• I could provide better answers i I knew which topics made you feel stuck or confused. I usually am a fan of understanding more and memorizing less, but with trig you do need to memorize the basic definitions. – Sue VanHattum Aug 6 '19 at 14:48
• @JessicaB The reason I prefer newer textbooks is because I like the style of new textbooks. I tried some from back to the 1900s, and it turned out the language and visual aids weren't as appealing or fitting as I hoped. Not to mention newer textbooks often have neat features older textbooks don't have online, such as hyperlinks (especially a hyperlinked table of contents), font adjustments, interactive visuals, etc. – Lex_i Aug 6 '19 at 16:43

## 4 Answers

Schaum's outlines are very practical in general and cheap. Well suited to an older learner. Often the answers are right after the problems versus at the end. And you get all the answers, not the odd/even gyp. Thus suited to self learning.

I like this one, overall and own it: https://www.amazon.com/gp/product/0070026505/ref=dbs_a_def_rwt_hsch_vapi_taft_p1_i10

It is from 1960ish, so the language is not archaic, but it's not "new". Not sure what benefit other than language you want from newer versions, but if you want a newer one, they have a recent 4th edition College Math you can get instead.

Note, this is a general pre-calc book (and probably what you need). But if you just want a trig primer, Schaum's has that as well. Obviously more trig problems in the trig book than the precalc book (which has all normal high school courses covered).

P.s. It would be easier to advise you if had told us what books failed you. Like did I write a long answer in vain?

P.s.s. I'm not sure why trig is so much a hurdle to people. But I do recommend to first think about the sin and cos and such in the context of the unit circle, not ratios of sides of triangles. It's just a little simpler concept and without a ratio to keep track of.

https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:trig/x2ec2f6f830c9fb89:unit-circle/v/unit-circle-definition-of-trig-functions-1 (Kahn makes it a little more complex here by talking about ratios. But when I learned it the big benefit was a very first introduction with no ratios...just x and y axes of the unit circle.

• Thanks for the reply! And you're right, I should have mentioned what books. The 3 books are Trigonometry, 5th edition by Lial, Miller, Hornsby, 1993., Trigonometry Workbook for Dummies by Mary Sterling, and College Trigonometry by Stitz and Zeager, 2013. I'll be starting precalc at uni once summer ends, and i'm sure i'll grow comfortable with trig soon enough. I just hope to learn enough in the mean time so i finish my first course without too many bumps down the road. – Lex_i Aug 3 '19 at 2:41
• Make sure you work lots of problems. You might be feeling "I'm not getting it". But if you work large amounts of problems, it will just get grooved into your head. And working problems means covering the answer, working the problem all the way. Checking your answer. Repeating (entirely) any problems missed from scratch (even for silly sign errors). Treat it like physical training for a sport or learning a musical instrument. Be diligent. – guest Aug 3 '19 at 3:53
• @RustyCore Just to be clear, I'm transferring from a local college. What I majored in college was unrelated to math and had very little math requirements, hence my first math class at uni being precalc. – Lex_i Aug 6 '19 at 16:54
• @guest, I understand. But I do think Rusty was presumptuous and rude. I'm fully aware getting this degree will probably be the most challenging and stressful time of my life, but I don't really want to shut myself from it just because I'm having a hard time with one subject. Most people quit and say they're just not math people when they face a roadblock and immediately shut themselves off from further math or from the basics they need a refresher of. I'm trying to avoid that because I did exactly that previous years. – Lex_i Aug 7 '19 at 13:36
• @Lex_i, you sound like a mature student, and I have had plenty of students like you who excel. I hope your adventures in math bring you joy. – Sue VanHattum Aug 16 '19 at 16:32

Maybe a visual approach could supplement your study? There are many such resources available on the web, not in textbooks. E.g., Trig Intuitively:

Note: the labels show where each item "goes up to."

Another: Interactive Unit Circle. Another: Inverse Trig Functions.

• it's a useful diagram. I would add a disclaimer that the concept of similar triangles is being used, in order to prevent confusion. – K.M Aug 3 '19 at 16:46
• I think the diagram would be more helpful if it showed the angle and what all the functions are a function of. It looks like it's designed for remembering what you already know, not for learning trig from scratch. – Jessica B Aug 6 '19 at 15:51
• @JessicaB: 1st, it is not my diagram :-). 2nd, there is a narrative that goes along with it; it is not intended to stand alone. 3rd, I do find it useful to see, for example, that $\sin \le \tan$ and $\sec \ge \tan$ and $\tan$ can be unbounded, etc. – Joseph O'Rourke Aug 6 '19 at 15:56
• @JessicaB: PS. The angle is the angle at the center of the circle, which circle is unfortunately nearly invisible in my snapshot. – Joseph O'Rourke Aug 6 '19 at 15:58
• @JosephO'Rourke I know you didn't draw it. And I know now that the angle is the one at the centre, because I know trig. But when I first came across it I got very confused because I hadn't picked up the relationship to the angle. – Jessica B Aug 6 '19 at 20:47

I personally would prefer a textbook recommendation I can download or pick up that is [preferably] not old and does not make trigonometry intimidating to approach (especially one that emphasizes understanding proofs behind properties/theorems).

I don't have textbooks to recommend, but I can recommend an approach to doing trigonometry that facilitates mathematical understanding of it by crystallizing the logical foundation of trigonometry and algebraic structure of trigonometric expressions. There are two 'levels' to this, depending on whether you want to go straight to complex numbers or stay within real trigonometry. In either case, the focus is on identifying the intrinsic core of trigonometry and reducing everything else to that.

### Real trigonometry

The key quantities are $$\cos(t)$$ and $$\sin(t)$$, which are the $$x$$ and $$y$$ coordinates of the point $$P_t$$ on the unit circle that subtends an arc of length $$t$$ anti-clockwise from the $$x$$-axis, as depicted in the image from wikipedia:

Here arc length is measured along the unit circle, and $$π$$ is defined as the arc length of the semicircle, so $$2π$$ is $$360°$$. (This way of measuring angles is often called measuring them in "radians", but I personally think it is an unnecessary term.) Note that $$P_t = P_{t+2πk}$$ for any integer $$k$$, because $$2πk$$ would be an integer multiple of full rounds. Also note that increasing $$t$$ moves $$P_t$$ anti-clockwise, while decreasing $$t$$ moves $$P_t$$ clockwise. Related to that, $$P_{-t}$$ is the reflection of $$P_t$$ across the $$x$$-axis.

Note that the signs of $$\cos(t)$$ and $$\sin(t)$$ match exactly the signs of the $$x$$ and $$y$$ coordinates of the point on the circle. (Do not listen to people who tell you to memorize something to determine which of them is positive in which quadrant.)

And just by definition, $$\cos(t)^2+\sin(t)^2 = 1$$ for every real $$t$$. This is the first key algebraic fact.

Next, $$\tan(t)$$ is defined as $$\sin(t)/\cos(t)$$. (Historically, we also have defined $$\sec(t) := 1/\cos(t)$$ and $$\csc(t) := 1/\sin(t)$$ and $$\cot(t) := 1/\tan(t)$$, but frankly there is little benefit to have so many when $$\cos,\sin$$ alone suffice.) Whenever you wish to simplify any trigonometric expression involving $$\cos,\sin,\tan,\sec,\csc,\cot$$, you probably should perform the standard mathematical technique of rewriting in canonical form, which in this case means rewriting in terms of $$\cos,\sin$$ alone, while taking note of where the original expression is not defined (for instance, $$1/\csc(t) = \sin(t)$$ for any real $$t$$ only when $$t$$ is not a multiple of $$π$$).

The other key algebraic facts arise from considering rotation matrices applied to vectors. (If you are unfamiliar with matrices as operators on vectors, please read this first. For an introduction to vectors in euclidean space, see here.) Let $$R$$ be any rotation about the origin in the plane. Then $$R$$ satisfies three properties:

1. $$R(u+v) = R(u)+R(v)$$ for any vectors $$u,v$$ (i.e. summing two vectors then rotating the result gives the same as rotating the two vectors first before summing them).
2. If $$R,S$$ are anti-clockwise rotations of angles $$t,u$$ respectively, then $$R∘S$$ is an anti-clockwise rotation of angle $$t+u$$.
3. If $$R$$ is an anti-clockwise rotation of angle $$t$$, then:
a. $$R(⟨x,0⟩) = ⟨x·\cos(t),x·\sin(t)⟩$$ for any real $$x$$.
b. $$R(⟨0,y⟩) = ⟨-y·\sin(t),y·\cos(t)⟩$$ for any real $$y$$.

We can take these properties as axioms (assumption) about rotations. After all, if $$R$$ does not satisfy them then we would not call $$R$$ a rotation to begin with. To see why, property (1) captures the intuition that rotating two connected rods will rotate both rods by the rotation angle while preserving where they connect. Property (2) is only needed in conjunction with property (3). Property (3a) follows from the definition of $$\cos,\sin$$, and property (3b) follows from the same definition rotated $$90°$$ anti-clockwise.

Properties (1) and (3) yield the matrix form of a 2d rotation:

If $$R$$ is an anti-clockwise rotation of angle $$t$$, then $$R = \small \pmatrix{ \cos(t) & -\sin(t) \\ \sin(t) & \cos(t) }$$.

And then using property (2) we get:

$$\small \pmatrix{ \cos(t+u) & -\sin(t+u) \\ \sin(t+u) & \cos(t+u) } = \pmatrix{ \cos(t) & -\sin(t) \\ \sin(t) & \cos(t) } \pmatrix{ \cos(u) & -\sin(u) \\ \sin(u) & \cos(u) }$$ for any reals $$t,u$$.

Multiplying the matrix product on the right and comparing with the matrix on the left immediately gives the angle-sum identities:

$$\cos(t+u) = \cos(t)·\cos(u) - \sin(t)·\sin(u)$$ for any reals $$t,u$$.

$$\sin(t+u) = \cos(t)·\sin(u) + \sin(t)·\cos(u)$$ for any reals $$t,u$$.

Whenever you wish to simplify expressions involving trigonometric functions on sums of angles, you should consider using these identities to reduce the expression to be in terms of $$\cos,\sin$$ of as few angles as possible.

In fact, all trigonometric identities involving only arithmetic operations and trigonometric functions can be proven using just the above definitions and key algebraic facts. A bit curiously, even the symmetry properties can be proven algebraically as follows.

Given any real $$t$$:

$$1 = \cos(t+(-t)) = \cos(t)·\cos(-t) - \sin(t)·\sin(-t)$$.   [angle-sum]

$$0 = \sin(t+(-t)) = \cos(t)·\sin(-t) + \sin(t)·\cos(-t)$$.   [angle-sum]

$$\cos(t) = \cos(t)^2·\cos(-t) - (\cos(t)·\sin(-t))·\sin(t)$$

$$= \cos(t)^2·\cos(-t) + (\sin(t)·\cos(-t))·\sin(t)$$

$$= (\cos(t)^2+\sin(t)^2)·\cos(-t)$$

$$= \cos(-t)$$.

$$\sin(t) = (\sin(t)·\cos(-t))·\cos(t) - \sin(t)^2·\sin(-t)$$

$$= -(\cos(t)·\sin(-t))·\cos(t) - \sin(t)^2·\sin(-t)$$

$$= -\sin(-t)·(\cos(t)^2+\sin(t)^2)$$

$$= -\sin(-t)$$.

Going on to real analysis, we would need the following facts, which can be taken as axioms for now (and justified separately later):

1. $$\sin' = \cos$$.
2. $$\cos' = -\sin$$.

As before, everything can be reduced to these, so there is no real need to memorize anything more (even though it may be convenient to do so).

### Complex trigonometry

Personally, I think it is best to go straight to the complex-valued trigonometric functions, if one desires a complete and rigorous foundation for the mathematical field of analysis. One simply defines: $$\def\rr{\mathbb{R}} \def\cc{\mathbb{C}} \def\lfrac#1#2{{\large\frac{#1}{#2}}}$$

$$\exp(z) := \sum_{k=0}^∞ \lfrac{z^k}{k!}$$ for every complex $$z$$ (after proving that the sum converges).

$$\cos(z) := \lfrac{\exp(iz)+\exp(-iz)}{2}$$.

$$\sin(z) := \lfrac{\exp(iz)-\exp(-iz)}{2i}$$.

$$π$$ is twice the first positive root of $$\cos$$ (after proving that it exists).

The motivation is that we want $$\exp : \cc→\cc$$ such that $$\exp' = \exp$$ and $$\exp(0) = 1$$, to be able to solve general linear differential equations, and we want $$\cos,\sin : \rr→\rr$$ such that $$\cos'' = -\cos$$ and $$\sin'' = -\sin$$ and $$⟨\cos(0),\cos'(0)⟩ = ⟨1,0⟩$$ and $$⟨\sin(0),\sin'(0)⟩ = ⟨0,1⟩$$, to be able to solve simple harmonic motion, and Taylor expansion brings us to the above definitions for $$\exp,\cos,\sin$$, which we can prove to converge on the entire complex plane. The above definition of $$π$$ is the easiest one that I know of that does not depend on any geometry. (For more details on this motivation, see this post.)

Suffice to say that with these definitions, we can prove by basic analysis that $$\exp,\cos,\sin$$ satisfy the desired motivating properties as well as another key property of $$\exp$$:

$$\exp(z+w) = \exp(z)·\exp(w)$$ for any complex $$z,w$$.

Using this property, we can prove all the trigonometric identities via algebraic manipulation alone (and they hold for complex variables and not just real variables).

For instance, given any complex $$z$$:

$$\cos(z)^2+\sin(z)^2 = \lfrac{(\exp(iz)+\exp(-iz))^2}{4} - \lfrac{(\exp(iz)-\exp(-iz))^2}{4}$$

$$= \exp(iz)·\exp(-iz) = \exp(0) = 1$$.

Nevertheless, it is often still easier to first prove the same key algebraic facts for $$\cos,\sin$$ and then use them to prove other identities, than to reduce everything to $$\exp$$.

• For further mathematical inquiry, you are welcome to this chat-room. – user21820 Aug 16 '19 at 8:16

Do Saylor Academy or edX have anything that will help you? They are both free platforms with math courses. Saylor Academy is almost exclusively uses a textbook - you can actually get credit through them. Modernstates.org also may help you - they have a self-guided course with videos to teach it. Rootmath may be a good resource too. Are you planning on getting credit for this course through the Clep?