Ratios defining trigonometric functions

[hobz]

I am revisiting some of the ol' trigonometry, and while I was reading about in various places, it occurred to me that the authors fairly quickly introduce definitions of the trigonometric function.

 

I am left with definitions like, [math]\sin \theta = \frac{opp.}{hyp.}[/math], which stems from the fact that in a right angled triangle, the ratio of the triangle ABC (with the hypotenuse AB = 1 and angle between AB and AC = [math]\theta[/math]) relates to the ratio of any scaled version of the triangle having the same angle. If we assume that there is a larger triangle ADE (hypotenuse = AD), then from Euclid (if I remember correctly) the relation [math]\frac{BC}{AB} = \frac{DE}{AD}[/math] defines [math]\sin \theta[/math] as BC with AB being 1.

 

Now what I don't understand is, why could [math]\sin \theta[/math] not have been chosen to be the inverse ratio? And likewise with cosine? What is so special about the ratios where we divide by the hypotenuse, contra the ratio where we divide by the "end"-height?

 

Thanks!

[Riogho]

Trigonometric functions are also defined as the sum of an infinite series, look em up on wiki.

[D H]

Now what I don't understand is, why could [math]\sin \theta[/math] not have been chosen to be the inverse ratio? And likewise with cosine? What is so special about the ratios where we divide by the hypotenuse, contra the ratio where we divide by the "end"-height?

 

Thanks!

 

Nothing is so special about which is the numerator and which is the denominator. With three sides in a triangle, there are six ways to construct ordered pairs of sides. Each of these different ordered pairs corresponds to a trigonometric function: sine, cosine, tangent, cosecant, secant, and cotangent.

[the tree]

Now what I don't understand is, why could [math]\sin \theta[/math] not have been chosen to be the inverse ratio? And likewise with cosine?

No particular reason. You could use the secant (hyp/adj) cosecant (hyp/opp) and cotangent (adj/opp) functions just as well for the same purposes as you would use the standard three, but the convention is there and sticking to it makes for less confusion and easier communication. Think of it this way: there's no particular reason to drive on either side of the road over another, so long as people agree on which one they're going to chose.

Trigonometric functions are also defined as the sum of an infinite series, look em up on wiki.

As are all infinitely differentiable functions, including the ratios that the OP suggested, this doesn't make them anything special.

[Kyrisch]

Trigonometric functions are also defined as the sum of an infinite series, look em up on wiki.

 

Actually any differentiable function can be defined using infinite Taylor or Maclaurin series:

 

[math] f(x) = \frac{f©}{0!} + \frac{f'© (x - c)}{1!} + \frac{f''© (x-c)^2}{2!} + \frac{f^n©(x-c)^n}{n!} ... [/math] for any c.

 

The series is actually, conversely, defined by the function.

[the tree]

Actually any infinitely differentiable function can be defined using infinite Taylor or Maclaurin series

fix'd.

 

The series is actually, conversely, defined by the function.

Which way round depends on the context and the function in question, but in this case I'd agree, trig functions define their Taylor series and not the other way around.

[hobz]

Thanks for clarifying! I would seem out of the six different possibilities, it just happens to be the sine, cosine and tangent that are most often used?

[hobz]

A note.

 

The Taylor expansion of e.g. sine, requires the n'th derivative to be known.

Thus before the sine can accurately be constructed (using Taylor), we must know beforehand that the first few derivatives are cos x, - sin x, - cos x, sin x, cos x. But how can we know this (exactly) when we not yet have defined any way of calculating these functions (using Taylor)?

[Country Boy]

A note.

 

The Taylor expansion of e.g. sine, requires the n'th derivative to be known.

Thus before the sine can accurately be constructed (using Taylor), we must know beforehand that the first few derivatives are cos x, - sin x, - cos x, sin x, cos x. But how can we know this (exactly) when we not yet have defined any way of calculating these functions (using Taylor)?

The nth Taylor polynomial requires the nth derivative. The "Taylor expansion" or "Taylor series" requires all of them.

 

One answer to your question is that if we define sine and cosine by their Taylor series, we can get the nth derivative by differentiating the series term by term:

If sin(x) is defined by sin(x)= x- x^3/3!+ x^5/5!+ ..., then it is obvious that sin(0)= 0, sin'(x)= 1- 3x^2/3!+ 5x^4/5!+...= 1- x^2/2!+ x^4/4!= cos(x), sin"(x)= -2x/2!+ 4x^3/4!+ ...= -(x- x^3/3!+...)= -sin(x) etc. You could then show that y= sin(x) satisfies the differential equation y"= -y as well as the initial condition sin(0)= 0, sin'(0)= 1 and so, by the existance and uniqueness" theorem for initial value problems, is identical to the "traditional" sine function.

 

Actually any differentiable function can be defined using infinite Taylor or Maclaurin series:

 

[math] f(x) = \frac{f©}{0!} + \frac{f'© (x - c)}{1!} + \frac{f''© (x-c)^2}{2!} + \frac{f^n©(x-c)^n}{n!} ... [/math] for any c.

 

The series is actually, conversely, defined by the function.

Given any infinitely differentiable function we can define its Taylor's series at, say, 0 (the "McLaurin series"). It does NOT follow that the series converges to that function. For example, the function

[math]f(x)= e^{-\frac{1}{x^2}}[/math] for [math]x\ne 0[/math], f(0)= 0 is infinitely differentiable at 0. All derivatives are equal to 0 at x= 0 so its Taylor's series about x= 0 is identically equal to 0. That converges, of course, for all x but is equal to f(x) only for x= 0.

[hobz]

The nth Taylor polynomial requires the nth derivative. The "Taylor expansion" or "Taylor series" requires all of them.

 

One answer to your question is that if we define sine and cosine by their Taylor series, we can get the nth derivative by differentiating the series term by term:

If sin(x) is defined by sin(x)= x- x^3/3!+ x^5/5!+ ..., then it is obvious that sin(0)= 0, sin'(x)= 1- 3x^2/3!+ 5x^4/5!+...= 1- x^2/2!+ x^4/4!= cos(x), sin"(x)= -2x/2!+ 4x^3/4!+ ...= -(x- x^3/3!+...)= -sin(x) etc. You could then show that y= sin(x) satisfies the differential equation y"= -y as well as the initial condition sin(0)= 0, sin'(0)= 1 and so, by the existance and uniqueness" theorem for initial value problems, is identical to the "traditional" sine function.

 

 

Given any infinitely differentiable function we can define its Taylor's series at, say, 0 (the "McLaurin series"). It does NOT follow that the series converges to that function. For example, the function

[math]f(x)= e^{-\frac{1}{x^2}}[/math] for [math]x\ne 0[/math], f(0)= 0 is infinitely differentiable at 0. All derivatives are equal to 0 at x= 0 so its Taylor's series about x= 0 is identically equal to 0. That converges, of course, for all x but is equal to f(x) only for x= 0.

 

This solution requires knowledge of how to construct the Taylor series in the first place. That is, knowing that sine and cosine are closely related in calculus. Historically one must have had a feeling or hunch that sin'(x) = cos(x).

[Bignose]

Historically one must have had a feeling or hunch that sin'(x) = cos(x).

 

No, you can derive [math]\frac{d}{dx}\sin{x}=\cos{x}[/math] just from the definition of the derivative and the angle sum formulas.

 

i.e. using [math]\frac{df(x)}{dx} = \lim{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}[/math]

 

and

 

[math]\sin{a\pm b} = \sin{a}\cos{b}\pm\cos{a}\sin{b}[/math]

and

[math]\cos{a\pm b} = \cos{a}\cos{b} \mp \sin{a}\sin{b}[/math]

[hobz]

No, you can derive [math]\frac{d}{dx}\sin{x}=\cos{x}[/math] just from the definition of the derivative and the angle sum formulas.

 

i.e. using [math]\frac{df(x)}{dx} = \lim{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}[/math]

 

and

 

[math]\sin{a\pm b} = \sin{a}\cos{b}\pm\cos{a}\sin{b}[/math]

and

[math]\cos{a\pm b} = \cos{a}\cos{b} \mp \sin{a}\sin{b}[/math]

 

Ahh, cool stuff. Thanks! Will look into it.

[Dave]

No, you can derive [math]\frac{d}{dx}\sin{x}=\cos{x}[/math] just from the definition of the derivative and the angle sum formulas.

 

i.e. using [math]\frac{df(x)}{dx} = \lim{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}[/math]

 

To do this, you will also need to prove that:

 

[math]\lim_{x\to 0} \frac{\sin x}{x} = 1[/math]

 

which is a little tricky.