Fundamentals of Trigonometry (No Graphing/Inverse Trig)

[yourdadonapogostick]

it was a one semester class, so it is basic. note: i will not cover graphing and inverse trig function.

i will begin by introducing a unit of measurement of angles, because i find them much easier to work with. that unit is radian. the name will make sense after the description. we have a circle whose center is at the origin. as we all know, the circumference of a circle is $$C=2{\pi}r$$. assume the radius of the circle is one. the circumference can be thought of as the full rotation of the radius, so a full rotation is $$2{\pi}=360^0$$. half a rotation is $$\pi=180^0$$. a forth of a rotation is $$\frac{\pi}{2}=90^0$$ and so on. angles are measured from the positive x-axis(initial side) in a counter clockwise manner to the terminal side. negative angles are clockwise. to convert radians to degrees, multiply the radian measurement by $$\frac{180}{\pi}$$. to convert from degrees to radians multpily the degree measurement by $$\frac{pi}{180}$$

every angle has a reference angle($$\alpha\angle$$. a reference angle is the smallest positive acute angle made by the terminal side of $$\theta$$ and the x-axis. in the first quadrant, $$\alpha\angle=\theta$$. in the second quadrant, $$\alpha\angle=\pi-\theta$$. in the third, $$\alpha\angle=\theta-\pi$$. in the fourth, $$\alpha\angle=2\pi-\theta$$. trig functions of $$\theta=\underline{+}same function of \alpha\angle$$

each angle also has an infinite number of coterminal angles. coterminal angles are angles that have the same terminal side(kinda makes sense, huh).coterminal angle=$$\theta\underline{+}n2\pi$$

the trig functions: sin, cos, tan, csc, sec, cot are all ratios of the sides of a right triangle. each angle has a specific value for each of the trig functions.
sin and cos, sec and csc, tan and cot are what are called cofunctions. cofunctions are positive in the same quadrant. in the first quadrant, all functions are positive. in the second, sin and csc are positive. in the third, tan and cot are positive. in the fourth, cos and sec are positive. the trig function of any acute angle equals the cofunction of said angle's complement.

sin and csc, cos and sec, tan and cot are reciprocal functions. that will make sense once you see their definitions and identities

$$sin=\frac{opposite side}{hypotenuse}$$
$$cos=\frac{adjacent side}{hypotenuse}$$
$$tan=\frac{opposite side}{adjacent side}$$
$$csc=\frac{hypotenuse}{opposite side}$$
$$sec=\frac{hypotenuse}{adjacent side}$$
$$cot=\frac{adjacent side}{opposite side}$$

reciprocal identities
$$sin\theta=\frac{1}{scs\theta}$$
$$csc\theta=\frac{1}{sin\theta}$$
$$cos\theta=\frac{1}{sec\theta}$$
$$sec\theta=\frac{1}{cos\theta}$$
$$tan\theta=\frac{1}{cot\theta}$$
$$cot\theta=\frac{1}{tan\theta}$$

ratio identites
$$tan\theta=\frac{sin\theta}{cos\theta}$$
$$cot\theta=\frac{cos\theta}{sin\theta}$$

pythagorean identities
$$sin^2\theta+cos^2\theta=1$$
$$1+tan^2\theta=sec^2\theta$$
$$1+cot^2\theta=sec^2\theta$$

cofunction identities
$$sin(\frac{\pi}{2}-\theta)=cos\theta$$
$$cos(\frac{\pi}{2}-\theta)=sin\theta$$
$$cos(\frac{\pi}{2}-\theta)=sin\theta$$
$$tan(\frac{\pi}{2}-\theta)=cot\theta$$
$$cot(\frac{\pi}{2}-\theta)=tan\theta$$
$$sec(\frac{\pi}{2}-\theta)=csc\theta$$
$$scs(\frac{\pi}{2}-\theta)=sec\theta$$

 

[Jameson]

My browser cannot load half of the Latex graphics. I don't know if that's just myself.

I like how you are going over basic trig, but it seems to be a lot of formulas and no so much explaining. Perhaps if you derived the formulas, it would be more helpful. Keep it up.

Jameson

EDIT: When writing plain text in Latex, use this: \mbox{} . It looks nicer.

 

[yourdadonapogostick]

maybe i should break it up.

with formulas you don't really need a lot of explaining. if anyone has questions, just ask.

 

[yourdadonapogostick]

even/odd identities
$$sin(-\theta)=-sin\theta$$
$$cos(-\theta)=cos\theta$$
$$tan(-\theta)=-tan\theta$$
$$csc(-\theta)=-csc\theta$$
$$sec(-\theta)=sec\theta$$
$$cot(-\theta)=-cot\theta$$

solving triangles(side a is opposite anlge alpha; side b is opposite angle beta; side c is opposite angle gamma)
law of sines-$$\frac{a}{sin\alpha\frac{b}{sin\beta}=\frac{c}{sin{\gamma}}$$
law of cosines-$$c^2=a^2+b^2-2abcos\gamma$$ *note: when gamma is a right angle, law of cosines turns into pythagorean theorem*

area of triangles
$$A=\frac{1}{2}absin\gamma$$
$$A=\sqrt{s(s-a)(s-b)(s-c)}$$, when $$s=\frac{a+b+c}{2}$$

now that you have all of that, here are some formulas

double angle formulas
$$sin2\theta=2sin\thetacos\theta$$
$$cos2\theta=cos^2\theta-sin^2\theta$$
$$cos2\theta=1-2sin^2\theta$$
$$cos2\theta=2cos^2\theta-1$$
$$tan2\theta=\frac{2tan\theta}[1-tan^2\theta}$$

half angle formulas
$$sin\frac{\theta}{2}=\sqrt{\frac{1-cos\theta}{2}}$$
$$cos\frac{\theta}{2}=\sqrt{\frac{1+cos\theta}{2}}$$
$$tan\frac{\theta}{2}=\sqrt{\frac{1-cos\theta}{1+cos\theta}}$$
$$tan\frac{\theta}{2}=\frac{1-cos\theta}{sin\theta}$$
$$tan\frac{\theta}{2}=\frac{sin\theta}{1+cos\theta}$$

power reducing formulas
$$sin^2\theta=\frac{1-cos2\theta}{2}$$
$$cos^2\theta=\frac{1+cos2\theta}{2}$$
$$tan^2\theta=\frac{1-cos2\theta}{1+cos2\theta}$$

sum and difference formulas
$$sin(\alpha+\beta)=sin\alphacos\beta+cos\alphasin\beta$$
$$sin(\alpha-\beta)=sin\alphacos\beta-cos\alphasin\beta$$
$$cos(\alpha+\beta)=cos\alphacos\beta-sin\alphasin\beta$$
$$cos(\alpha-\beta)=cos\alphacos\beta+sin\alphasin\beta$$
$$tan(\alpha+\beta)=\frac{tan\alpha+tan\beta}{1-tan\alphatan\beta}$$
$$tan(\alpha-\beta)=\frac{tan\alpha-tan\beta}{1+tan\alphatan\beta}$$

sum and difference to product formulas
$$sin\alpha+sin\beta=2sin\frac{1}{2}(\alpha+\beta)cos\frac{1}{2}(\alpha-\beta)$$
$$sin\alpha-sin\beta=2cos\frac{1}{2}(\alpha+\beta)sin\frac{1}{2}(\alpha-\beta)$$
$$cos\alpha+cos\beta=2cos\frac{1}{2}(\alpha+\beta)cos\frac{1}{2}(\alpha-\beta)$$
$$cos\alpha-cos\beta=-2sin\frac{1}{2}(\alpha+\beta)sin\frac{1}{2}(\alpha-\beta)$$

 

[yourdadonapogostick]

product to sum and difference formulas
$$sin\alphasin\beta=\frac{1}{2}[cos(\alpha-\beta)+cos(\alpha+\beta)]$$
$$cos\alphacos\beta=\frac{1}{2}[cos(\alpha-\beta)+cos(\alpha+\beta)]$$
$$sin\alphacos\beta=\frac{1}{2}[sin(\alpha+\beta)+sin(\alpha-\beta)]$$

 

[Jameson]

I've already taken trig... I don't need the explaining. However, I find that memorizing a formula and not really understanding why isn't good for the student. How do you even konw that these formulas are true? Prove it.

I'm not saying you can't, I just know I like it when my teachers give a little more explanation.

Jameson

 

[lurflurf]

The thing I disliked about basic trig is a complete lack of rigor. There are several ways to define trig functions
1. as solutions to differential equations
2. as infinite series
3. as ratios of triangles
4. as functions of arc lenth traversed on a circle
5. as functions satisfying certain functional relations
1,2 require too much calculus
3,4 require too much geometry and are at great risk of losing rigor in how they define angle measure.
Clearly 5 is the best despite being least common.
Among several ways of presenting 5 a particularlly nice one is
Let sin and cos be the functions R->R which have the following properties
1) sin(x+y)=sin(x)cos(y)+cos(x)sin(y) all x,y in R
2) cos(x+y)=cos(x)cos(y)-sin(x)sin(y) all x,y in R
3) 1=(sin(x))^2+(cos(x))^2 all x in R
4) lim x->0 sin(x)/x=1

Theorem 1 sin and cos exist and are unique
Theorem 2 there exit at least 1 x in R such that sin(x)=cos(x)
Definition pi/4= the smallest positive value such that sin(x)=cos(x)
From here other functions can be defined, other identities derived, and values can be found such as sin(pi/10), cos(pi/12). And all is good in trig land.

 

[krab]

LaTeX notes: (1) precede trig functions with a backslash. LaTeX otherwise cannot tell that sin is not s times i times n. (2) If you have non-math, like "hypotenuse" in a formula, same thing applies: LaTeX thinks it is h times y times... In these cases, put the English in \mbox{...} (put it where the dots are). Then it is formatted as prose instead of math and in particular, spaces and punctuation are observed. So for example,
$$\sin\theta=\frac{\mbox{opposite side}}{\mbox{hypotenuse}}$$

 

[Jameson]

For the pythagorean identities, you could show that the second two can be derived from the first.

$$\sin^2\theta+\cos^2\theta=1$$

$$\frac{\sin^2{\theta}}{\sin^2{\theta}}+\frac{\cos^2{\theta}}{\sin^2{\theta}}=\frac{1}{sin^2{\theta}}$$

Which leads into the third identity you listed.

$$1+\cot^2{\theta}=csc^2{\theta}$$

Divide the first by $\cos^2{\theta}$ for the second.

 

[yourdadonapogostick]

The thing I disliked about basic trig is a complete lack of rigor. There are several ways to define trig functions
1. as solutions to differential equations
2. as infinite series

it said BASIC...do you know what that means?

 

[Jameson]

Ok, Ok, fair enough. You did say basic.

Also, I cannot view the Latex for the Law of Sines. Perhaps there is an error.

 

[yourdadonapogostick]

law of sines
$$\frac{a}{\sin{\alpha}}=\frac{b}{\sin{\beta}}=\frac{c}{\sin{\gamma}}$$

 

[lurflurf]

it said BASIC...do you know what that means?

It could mean several things
such as
A programming language.
a pH>7.0
a certain type of rock
I will consider those meanings most likely to apply
The most simple complete form
Who can say which forms are complete or most simple. I would argue that given the geometry background of most students of trigonometry a geometric definition is not complete or simple.
primary importance
The primary importance in trig is debatable, but for me triangle ratios while an interesting and useful application are not primarily important.
a starting point
Certainly any definition could be used as a starting point and hence would be basic.

Basic does not mean easy. Though it is questionable if your style of presenting trig is the easiest (for a typical student), for the sake of argument say it is 10% easier than other methods. That does not mean it is best as other methods might offer a greater than 10% gain in return for the 10% increase in effort.

My key point is that geometric definitions require a notion of arc lenth and/or one of angle measure. These notions are quite obscure at the level of intro to trig. Some would say that drawing a triangle and holding a protrator up to it is sufficient to define angle measure. I am not one of those people. Also there is something to be said for learning something right the first time. Often in eduction there is a push to try and simplify ideas too much. In the end more effort can be expended learning several versions of something than is needed to learn a good version first. These thing need to be carefully considered I do not necessarily think Lebesgue Measure needs to be taught to high school students.

 

[yourdadonapogostick]

it's called context...please stop the pointless flame...

well, you seem to already have sead notions down, so it doesn't really matter, now does it?

 

[Cincinnatus]

"Theorem 1 sin and cos exist and are unique
Theorem 2 there exit at least 1 x in R such that sin(x)=cos(x)
Definition pi/4= the smallest positive value such that sin(x)=cos(x)
From here other functions can be defined, other identities derived, and values can be found such as sin(pi/10), cos(pi/12). And all is good in trig land."

Lurflurf, how would you prove these theorems from the properties you stated?

 

[lurflurf]

Lurflurf, how would you prove these theorems from the properties you stated?

Thats the best part!
I will outline it hopefully anyone interested can provide themselves with the details.
Exististance is easy is you cheat a little by considering a pair of functions for a dense subset of the reals defined by f(0)=1 f(pi/2)=0 g(0)=1 g(pi/2)=0 0<f(x),g(x)<1 when 0<x<pi/2. And satisfation of f(x+y)=f(x)f(y)-g(x)g(y) g(x+y)=g(x)f(y)=(f(x)g(y) f(x)^2+g(y)^2=1 will be consistant. Next show lim x->0 f(x)/x=1. Now the limit implies continuity so we define a function on all real by filling in the gaps. In particular we can use n*pi/2^m for n,m integers as the dense subset. For uniquness Assume the pair S(x),C(X) also meet the definition. since lim s(x)/x=1 and lim sin(x)/x=1 lim sin(x)/S(x)=1 consider x> 0 so sin(x) and S(x) do not agree then consider repeated use of half angle identity sin(x/2^n)/S(x/2^n) will not be close to 1 for large n -><-. sin(0)=0 cos(x)=1 and sin(x)^2+cos(x)^2=1 this implies that sin increases and cos decreases at certain rate in an interval (0,b). If they are never equal at some point cos(x)<sin(x) but this cannot happen without them having been equal prior. We can get values for pi/10 and pi/12 by considering sin(5x) and sin(3x) type identities.

 

[Cincinnatus]

Sorry, I don't really understand this proof.
I have never proved this kind of statement before and I am not really sure what must be shown in order to make this kind of conclusion...

Perhaps I don't really know what is meant when you say the functions exist in this sense...

could you perhaps give an example of a set of properties that would define a function that could not exist?

 

[lurflurf]

Sorry, I don't really understand this proof.
I have never proved this kind of statement before and I am not really sure what must be shown in order to make this kind of conclusion...

Perhaps I don't really know what is meant when you say the functions exist in this sense...

could you perhaps give an example of a set of properties that would define a function that could not exist?

A function that does not exist would be one whose properties are contradictory. A silly example might be a function where x<f(x) for all x that also has f(x)<1 for all x, 2<f(2)<1 which cannot happen. A slightly less silly example might be |f(x)-f(y)|<(x-y)^2 for all x and y (where x and y are not equal )together with f(2)=1+f(1). A good definition defines something that exist in the sense that it is possible for something to satify all the requirements and is unique in the sense that the definition does not cause confusion in that it implies several non identiacl thins are identical. To prove sin and cos are unique you can start by defining a function that meets some of the requirements. You can cheat a little by using known properties of sin and cos such as sin(pi/4)=cos(pi/4) since for the existence part one is trying to show that it is possible for a functon to have these properties. lim x->0 sin(x)/x=1 implies that sin is continuous so we can define a function on a dense subset of the reals. If two continuous functions on agree on a dense subsets of the reals they agree at all real numbers. So we can define functions on real numbers of the form n*pi/2^m where n and m are integers and this will be a dense subset. We do this by letting C(0)=S(pi/2)=1 and C(pi/2)=S(0)=1 where C(x) and S(x) are prototypes of trig functions in a sense. Then the relations
C(x+y)=C(x)C(y)-S(x)S(y)
S(x+y)=S(x)C(y)+C(x)S(y)
(S(x))^2+(C(x))^2=1
are used do define the functions are all real numbers of the form n*pi/2^m
Since these relations were used in defining the functions they are consistent. Finally it is shown that lim x->0 sin(x)/x=1
Finally we extend the definition to all real numbers filling in gaps that is by requireing the functions to be continuous. This is the same method used for introducing irrational exponents in high school algebra. x^(m/n) is defined for n,m integers (n>0) thus making sure x^r is continuous defines x^r for r a real number. An interesting note is pi is used about but an unknow constant could be used, then during the work one would learn that the constant is pi.

 

[pl_terranine]

i have 2 questions. how were trigonometric tables made back then? I couldn't just divide sin(45) in half to get sin(22.5) so i was wondering what techniques they used.

 

[lurflurf]

i have 2 questions. how were trigonometric tables made back then? I couldn't just divide sin(45) in half to get sin(22.5) so i was wondering what techniques they used.

What do you mean by back then? If you know the values for 18 degrees and 15, then you can find the values for multiples of 3 degrees. You can use half angle formula to get values for multiples of 1.5 degrees. Then you can write values in terms of known values and small values and use.
for small x
sin(x)~x-x^3/6+x^5/120+...
cos(x)~1-x^2/2+x^6/720+...
This assumes that you can calculate the roots in exact values, otherwise you can build up values from small values. Also you could draw large triangles and try to measure it as closely as possible. Of course now we have computers...

 

[Cincinnatus]

If two continuous functions on agree on a dense subsets of the reals they agree at all real numbers.

I didn't know this fact, actually I've never heard the word dense used, but I've looked it up now and I guess I understand. I'll have to mull over the rest of your explanation a little longer though.

thanks

 

[mathwonk]

here is a brief outline of the first basic trig cousre i ever had. i recommend it.

define e^z for complex z by the infinite series: 1 + z + z^2/2! + z^3/3! + ...

deduce that d/dz (e^z) = e^z.

then prove that e^(z+w) = e^z e^w; and [e^z]^w = e^(zw).

then define cos(z) = (1/2) [e(iz) + e^(-iz)], and sin(z) = (1/2i)[e^(iz) - e^(-iz)].

then deduce that sin(z+w) = sin(z)cos(w) + cos(z)sin(w),

and cos(z+w) = cos(z)cos(w) - sin(z)sin(w).

deduce that cos' = sin and sin' = -cos, and cos^2 + sin^2 = 1, and cos, sin both satisfy f'' + f = 0.

hence e^(ix) = cos(x) + isin(x), and thus for some value of x, e^ix = 1. define the smallest positive such x to be 2pi.

then one is pretty much done.

 

[Cincinnatus]

define e^z for complex z by the infinite series: 1 + z + z^2/2! + z^3/3! + ...

deduce that d/dz (e^z) = e^z.

then prove that e^(z+w) = e^z e^w; and [e^z]^w = e^(zw).

then define cos(z) = (1/2) [e(iz) + e^(-iz)], and sin(z) = (1/2i)[e^(iz) - e^(-iz)].

then deduce that sin(z+w) = sin(z)cos(w) + cos(z)sin(w),

and cos(z+w) = cos(z)cos(w) - sin(z)sin(w).

deduce that cos' = sin and sin' = -cos, and cos^2 + sin^2 = 1, and cos, sin both satisfy f'' + f = 0.

hence e^(ix) = cos(x) + isin(x), and thus for some value of x, e^ix = 1. define the smallest positive such x to be 2pi.

Ok, I can do all of that up till the last line where it becomes a bit fuzzy.
Why can we just declare that value to be 2pi like that?

 

[lurflurf]

Ok, I can do all of that up till the last line where it becomes a bit fuzzy.
Why can we just declare that value to be 2pi like that?

Several of the possible definitions of sin and cos fully define the functions, but do not explictly give the period. At what ever stage addition laws are developed it becomes clear that any non zero number number for which sin(x)=0 and cos(x)=1 will be a period then we "define" 2pi as the smallest positive number with that property. That the 2pi we define is the same as the two pi that comes from some other definition of pi has to be established. Mathwonks definition is one of the five main definitions that are used, but I question if it is the best one for beginers. That definition requires knowledge of complex numbers, infinite series of complex numbers, and differential calculus. It also does not make all the most important features of the functions obvious. It is can be a good definition when one has the needed back ground. The fact that similar definitions are used in Landau and in Rudin says alot.

 

[Cincinnatus]

So how can we show that our "2pi" agrees with some other definiton of 2pi?


I've actually heard of all the definitions you mentioned earlier before (though not in so much detail). I asked about your function definition since it was most unfamiliar to me.

 

[lurflurf]

So how can we show that our "2pi" agrees with some other definiton of 2pi?


I've actually heard of all the definitions you mentioned earlier before (though not in so much detail). I asked about your function definition since it was most unfamiliar to me.

Which "other" definition did you have in mind? As a matter of taste I prefer to define pi/4 because the fact that sin(pi/4)=cos(pi/4) speaks to me personally, but it does not really matter. Consider that
$$\lim_{x\rightarrow\infty}\frac{\sin(x)}{x}=1$$
thus
$$\lim_{n\rightarrow\infty}2^n\sin(\frac{\pi}{2^n})=\pi$$
The fact that both sides of the equation involve pi could be a complication, but we note sin(pi/4)=sqrt(2)/2 and use the half angle identity many times to yeild
$$\frac{\pi}{4}=\lim_{n\rightarrow\infty}2^n\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{...+\sqrt{2}}}}}}}}$$
where the number of square roots used is n+2
thus giving another definition of pi, more importantly one that can be used to calculate it.

 

[mathwonk]

it depends on your definition of beginner. i liked the approach via infinite series because it was precise and not boring like the high school approach. in that approach one has no rigorous definition of angle measure.

i found trig so tedious in high school i skipped school durting the whole 6 weeks of it. all they did was compute angles using numerical equations.

to me, the series approach using derivatives does also make the most basic properties "obvious" at least if you believe in proofs. and i had this course in freshman year calculus, so it came about as early as necessary.

remember euler taught infinite series in his precalculus book as a reasonable precursor to calculus. i think euler was right. i.e. infinite series are a very intuitive tool, that is taught much too late.

even naive students who do not appreciate rpofos, can appreciate and use infinite series calculations very early.

the best way to persuade a student that the derivative of e^x is really e^x is to show them the series and let them differentiate it.

failing to understand the full theory of convergencve of seris is no worse than failing to understahnd the continuity and limit theory assuemd in calculus.

and the bonus is that afterwards a student can actually use infinite series, whereas most students having been presented with the fundamental theorem of calculus, still cannot understand it at all.

i.e. it is much easier for most students to believe thjat an infinite series defines a function than that an indefinite integral does.

indeed we were first presented with an infinite series in 8th grade, the geometric series. i never forgot it.

but maybe others enjoy trig presented in the usual imprecise way.

if you think about it, sin and cos are inverse functions of arclength. and arclength is quite challenging to define.

 

[mathwonk]

oh, and the fact that e^ix = cos(x) + isin(x) makes it rather clear that this definition of 2pi is the usual one.

 

[lurflurf]

it depends on your definition of beginner. i liked the approach via infinite series because it was precise and not boring like the high school approach. in that approach one has no rigorous definition of angle measure.

The definition I advocated was
sin(x),cos(x):R->R and have the following 4 properties.
1. sin(x+y)=sin(x)cos(y)+cos(x)sin(y) for all x,y
2. cos(x+y)=cos(x)cos(y)-sin(x)sin(y) for all x,y
3. (sin(x))^2+(cos(x))^2=1 for all x
4. $$\lim_{x\rightarrow 0}\frac{sin(x)}{x}=1$$

I fell this definition offers the best balance between amount of rigor, required base knowlege, and illumination of the functions most important property. I admit this is largely a matter of taste. I did not advocate the so called standard definition that defines Arcsin in terms of integrals and defines sin as a periodic extension of the inverse of those integrals. My perfered definition is fairly rigorous at the intended level, does not appeal to results from geometry and requires little background. Understanding this definition requires knowledge of real numbers, limits, and continuous functions. It is a separate matter of taste when to introduce infinite series, this definition allows the trigonometric functions to be defined without them. Where ever infinite series are introduced this definition allows the circular functions to be defined and used at an early stage. I personally feel that defining these functions (as a first definition) using things like integrals, infinite series, or differential equations either obscures their meaning, or delays their introduction. Many authors end up writing things like "this sin(x) thing is what you think it is, I promise to be rigorous later" in order to shoe horn trig function into early chapters.

 

[mathwonk]

you are probably right lurflurf. i like your approach. i was just reporting on what worked for me. having skipped trig in high school iw as rather glad it was treated from scratch in college.