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Krazie

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Krazie

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CAH

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there I taught you trig

lol, jk

I do believe it is possible to teach yourself trig, although like with anything it will take practice and dedication

However the basic applications of trig should be very easy to pick up-- of course in hindsight i guess most things seems easy once you know them

If you have any questions about trig just ask and I will see what i can do

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Krazie

Next semester I am taking a College Algebra and Trig combination class, but I am afraid that I will not learn everything that I need to know, about the trig. I have a feeling that the class will be mostly college algebra with a little trig. I already know most of college algebra now, but I have to take that class to get the credit. I just really want to be ready for calculus. I take calculus 1 this summer.

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Integral

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Krazie

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because if you are worried about being prepared for calc

skim through a precalc book and see if you understand the material after you study some trig

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Krazie

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Start learning trig then see if you can do the stuff in a precalc book

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Krazie

That is exactly what I will do. Thanks for the advice Tom, it is greatly apreciated.

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mathwonk

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i have taught calculus in college for over 35 years and no one has ever failed for not knowing trig. the main problem is algebra, and then geometry. the most important thing by far is to be good at algebra.

one other thing that also holds many people back is not knowing what a function is. i.e. most people think a function is a formula, rather than a pair, namely a domain, plus a rule for assigning values to each element of the domain.

so f(x) = 1/x defines a function on the domain of positive reals. and the additional rule that f(0) = 4, extends that to a function g defined on the non negative reals. I.e. then g(x) = 1/x if x is positive, and g(0) = 4. some people can never get it through their heads that it is ok to define a function anyway you want like this, using possibly several different rules for different parts of the domain.

but the main thing to know, above all things, is algebra.

trig takes about 5 minutes to learn. you only need to know the definitions of the two basic functions sin and cos, their relation to points and arcs on the circle, and then it helps to know the basic formulas, sin^2 + cos^2 = 1, and sin(x+y) = sin(x)cos(y) + sin(y)cos(x), and cos(x+y) = cos(x)cos(y) - sin(x)sin(y).

most people who have had trig do not even know these last two, (much less esoteric matter like the "law of cosines"),

just the basic one: sin^2 + cos^2 = 1.

but stuff like: why is [1/x - 1/a]/(x-a) = [a-x]/[(x-a)ax]? that is what stumps people.

or why is [a^(1/3) - b^(1/3)]/(a-b) =

[a-b]/[(a-b)(a^(2/3) + (ab)^(1/3) +b^(2/3)]?

its always the algebra. and this is needed every day in calculus. for example what good does it do to take a derivative and set it equal to zero, if you cannot solve it afterwards?

after the algebra it would help to know a couple basic things from geometry like similar triangles have proportional sides, and the pythagorean theorem. that's about it for geometry. of course it is nice to know a couple basic area and volume formulas, like the area of a circle, rectangle, cylinder, sphere, and triangle, but we usually review that stuff. the algebra never goes away. including the algebra of exponents.

like what is a^(1/3)? and what is (a^3 - b^3)/(a-b) ?

or (a^b)^c = ? or log(x^(10)) ?

from trig, i always have many people every year who fail to notice that since tan(x) = sin(x)/cos(x) that it is unbounded near where cos = 0. you do not need much trig to know that, just basics like do not divide by zero unless you expect something odd to happen. these people have had trig, they are just not thinking.

one other thing that also holds many people back is not knowing what a function is. i.e. most people think a function is a formula, rather than a pair, namely a domain, plus a rule for assigning values to each element of the domain.

so f(x) = 1/x defines a function on the domain of positive reals. and the additional rule that f(0) = 4, extends that to a function g defined on the non negative reals. I.e. then g(x) = 1/x if x is positive, and g(0) = 4. some people can never get it through their heads that it is ok to define a function anyway you want like this, using possibly several different rules for different parts of the domain.

but the main thing to know, above all things, is algebra.

trig takes about 5 minutes to learn. you only need to know the definitions of the two basic functions sin and cos, their relation to points and arcs on the circle, and then it helps to know the basic formulas, sin^2 + cos^2 = 1, and sin(x+y) = sin(x)cos(y) + sin(y)cos(x), and cos(x+y) = cos(x)cos(y) - sin(x)sin(y).

most people who have had trig do not even know these last two, (much less esoteric matter like the "law of cosines"),

just the basic one: sin^2 + cos^2 = 1.

but stuff like: why is [1/x - 1/a]/(x-a) = [a-x]/[(x-a)ax]? that is what stumps people.

or why is [a^(1/3) - b^(1/3)]/(a-b) =

[a-b]/[(a-b)(a^(2/3) + (ab)^(1/3) +b^(2/3)]?

its always the algebra. and this is needed every day in calculus. for example what good does it do to take a derivative and set it equal to zero, if you cannot solve it afterwards?

after the algebra it would help to know a couple basic things from geometry like similar triangles have proportional sides, and the pythagorean theorem. that's about it for geometry. of course it is nice to know a couple basic area and volume formulas, like the area of a circle, rectangle, cylinder, sphere, and triangle, but we usually review that stuff. the algebra never goes away. including the algebra of exponents.

like what is a^(1/3)? and what is (a^3 - b^3)/(a-b) ?

or (a^b)^c = ? or log(x^(10)) ?

from trig, i always have many people every year who fail to notice that since tan(x) = sin(x)/cos(x) that it is unbounded near where cos = 0. you do not need much trig to know that, just basics like do not divide by zero unless you expect something odd to happen. these people have had trig, they are just not thinking.

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JasonRox

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It is very important that you understand the fundamentals.

I taught myself Calculus, and now I am in 1st Year Calculus at a University. I learned a lot of knew stuff. I actually realized that I'm above average in the class, and some people are just lost. I also admit some parts I didn't know, and a function was one of them. I understand now. I knew that the derivative can be found using limits, while others completely forgot about First Principles.

I'm learning a lot, and it is unfortunate I skipped all my classes in high school.

Note: I also taught myself trigonometry, and physics. This is what I had to do because going back to high school wasn't an option. Now I'm a 1st year Physics/Math major. It's fun to see people brag about their math marks in high school, and I walk in and say "I HAD A SOLID 42%!".

The important thing is that, the one who takes the time to understand math (not just answer questions similiar to those in the book) will be the one who lasts.

PS. I hope I am that person. :)

Although you most likely will forget some trig identities, make sure you remember that you can derive them all using other identities you remember. Also, don't be discourage if you are a second year student and you forget something so simple like 0/0. Sometimes you are just thinking far too hard, 0/0 might seem like 1. A prof of mine said it is completely normal to just make a mistake like that.

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Alkatran

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I taught myself computer programming (go QBasic help! ) and was actually surprised how SIMPLE math functions were. I was used to having a function... DO stuff... and not just return a value. So it was really like a step down.mathwonk said:one other thing that also holds many people back is not knowing what a function is. i.e. most people think a function is a formula, rather than a pair, namely a domain, plus a rule for assigning values to each element of the domain.

So if you're taking calculus, I suggest learning to program first.

Oh... it helps with algebra, too.

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matt grime

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I'm an algebraist and I find writing programs most challenging: they do not tally.

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I disagree. I'm quite young, and haven't had loads of maths experience.programming will teach you bad habits, and even then it isn't a good idea.

I've taught myself DOS, HTML, JS, Java, C/C++ and am still learning more languages.

I found that programming has helped me ALOT. If I compare my mathematic ability now to before I learnt to program, I've improved dramatically. I also found that programming (especially the little bit of asm I've done) inspires me to do maths. :)

It also teached you about the use of different operands and functions and quite importantly; precedence.

Oh, and I also taught myseld trig (aswell as the sine and cosine rules). So, it definitely can be done :D

[r.D]

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mathwonk

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Saying trigonometry is a subject is like saying x^2 is a subject, or maybe that quadratic functions are a subject.

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HallsofIvy

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That strikes me as a peculiar thing to say. It is not unusual for secondary schools or even colleges to have courses name "trigonometry" and certainly there are many textbooks on "trigonometry". Perhaps you are using the word "subject" in a very precise sense?mathwonk said:

Saying trigonometry is a subject is like saying x^2 is a subject, or maybe that quadratic functions are a subject.

I would say that the subject, trigonometry, includes the 6 trig functions, sine, cosine, secant, cosecant, tangent, and cotangent (of course, all can be defined in terms of sine and cosine), methods of "solving" right triangles, methods of "solving" general triangles, and possibly some spherical geometry.

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Depending on the level of complexity desired you can select a good book(perhaps with the help of some of our PF friends here) for self-study. I still find trigonometry challenging at times because often, there are a lot of manipulations required which don't strike me at the right time (for instance, in an exam :uhh:).

Good luck

Cheers

Vivek

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Isn't about half of pre-calculus trig anyway? I'm not saying not to go through it all, but it's good to have a solid background in trig. Especially when you get into Caculus III and deal with vectors. I've had to do some review on trig when I did that.Tom McCurdy said:because if you are worried about being prepared for calc

skim through a precalc book and see if you understand the material after you study some trig

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mathwonk

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Halls of Ivy, I realize I said something not often said, but I intended to provoke some thought as to whether what I said is not in fact true. My point, more precisely, was that the content of so called trigonometry "courses" is so minimal or so trivial, that they do not deserve to be called courses.

As I think you yourself have pointed out elsewhere there are not really 6 essentially different trig functions, but at most one or two, or none, if you know about the exponential function. I quote you:"from the point of view of complex numbers, exponential, sine, and cosine are all the same function!"

Still perhaps I am unqualified to speak since I myself never had a trig course. We spent 6 weeks on it in high school and I stayed home those 6 weeks because school was getting so boring I needed a break.

Nonetheless, I have taught calculus for close to 40 years without much noticing the absence of that course. So perhaps I am an expert on the essential trig needed for calculus, since that is the only part I have had to learn. So I will attempt to provide a "brief trig course" here.

There are two basic trig functions called sin and cos. These are in fact inverse functions of the more natural "arclength" function. I.e. start from the point (1,0) on the unit circle and travel counterclockwise along the circle, to the point (x,y). Then for 0 <= y <= 1, the arclength traveled is a function of y, and sin is the inverse of this function. I.e. sin(t) = the y coordinate of the point reached after traveling along a circular arc of length t, in the counterclockwise direction. cos(t) is the x coordinate of the same point. These latter two functions are defined for all real t, even negative, by going on around the circle more than once, or by going clockwise for negative t.

The Pythagorean theorem implies then that for all t, cos^2(t) + sin^2(t) = 1.

This is the first big trig fact, and the most important one. Draw a picture and convince yourself of this before continuing.

The other trig functions apparently do not exist in all parts of the world, since they are just formed from dividing these, but we call sin/cos = tangent = tan, cos/sin = cotangent = cot, 1/sin = cosecant = csc, and 1/cos, = secant = sec.

Tangent is convenient because, since cos = 0 at -pi/2 and pi/2, tan goes to infinity at those points and provides a nice example of a one one correspondence between the finite interval (-pi/2, pi/2) and the whole real line.

You should now stop and graph all three functions, sin, cos, and tan and look at their graphs. You notice that they are all "periodic", i.e. repeating cyclically, in the sense that f(t+2pi) = f(t), for all t, and all trig functions f. The graphs also reveal that sin(t) = cos(t-pi/2).

A basic fact is that it is very hard to compute values of these functions. E.g. I myself do not know the value of cos(1). There are a few simple arguments whose trig values can be computed by geometry however and it helps to know these. By drawing coordinate triangles in the circle, you can see that the following versions of the trig functions are also valid:

We will say the arclength t on the unit circle cuts out the "angle t" at the origin, in terms of "radian" measure. Then if we draw a right triangle with acute angle t, we have sin(t) = opposite/hypotenuse, cos(t) = adjacent/hypotenuse, tan(t) = opposite/adjacent, etc..., where these ratios are of lengths of the correspondingly named sides of the triangle. Stop nopw and look at a picture of a unit circle to convince yourself these are the same as the original definitions of the trig functions. This approach is called "triangle trig."

From elementary triangle geometry you can deduce the values of the trig functions at pi/2. pi/3, pi/6, pi/4, and you should do so. E.g. cos(pi/2) = sqrt(2)/2 = sin(pi/2), and cos(pi/3) = 1/2, cos(pi/6) = sqrt(3)/2.

You now know about as much or more trig as the average entering calculus student, but I will continue with the "honors course" in trig. This consists of a few useful formulas which most students do not remember from their trig course.

Double angle formulas: sin(2t) = 2sin(t)cos(t). (A recent question on this site reveals that some current students of trig do not know this.)

cos(2t) = cos^2(t)-sin^2(t).

Once you know those you can bootstrap up to remembering their generalizations (addition formulas):

sin(s+t) = cos(s)sin(t) + sin(s)cos(t).

cos(s+t) = cos(s)cos(t)-sin(s)sin(t). Do you see the similarity?

These can be proved using triangle geometry, but few students know how to do this, I would guarantee. If you want a challenge, draw the picture for the trig values on the unit circle for the angle s+t, then rotate the picture down so that the arc of length s extends to one side of the point (1,0) and that of length t extends to the other. Then use pythagoras I guess, as I have not done it lately. Note these formulas imply the fact sin(t) = cos(t-pi/2), e.g.

This is plenty of trig for the first two courses in calculus. The only other formula I think you might need later, in vector situations, is the generalized Pythagorean theorem, called the "law of cosines". In a triangle with sides R,S,T and opposite angles r,s,t, this says that

R^2 + S^2 - -2RScos(t)= T^2. Notice if T = pi/2, this is Pythagoras. (I had to look this up myself.)

In Edwards and Penney, Calculus, there is a 6 page review of trig in the appendix, including a page of exercises, and then less than 3, pages of review of formulas from geometry and trig. That is more than you need for calculus.

Good luck.

As I think you yourself have pointed out elsewhere there are not really 6 essentially different trig functions, but at most one or two, or none, if you know about the exponential function. I quote you:"from the point of view of complex numbers, exponential, sine, and cosine are all the same function!"

Still perhaps I am unqualified to speak since I myself never had a trig course. We spent 6 weeks on it in high school and I stayed home those 6 weeks because school was getting so boring I needed a break.

Nonetheless, I have taught calculus for close to 40 years without much noticing the absence of that course. So perhaps I am an expert on the essential trig needed for calculus, since that is the only part I have had to learn. So I will attempt to provide a "brief trig course" here.

There are two basic trig functions called sin and cos. These are in fact inverse functions of the more natural "arclength" function. I.e. start from the point (1,0) on the unit circle and travel counterclockwise along the circle, to the point (x,y). Then for 0 <= y <= 1, the arclength traveled is a function of y, and sin is the inverse of this function. I.e. sin(t) = the y coordinate of the point reached after traveling along a circular arc of length t, in the counterclockwise direction. cos(t) is the x coordinate of the same point. These latter two functions are defined for all real t, even negative, by going on around the circle more than once, or by going clockwise for negative t.

The Pythagorean theorem implies then that for all t, cos^2(t) + sin^2(t) = 1.

This is the first big trig fact, and the most important one. Draw a picture and convince yourself of this before continuing.

The other trig functions apparently do not exist in all parts of the world, since they are just formed from dividing these, but we call sin/cos = tangent = tan, cos/sin = cotangent = cot, 1/sin = cosecant = csc, and 1/cos, = secant = sec.

Tangent is convenient because, since cos = 0 at -pi/2 and pi/2, tan goes to infinity at those points and provides a nice example of a one one correspondence between the finite interval (-pi/2, pi/2) and the whole real line.

You should now stop and graph all three functions, sin, cos, and tan and look at their graphs. You notice that they are all "periodic", i.e. repeating cyclically, in the sense that f(t+2pi) = f(t), for all t, and all trig functions f. The graphs also reveal that sin(t) = cos(t-pi/2).

A basic fact is that it is very hard to compute values of these functions. E.g. I myself do not know the value of cos(1). There are a few simple arguments whose trig values can be computed by geometry however and it helps to know these. By drawing coordinate triangles in the circle, you can see that the following versions of the trig functions are also valid:

We will say the arclength t on the unit circle cuts out the "angle t" at the origin, in terms of "radian" measure. Then if we draw a right triangle with acute angle t, we have sin(t) = opposite/hypotenuse, cos(t) = adjacent/hypotenuse, tan(t) = opposite/adjacent, etc..., where these ratios are of lengths of the correspondingly named sides of the triangle. Stop nopw and look at a picture of a unit circle to convince yourself these are the same as the original definitions of the trig functions. This approach is called "triangle trig."

From elementary triangle geometry you can deduce the values of the trig functions at pi/2. pi/3, pi/6, pi/4, and you should do so. E.g. cos(pi/2) = sqrt(2)/2 = sin(pi/2), and cos(pi/3) = 1/2, cos(pi/6) = sqrt(3)/2.

You now know about as much or more trig as the average entering calculus student, but I will continue with the "honors course" in trig. This consists of a few useful formulas which most students do not remember from their trig course.

Double angle formulas: sin(2t) = 2sin(t)cos(t). (A recent question on this site reveals that some current students of trig do not know this.)

cos(2t) = cos^2(t)-sin^2(t).

Once you know those you can bootstrap up to remembering their generalizations (addition formulas):

sin(s+t) = cos(s)sin(t) + sin(s)cos(t).

cos(s+t) = cos(s)cos(t)-sin(s)sin(t). Do you see the similarity?

These can be proved using triangle geometry, but few students know how to do this, I would guarantee. If you want a challenge, draw the picture for the trig values on the unit circle for the angle s+t, then rotate the picture down so that the arc of length s extends to one side of the point (1,0) and that of length t extends to the other. Then use pythagoras I guess, as I have not done it lately. Note these formulas imply the fact sin(t) = cos(t-pi/2), e.g.

This is plenty of trig for the first two courses in calculus. The only other formula I think you might need later, in vector situations, is the generalized Pythagorean theorem, called the "law of cosines". In a triangle with sides R,S,T and opposite angles r,s,t, this says that

R^2 + S^2 - -2RScos(t)= T^2. Notice if T = pi/2, this is Pythagoras. (I had to look this up myself.)

In Edwards and Penney, Calculus, there is a 6 page review of trig in the appendix, including a page of exercises, and then less than 3, pages of review of formulas from geometry and trig. That is more than you need for calculus.

Good luck.

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krab

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mathwonk

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As evidenced above, the trig functions cannot be rigorously defined until one has a way of defining arc length , i.e. angles, rigorously. Thus in fact trig is not a logical prerequisite for calculus, but the other way around. I.e some form of limiting process is required to define the trig functions.

One way is merely to fill the logical gap left above by defining circular arclength, which is of course done by the definite integral of 1/sqrt(1-x^2). The sin is the inverse function of the corresponding indefinite integral, and the discussion above goes through.

Another nice choice, the one made by my freshman calc prof, was to use infinite series. I.e. we began with limits of sequences and series then defined complex numbers and then the function e^z as the usual power series: sum of z^n/n! for n from 0 to infinity.

Then he defined cos(t) and sin(t) for real t, as the real and imaginary parts of e^(it). More generally, cos(z) = (1/2)[e^(iz) + e^(-iz)] and sin(z) = (1/2i)[e^(iz) - e^(-iz)].

Then one can deduce all the usual properties of sin and cos. In particular the sometimes hard to remember addition laws are just the real and imaginary parts of the law e^(z+w) = e^z e^w, applied to e^i(s+t).

So in fact my calculus course did not have any trig prerequisite, fortunately for me.

I repeat, trig is not a true subject, not an idea, not a method. It is a small collection of simple but important special functions with special properties. They have no role in preparing you for calculus. They provide examples for using calculus. Calculus rather is useful in studying these special functions.

To reiterate, the opposite is true for algebra. Algebra is a subject, and its methods are used every day in calculus. I advise reading say the 6 page appendix in edwards and penney's calculus book on trig, and working the exercises, as a minimal preparation in trig, hopefully sufficient. I also suggest printing out my short course above and studying it. But the main prerequisite for calc is still algebra.

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mathwonk

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There is in fact an important subject inspired by the trig functions, namely fourier analysis. the point is that the trig functions, taken together, make a connection between linear motion and circular motion. In algebra this is a homomorphism from the translation group on the line, to the rotation group on the circle.

This is a fundamentally important method of transferring a problem from one group to another where it may be easier or more transparent.

The classical trig functions cos and sin taken separately lose this idea however since it is apparent only when they are taken together, in the guise of the compelx exponential e^(it) = cos(t) + isin(t). Then the unnatural addition laws of sin and cos become the natural homomorphism law of exp, i.e. e^(z+w) = e^z e^w, which visibly translates addition into multiplication.

Conceivably a classical education in trigonometry could lead someone to a greater interest in fourier analysis, and thus have a good outcome. perhaps my ignorance of trig is why I was slow to appreciate the beauties of fourier series and integrals.

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mathwonk

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"E.g. cos(pi/2) = sqrt(2)/2 = sin(pi/2)"

this should be cos(pi/4) = sqrt(2)/2 = sin(pi/4).

isn't it wonderful to have students who correct your mistakes?

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