where the dragons are

[allegansveritatem]

Here is the problem:


I confess that this stumped me to the point that I can't even figure out the solution that is given in the solutions manual which is as follows:

I can understand this to the following point:

What? How? Where do we get what comes after the third = sign?

 

[pka]

Here is the problem:
View attachment 19562

I confess that this stumped me to the point that I can't even figure out the solution that is given in the solutions manual which is as follows:
View attachment 19563
I can understand this to the following point:View attachment 19564

What? How? Where do we get what comes after the third = sign?

\(\begin{align*} \sin(4\beta)&=2\sin(2\beta)\cos(2\beta) \\&=2[2\sin(\beta)\cos(\beta)][\cos^2(\beta)-\sin^2(\beta)]\\&=4\sin(\beta)\cos^3(\beta)-4\sin^3(\beta)\cos(\beta) \end{align*}\)
Now divide through by \(4\)

BTW. Please complain to management about the failure of the site's LaTeX compiler.

 

[Dr.Peterson]

Your line (1) is irrelevant; are you thinking that identity is the double-angle or angle-sum identity they referred to? It isn't.

Here is pka's work with the troublesome part removed:

[MATH]\sin(4\beta)=2\sin(2\beta)\cos(2\beta) \\=2[2\sin(\beta)\cos(\beta)][\cos^2(\beta)-\sin^2(\beta)]\\=4\sin(\beta)\cos^3(\beta)-4\sin^3(\beta)\cos(\beta)[/MATH]The first line is the double-angle formula for sine, [MATH]\sin(2x) = 2\sin(x)\cos(x)[/MATH], applied to [MATH]x = 2\beta[/MATH].

The second line replaces the sine and cosine of the first line with their double-angle formulas.

The third line distributes.

 

[allegansveritatem]

\(\begin{align*} \sin(4\beta)&=2\sin(2\beta)\cos(2\beta) \\&=2[2\sin(\beta)\cos(\beta)][\cos^2(\beta)-\sin^2(\beta)]\\&=4\sin(\beta)\cos^3(\beta)-4\sin^3(\beta)\cos(\beta) \end{align*}\)
Now divide through by \(4\)

BTW. Please complain to management about the failure of the site's LaTeX compiler.

maybe mine is a failure to actually multiply matters out instead of leaving the multiplication implicit....if that makes any sense.I want to know where everything comes from especially m those pesky exponents come from.Thanks

 

[allegansveritatem]

Your line (1) is irrelevant; are you thinking that identity is the double-angle or angle-sum identity they referred to? It isn't.

Here is pka's work with the troublesome part removed:

[MATH]\sin(4\beta)=2\sin(2\beta)\cos(2\beta) \\=2[2\sin(\beta)\cos(\beta)][\cos^2(\beta)-\sin^2(\beta)]\\=4\sin(\beta)\cos^3(\beta)-4\sin^3(\beta)\cos(\beta)[/MATH]The first line is the double-angle formula for sine, [MATH]\sin(2x) = 2\sin(x)\cos(x)[/MATH], applied to [MATH]x = 2\beta[/MATH].

The second line replaces the sine and cosine of the first line with their double-angle formulas.

The third line distributes.

I had been working on identities for 2 hours when I wrote that top line and it was written with a flagging spirit. I don't know what I was thinking of. But I see now that I needed to write something with a sin(2A) in it somewhere. Anyway, I will study this post and the one above and try to come to some idea what is going on.I shall return....victorious I hope.

 

[Deleted member 4993]

Here is the problem:
View attachment 19562

I confess that this stumped me to the point that I can't even figure out the solution that is given in the solutions manual which is as follows:
View attachment 19563
I can understand this to the following point:View attachment 19564

What? How? Where do we get what comes after the third = sign?

In your work, from (2) to (3), only implied operation is

4*ß = 2 * (2*ß)

Is that your confusion?

 

[allegansveritatem]

In your work, from (2) to (3), only operation is

4*ß = 2 * (2*ß)

Is that your confusion?

Confused no more!

 

[allegansveritatem]

Your line (1) is irrelevant; are you thinking that identity is the double-angle or angle-sum identity they referred to? It isn't.

Here is pka's work with the troublesome part removed:

[MATH]\sin(4\beta)=2\sin(2\beta)\cos(2\beta) \\=2[2\sin(\beta)\cos(\beta)][\cos^2(\beta)-\sin^2(\beta)]\\=4\sin(\beta)\cos^3(\beta)-4\sin^3(\beta)\cos(\beta)[/MATH]The first line is the double-angle formula for sine, [MATH]\sin(2x) = 2\sin(x)\cos(x)[/MATH], applied to [MATH]x = 2\beta[/MATH].

The second line replaces the sine and cosine of the first line with their double-angle formulas.

The third line distributes.

I went back at it today and, consulting the posts in this thread and my laminated cheat sheet, I was able to conclude the business with gratifying ease thus:

 

[Dr.Peterson]

Unfortunately, at least the way you wrote this, you omitted some important parentheses, and then seem to have interpreted a line as meaning what you wrote instead of what it should have been.

Line 3 needs parentheses around the expansion of cos(2B). But if you put them in, and keep them in, then things that look like mistakes in subsequent lines become correct. So I can only hope that somehow you were thinking the right thing all the way through, and treating things as multiplication that are written as additions.

Please look though your work and insert those parentheses (in three places), so we can see your triumph as being real!

 

[pka]

I don't mean to be awful. But I simply cannot believe that someone who posted this OP would not understand that \(2(2\sin(x)\cos(x))=4\sin(x)\cos(x)\). But then there was the matter of "those pesky exponents". Does that mean how \(4\sin(\beta)\cos^3(\beta)-4\sin^3(\beta)\cos(\beta)\) works?

 

[JeffM]

@pka I have found that, for some reason I no longer remember, many students, even those who are intelligent and diligent, have a lot of trouble remembering that functions can be treated as simply numbers (at least at this level they can) when it is convenient to do so. A cosine cubed is something weird, not readily understandable in terms of the unit circle. While they ponder the mysterious nature of cubing an infinite series (something that sounds both tedious and complex), they forget that it also is just a plain old number cubed.

A student who will see in a microsecond that

[MATH]4uv^3 - 4u^3v = 4uv(v^2 - u^2)[/MATH] may struggle with

[MATH]4sin(\beta)cos^3(\beta) - 4sin^3(\beta)cos(\beta).[/MATH]
I do not know the reason for this psychological block, but it is real. And it comes up not just with trig functions but with log functions and hyperbolic functions as well. After a while, it seems just to evaporate until eventually you cannot remember what the problem was.

 

[allegansveritatem]

Unfortunately, at least the way you wrote this, you omitted some important parentheses, and then seem to have interpreted a line as meaning what you wrote instead of what it should have been.

Line 3 needs parentheses around the expansion of cos(2B). But if you put them in, and keep them in, then things that look like mistakes in subsequent lines become correct. So I can only hope that somehow you were thinking the right thing all the way through, and treating things as multiplication that are written as additions.

Please look though your work and insert those parentheses (in three places), so we can see your triumph as being real!

I looked through my working-out of this just now and see where I missed putting in parentheses in the third line...I guess the reason the answer came out right is that I had gone through this so many times by then that I considered the last two expressions in line three as a unit. Anyway, I will work it out again more carefully and report back

 

[allegansveritatem]

I don't mean to be awful. But I simply cannot believe that someone who posted this OP would not understand that \(2(2\sin(x)\cos(x))=4\sin(x)\cos(x)\). But then there was the matter of "those pesky exponents". Does that mean how \(4\sin(\beta)\cos^3(\beta)-4\sin^3(\beta)\cos(\beta)\) works?

well, as you put it here I can only say that I do understand it the way you have it down...but in the heat of the working-out of these identities often and often little things go awry. I recall a saying of Francis Bacon: The study of mathematics makes a man exact, or something like that. I hope someday it will be so with me.

 

[allegansveritatem]

@pka I have found that, for some reason I no longer remember, many students, even those who are intelligent and diligent, have a lot of trouble remembering that functions can be treated as simply numbers (at least at this level they can) when it is convenient to do so. A cosine cubed is something weird, not readily understandable in terms of the unit circle. While they ponder the mysterious nature of cubing an infinite series (something that sounds both tedious and complex), they forget that it also is just a plain old number cubed.

A student who will see in a microsecond that

[MATH]4uv^3 - 4u^3v = 4uv(v^2 - u^2)[/MATH] may struggle with

[MATH]4sin(\beta)cos^3(\beta) - 4sin^3(\beta)cos(\beta).[/MATH]
I do not know the reason for this psychological block, but it is real. And it comes up not just with trig functions but with log functions and hyperbolic functions as well. After a while, it seems just to evaporate until eventually you cannot remember what the problem was.

Part of the thing with trigonometry, seems to me, is it takes a long time to, as it were, get penetrated with the spirit of the unit circle and how the mysterious mathematical entities that live on it never seem to increase beyond a certain point no matter how mightily they are exponentiated (?). Or am I wrong in assuming this?I think I need to see more clearly how the graphs of the basic trig functions are generated with regard to the unit circle, that is: to study the connection between what the sine means and why it looks like a wavy line when graphed, and why tanx looks the way it does and secx and cscx etc. On the face of it, to a new student like me, the pictures are uncanny. This is due to, as far as I can see, an imperfect understanding of the dynamic interaction of the unit circle and the rational entities that inhabit it. I am working on tightening my grasp on these matters.

 

[allegansveritatem]

Unfortunately, at least the way you wrote this, you omitted some important parentheses, and then seem to have interpreted a line as meaning what you wrote instead of what it should have been.

Line 3 needs parentheses around the expansion of cos(2B). But if you put them in, and keep them in, then things that look like mistakes in subsequent lines become correct. So I can only hope that somehow you were thinking the right thing all the way through, and treating things as multiplication that are written as additions.

Please look though your work and insert those parentheses (in three places), so we can see your triumph as being real!

well, I went at it again today and I think I corrected myself. Here is what I came up with:

 

[Dr.Peterson]

Perfect.

You added a pair you didn't really need, but that doesn't hurt!

 

[JeffM]

Part of the thing with trigonometry, seems to me, is it takes a long time to, as it were, get penetrated with the spirit of the unit circle and how the mysterious mathematical entities that live on it never seem to increase beyond a certain point no matter how mightily they are exponentiated (?). Or am I wrong in assuming this?I think I need to see more clearly how the graphs of the basic trig functions are generated with regard to the unit circle, that is: to study the connection between what the sine means and why it looks like a wavy line when graphed, and why tanx looks the way it does and secx and cscx etc. On the face of it, to a new student like me, the pictures are uncanny. This is due to, as far as I can see, an imperfect understanding of the dynamic interaction of the unit circle and the rational entities that inhabit it. I am working on tightening my grasp on these matters.

Let's start historically as Morris Kline suggested well over a half century ago.

"Polygon" means "many sides;" a "trigon" was simply a triangle, and trigonometry grew directly out of the facts about similar triangles. So, to use completely anachronistic in language, if you had been studying trigonometry in Alexendria in the third century BCE, your teacher would have said

[MATH]\text {The domain of the sine function is numbers measuring less than the sum of two right angles.}[/MATH]
Cyclic behavior would not have been apparent.

Remember that the Greeks had no concept of the number zero, but they had become fully aware that there were irrational numbers. Archimedes proved by the sandwich theorem a very good approximation of pi.

For centuries, the primary uses of trigonometry outside of astronomy were practical applications in building and surveying and, much later, in navigation. Calculations by hand of a function involving even just the sine function were tedious and ugly. As late as sixty years ago, we used a slide rule for rough approximations and tables for more accruate ones. I had to buy both a slide rule and a book called "Standard Mathematical Tables" when I was in secondary school. I still have the book. The trig tables alone run from page 93 through 135. Back then, you did the arithmetic by hand. To help you, there was a supplementary table of the logs of trig functions. Trig identities allowed you to do a bit of algebra and thereby minimize tedious and error prone arithmetic computations. The hand calculator has rendered all those tables and all that use of trig identities as obsolete as the neolithic knowledge of how best to hunt a mammoth. I do not know how to hunt mammoths, and my grandson will be ignorant of both that and of what a slide rule was. But using those tables totally demystified trig functions: they are numbers, whose approximations can be found from pages 93 through 135.

I shall let the mathematicians explain why trig identities are still in the curriculum. I can think of two possible reasons. One is that many integrals, which are operations on functions rather than numbers, result in trigonometric functions, and to understand what those integrals mean, it may be helpful to restate them in a different trigonometric form. The second is Fourier analysis, which restates many functions into trig equivalents. Any repetitive phenomenon may be modeled by trig functions.

The unit circle is clearly a recent concept, late 17th century at the earliest. It transforms the sine and cosine functions from propositions about triangles into propositions about circles. Mark a circle off in your back yard. Walk around it once. Where are you? Right back where you started. Walk around it twice. Where are you? Right back where you started. The circle and the trig functions provide the perfect mathematical model for repetitive behavior. Why does the sine repeat? Why does walking in a circle 50 times get you nowhere? To get the repetitive nature of the sine,, you must turn your mind off numbers and think about walking in a circle, and the mystery disappears.[/MATH]

 

[allegansveritatem]

Let's start historically as Morris Kline suggested well over a half century ago.

"Polygon" means "many sides;" a "trigon" was simply a triangle, and trigonometry grew directly out of the facts about similar triangles. So, to use completely anachronistic in language, if you had been studying trigonometry in Alexendria in the third century BCE, your teacher would have said

[MATH]\text {The domain of the sine function is numbers measuring less than the sum of two right angles.}[/MATH]
Cyclic behavior would not have been apparent.

Remember that the Greeks had no concept of the number zero, but they had become fully aware that there were irrational numbers. Archimedes proved by the sandwich theorem a very good approximation of pi.

For centuries, the primary uses of trigonometry outside of astronomy were practical applications in building and surveying and, much later, in navigation. Calculations by hand of a function involving even just the sine function were tedious and ugly. As late as sixty years ago, we used a slide rule for rough approximations and tables for more accruate ones. I had to buy both a slide rule and a book called "Standard Mathematical Tables" when I was in secondary school. I still have the book. The trig tables alone run from page 93 through 135. Back then, you did the arithmetic by hand. To help you, there was a supplementary table of the logs of trig functions. Trig identities allowed you to do a bit of algebra and thereby minimize tedious and error prone arithmetic computations. The hand calculator has rendered all those tables and all that use of trig identities as obsolete as the neolithic knowledge of how best to hunt a mammoth. I do not know how to hunt mammoths, and my grandson will be ignorant of both that and of what a slide rule was. But using those tables totally demystified trig functions: they are numbers, whose approximations can be found from pages 93 through 135.

I shall let the mathematicians explain why trig identities are still in the curriculum. I can think of two possible reasons. One is that many integrals, which are operations on functions rather than numbers, result in trigonometric functions, and to understand what those integrals mean, it may be helpful to restate them in a different trigonometric form. The second is Fourier analysis, which restates many functions into trig equivalents. Any repetitive phenomenon may be modeled by trig functions.

The unit circle is clearly a recent concept, late 17th century at the earliest. It transforms the sine and cosine functions from propositions about triangles into propositions about circles. Mark a circle off in your back yard. Walk around it once. Where are you? Right back where you started. Walk around it twice. Where are you? Right back where you started. The circle and the trig functions provide the perfect mathematical model for repetitive behavior. Why does the sine repeat? Why does walking in a circle 50 times get you nowhere? To get the repetitive nature of the sine,, you must turn your mind off numbers and think about walking in a circle, and the mystery disappears.[/MATH]

Yes, it is all about cycles. I see that. I don't like the identities but they are good algebra practice and I suppose it is good to know how..., what?... slippery these are. They are the original shape shifters. I have thought about getting a book of trig tables...or maybe a graphic encased in plastic of the unit circle with every degree marked with its sine and cosine. So far I haven't been able to find one. I know the tables go from one to ninety degrees but tables for dummies that give the functions for every degree of the 360 is what I want!

 

[allegansveritatem]

Perfect.

You added a pair you didn't really need, but that doesn't hurt!

I wasn't taking any chances.

 

[JeffM]

Yes, it is all about cycles. I see that. I don't like the identities but they are good algebra practice and I suppose it is good to know how..., what?... slippery these are. They are the original shape shifters. I have thought about getting a book of trig tables...or maybe a graphic encased in plastic of the unit circle with every degree marked with its sine and cosine. So far I haven't been able to find one. I know the tables go from one to ninety degrees but tables for dummies that give the functions for every degree of the 360 is what I want!

I looked on Google. There is a used copy of the 11th edition of Standard Mathematical Tables available for under 6 dollars. My edition is the 12th, but it has the same awful brown hardcover. I presume it has virtually the same trig tables on almost the same pages.

So I looked more carefully at my edition. I was wrong yesterday; the table of sines, cosines, tangents, and cotangents only runs from pages 93 through 116. In one sense, it provides more than you want because it gives their value to four decimal places for each minute of a degree from 0 to 60. In another sense, it provides less than you want because it only runs from 0 degrees to 44 degrees 60 minutes. Of course that is all you need if you are prohibited by religion from using a hand calculator. And it will firmly persuade you that the trig functions represent numbers (or, to be more modern in terminology, represent real numbers that can be approximated by rational numbers to any level of accuracy required).

Why is that all you need? Because, working in degrees,

[MATH]0 \le \theta \le 45 \implies sin(\theta) = cos(90 - \theta) \text { and } cos(\theta) = sin(90 - \theta).[/MATH]
So if you need the sine of 60 degrees, you just look up the cosine of 30 degrees. That comes right out of the geometric definition of sine and cosine in terms of right triangles. And it explains the similarity of the names.

If you are willing to waste paper and ink, you can get a desktop computer to print out a table of sines and cosines for 0 to 360 degrees; nowadays you do not need a book; you can do it yourself

What I suspect would really help you is to go back and work out for yourself the relationship between the right angle definition and unit circle definitions of the sine and cosine of an arbitrary theta greater than 0 degrees but less than 90, of theta plus 90 degrees, of theta plus 180 degrees, and theta plus 270 degrees. You will find that the circle definition merely affects the signs of the functions, not their magnitude.

And then do in a right triangle the geometry of the sum of angle fotmulas. The other trig functions are just elaborations of sine and cosine and the sine and cosine just represent shifts in the same graph. If you fully feel the sine curve and its connection to the other functions, it becomes mechanics.

Now to understand all this in terms of the modern infinite series definitions is probably easier algebraically, but you lose any connection to a concrete geometric object that you can see.