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### Deriving Trig Functions; Taylor Series

```
Date: 05/01/2001 at 23:30:23
From: Harold Brochmann
Subject: Deriving trig functions

Suppose I want to find, *from first principles* - no tables, no
calculator - for example 32 degrees... how would I go about it?

If I use a formula, how is *it* derived?

I have asked many math teachers. They do not know. I have never seen
it in a textbook. Where to look?
```

```
Date: 05/07/2001 at 15:43:31
From: Doctor Douglas
Subject: Re: Deriving trig functions

Hi Harold,

I think you are asking how one computes quantities such as

sin(32 degrees)

without using a calculator. Is that correct? A straightforward way to
do this is to use the power series representation of sin(x):

sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ...

where the exclamation points mean factorial, e.g. 5! = 5*4*3*2*1 = 120
and the x is a number in radians, not degrees. The conversion from

x = (number of degrees)* pi / 180.

So if you plug in 32 degrees for number-of-degrees in this formula,
and you know pi, you can obtain x=0.55850536 radians. Then, plugging
this value for x into the series expression for sin(x), we can compute
it to any desired accuracy, provided we take enough terms, and that we
know x to the requisite number of digits.

There is a corresponding equation for cos(x):

cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! + ...

and one for tan(x):

tan(x) = x + x^3/3 + 2x^5/15 + 17x^7/315 + ...

or if you like, you can obtain it from tan(x) = sin(x)/cos(x).

Real calculators could use this type of method, but it is
more common to implement special routines that start from data in
a "table" stored in electronic memory and interpolate between
points when the x-value isn't exactly equal to an entry already
in the table, or evaluate the trigonometric functions by
exploiting their close relationship with complex numbers ("CORDIC"
algorithms).  Both of these methods can get intricate, especially
if you have a demanding application that tries to optimize speed,
accuracy, or both.

I hope this helps answer your question. It's not exactly a complete
"first-principles" calculation, since that would involve how we
actually define the sine and cosine functions, but it does show you
how one could compute values of these functions using enough pencil
and paper and only the mathematical operations of addition/
subtraction, and multiplication/division.

To explain the connection between the trigonometric functions
and these series representations usually requires differential
from there.

- Doctor Douglas, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 05/07/2001 at 21:14:33
From: Harold Brochmann
Subject: Re: Deriving trig functions

I would appreciate that.

Harold Brochmann
```

```
Date: 05/09/2001 at 06:56:33
From: Doctor Douglas
Subject: Re: Deriving trig functions

Hi again, Harold, and thanks for writing back.

The connection between the trigonometric functions and their series
representation comes from the Taylor expansion of these functions.
In what I say below, let's measure x in radians.

Consider f(x) = cos(x). Suppose we are trying to compute this function
near a = 0. We know that cos(a=0)=1, but what about a number such as
cos(x=0.1)? It's probably a bit less than 1, but how close is it to 1?

The answer lies in estimating the function f(x) by its Taylor
expansion. We want a power series in factors of (x-a), where we can
easily obtain the coefficients of (x-a)^k. If we can compute these
coefficients, then we can (with a calculator, even if it doesn't have
a "cos" button) compute f(x) to any desired accuracy we wish.

Let the series be given by p(x):

p(a) = f(a)     if x = a, then p(a) better be f(a)

p'(a) = f'(a)   if x is near a, then make the slope of the polynomial
p(x) have the same slope as f(x) there.

p''(a) = f''(a) if x is near a, make the second derivatives of p(x)
and f(x) equal at x=a

If you extend this argument to higher and higher order derivatives,
then you can see that the following power series

p(x) = f(a) + f'(a)(x-a)^1 / 1! + f''(a)(x-a)^2 / 2! +
f'''(a)(x-a)^3 / 3! + ...

has the correct derivatives - i.e., p(a) = f(a), p'(a) = f'(a), and so
on. This polynomial is called the Taylor's series for f(x) at x = a.

In the case of f(x) = cos(x), f'(x) = -sin(x), f''(x) = -cos(x),
f'''(x) = sin(x), and so on. For x = a = 0, all of the odd derivatives
vanish, and we are left with

f(0) = 1        f'(0) = 0
f''(0) = -1     f'''(0) = 0
f''''(0) = 1    f'''''(0) = 0
f^6(0) = -1
f^8(0) = 1

and so on.  We see how easy it is to compute these coefficients at
x = a = 0, because of the nice properties of sine and cosine.
Plugging these values back into the expression for p(x) yields

p(x) = 1 + 0 + (-1)(x-0)^2/2! + 0 + (1)(x-0)^4/4! + 0 + (-1)(x-0)^6/6!
+ ...

= 1 - x^2/2! + x^4/4! - x^6/6! + x^8/8! - ...

as the Taylor series for f(x)=cos(x). The derivation of p(x) for
sin(x) at x = 0 is similar. Of course at x = 0, the function sin(x)
and its even derivatives vanish, so that the Taylor series for sin(x)
contains only terms with odd powers in x.

The question of whether or not f(x) = p(x) for all x is a slightly
delicate issue. There are ways of estimating the error
Rn(x) = f(x)-p[n](x) that is obtained by truncating the series p(x) to
p[n](x) after n+1 terms. As long as this error can be made to go to
zero (say in an interval around x = a), then we say that p(x) = f(x)
in this interval. In estimating f(x), we say that p[n](x) estimates
f(x) with an error no bigger than Rn(x) in this interval.

You can see that, subject to this issue about making the error Rn(x)
small, we can make numerical computations of f(x) near x = a using
p[n](x). It's instructive to illustrate this for the first few n, say
for x = 0.1 and f(x) = cos(x):

f(x) = cos(0.1) = 0.995004165          (from a calculator with "cos")
p[0](x) = 1                            well that was easy
p[1](x) = 1 - (0.1)^2/2! = 0.995       pretty close already!
p[2](x) = 1 - (0.1)^2/2! + (0.1)^4/4!
= 0.995 + 0.0001/24
= 0.995004167                  good to eight digits already!

This shows how we can compute f(x) with fairly high accuracy without
necessarily using a large number of terms, and without using anything
more complicated than raising a number to a power and elementary math
operations such as addition/subtraction and multiplication/division.

Remark: In some cases, a single value for a suffices for the entire
domain of x.  This is true for f(x)=cos(x) above.  The series
above represents cos(x) for any x (even x not close to a=0) as
long as you carry out the sum to a sufficient number of terms.
Again, the issue is how large the error is after that number
of terms. It turns out that for cos(x) the error can be made
to go to zero as n increases to infinity, no matter what x is.

Remark: You can choose different values for a.  If you are looking for
sin(pi/2 + 0.001), then you should choose a=pi/2, and compute
all of the derivatives there, rather than a=0.  This will make
the sequence of polynomials p[n](x) converge to f(x) much more
rapidly than if you had expanded at a = 0.

- Doctor Douglas, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 05/09/2001 at 11:31:26
From: Harold Brochmann
Subject: Re: Deriving trig functions

> The connection between the trigonometric functions and their series
> representation comes from the Taylor expansion of these functions.
..... etc.

Thank you for taking the trouble.

It's going to take me a while to digest the setting up of the Taylor
series... but this is at least a place to start.

You're doing fine work.

:-)
```

```
Date: 05/13/2001 at 09:17:34
From: Harold Brochmann
Subject: Re: Deriving trig functions

to find:

Taylor Series - Why We Want Them and How We Find Them - James A.
Sellers, Dept. of Science and Mathematics, Cedarville College
http://www.krellinst.org/UCES/archive/resources/taylor_series/

which contains a terse yet more detailed explanation. You might wish
to refer others with similar questions there.

Thanks again.

Harold Brochmann
```
Associated Topics:
High School Calculus
High School Sequences, Series
High School Trigonometry

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