The approximation sin(x) = x - (1/6)x^3
Dave L. Renfro
22 April 2008 in another group, is intended
in the same spirit as the 2 most recent posts
I've made, which are archived at
http://groups.google.com/group/sci.math/msg/1472c60d9d643639
http://groups.google.com/group/sci.math/msg/dfb992fe3d16fc49
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[URL for archived version of the original post]
http://mathforum.org/kb/message.jspa?messageID=6189492
Dave L. Renfro wrote (in part):
http://mathforum.org/kb/message.jspa?messageID=6187665
> [x - (sin x)] / x^3.
>
> However, if there is a simple way of determining this
> geometrically or algebraically, only making use of
> (sin x)/x --> 1 as x --> 0, then it's escaped all the
> 19th century textbook authors who gave involved geometric
> derivations of the limit [x - (sin x)] / x^3 (Todhunter,
> Hobson, Loomis, Legendre, De Morgan, Serret, etc.).
I looked through some of my things this morning
to see how these authors did this and decided to
write a summary of what I found and what I managed
to come up with. This post is essentially a sequel
to one that I made here, back in 1999 (!), which began
with an e-mail I got from Richard Askey, who asked
me to post his comments:
ap-calculus (5 December 1999)
http://mathforum.org/kb/message.jspa?messageID=658399
I looked through several 19th century references
on trigonometry and essentially found only these
results (for x slightly larger than 0):
x - (1/6)x^3 < sin(x) < x
1 - (1/2)x^2 < cos(x) < 1 - (1/2)x^2 + (1/24)x^4
These results can be proved without the use of
calculus methods. Near the end of this post
I'll give one way this can be done.
I have not been able to find, or to come up with,
a non-calculus way to prove that x - (1/6)x^3
is the correct 3rd order approximation to sin(x)
near x = 0. However, I was able to come up with
a way to show this using AB calculus methods.
What I came up with is probably in some books or
in some old papers, but at this point I don't
have a reference for the method I came up with.
I do have a number of references for the results
above, which are given at the end of this post.
In all that follows, the inequalities are only
designed and needed for values of x in some
right neighborhood of 0.
-----------------------------------
STEP 1: We show sin(x) < x
Let f(x) = sin(x) - x. Then f'(x) = cos(x) - 1 < 0,
so f is strictly decreasing. Since f(0) = 0, it
follows that for x in some right neighborhood
of 0 we have:
f(x) < 0
sin(x) - x < 0
sin(x) < x
-----------------------------------
STEP 2: We show cos(x) > 1 - (1/2)x^2
Using Step 1 we have:
sin(x/2) < x/2
(sin x/2)^2 < (x/2)^2 = (1/4)x^2
-2(sin x/2)^2 > -(1/2)x^2
1 - 2(sin x/2)^2 > 1 - (1/2)x^2
cos(x) > 1 - (1/2)x^2
In the last step I used the trig. identity
cos(2u) = 1 - 2(sin u)^2 with u = x/2.
-----------------------------------
STEP 3: We show sin(x) > x - (1/6)x^3
Let g(x) = sin(x) - x + (1/6)x^3. Then
g'(x) = cos(x) - 1 + (1/2)x^2. It follows from
Step 2 that g'(x) > 0. Hence, g is strictly
increasing. Since g(0) = 0, it follows that
for x in some right neighborhood of 0 we have:
g(x) > 0
sin(x) - x + (1/6)x^3 > 0
sin(x) > x - (1/6)x^3
-----------------------------------
STEP 4: We show cos(x) < 1 - (1/2)x^2 + (1/24)x^4
Using Step 3 we have:
sin(x/2) > x/2 - (1/6)(x/2)^3
sin(x/2) > (1/2)x - (1/48)x^3
(sin x/2)^2 > [ (1/2)x - (1/48)x^3 ]^2
(sin x/2)^2 > (1/4)x^2 - (1/48)x^4 + (1/2304)x^6
(sin x/2)^2 > (1/4)x^2 - (1/48)x^4 {x^6 term removed}
-2(sin x/2)^2 < -(1/2)x^2 + (1/24)x^4
1 - 2(sin x/2)^2 < 1 - (1/2)x^2 + (1/24)x^4
cos(x) < 1 - (1/2)x^2 + (1/24)x^4
In the last step I used the trig. identity
cos(2u) = 1 - 2(sin u)^2 with u = x/2.
-----------------------------------
STEP 5: We show sin(x) < x - (1/6)x^3 + (1/120)x^5
Let h(x) = sin(x) - x + (1/6)x^3 - (1/120)x^5.
Then h'(x) = cos(x) - 1 + (1/2)x^2 - (1/24)x^4.
It follows from Step 4 that h'(x) < 0. Hence,
h is strictly decreasing. Since h(0) = 0, it follows
that for x in some right neighborhood of 0 we have:
h(x) < 0
sin(x) - x + (1/6)x^3 - (1/120)x^5 < 0
sin(x) < x - (1/6)x^3 + (1/120)x^5
-----------------------------------
STEP 6: We show x - (1/6)x^3 is the correct 3rd order
approximation to sin(x).
Using Steps 3 and 5 we have:
x - (1/6)x^3 < sin(x) < x - (1/6)x^3 + (1/120)x^5
-(1/6)x^3 < sin(x) - x < -(1/6)x^3 + (1/120)x^5
-1/6 < [sin(x) - x]/x^3 < -1/6 + (1/120)x^2
It now follows from the squeeze theorem (also called
the sandwich theorem) that
[sin(x) - x]/x^3 approaches -1/6 as x --> 0+.
We get the same value for x --> 0- by using the
fact that [sin(x) - x]/x^3 is an even function.
-----------------------------------
NON-CALCULUS PROOF THAT sin(x) > x - (1/6)x^3
We start with the observation that sin(x) < x
for x in some right neighborhood of 0. This
is nothing more than the fact that the arc
length along the unit circle, between (1,0) and
(cos x, sin x),, is greater than the (vertical)
distance between (cos x, sin x) and (cos x, 0).
We will be using the triple angle formula a lot,
which can be obtained by expanding sin(2u + u):
sin(3u) = 3(sin u) - 4(sin u)^3
Letting u = x/3, this takes the form:
sin(x) = 3(sin x/3) - 4(sin x/3)^3
Using this last formula (with x/3 in place
of x) to rewrite this last formula, we get:
sin(x) = 3[3(sin x/9) - 4(sin x/9)^3] - 4(sin x/3)^3
= 9(sin x/9) - 4(sin x/3)^3 - (3)(4)(sin x/9)^3
Using the triple angle formula again, we get
9[3(sin x/27) - 4(sin x/27)^3] - 4(sin x/3)^3
- (3)(4)(sin x/9)^3
which, after some minor rearrangement, is
27(sin x/27) - 4(sin x/3)^3 - (3)(4)(sin x/9)^3
- (9)(4)(sin x/27)^3
Continue in this manner [the next step is to
apply the triple angle formula to 27(sin x/27)]
and, after n steps (n-1 steps? n+1 steps?), we get:
sin(x) = (3^n)(sin x/3^n) - 4[ (sin x/3)^3
+ 3(sin x/9)^3 + ... + 3^(n-1)(sin x/3^n)^3 ]
Using x > sin(x) in this last equation gives
the inequality
sin(x) > (3^n)(sin x/3^n) - 4/[ (x/3)^3
+ 3(x/9)^3 + ... + 3^(n-1)(x/3^n)^3 ],
which is equal to
(3^n)(sin x/3^n) - 4/[ x^3/3^3
+ x^3/3^5 + ... + x^3/3^(2n+1) ],
which is equal to
(3^n)(sin x/3^n) - (4x^3/3^3)[ 1 + 1/3^2
+ 1/3^4 + ... + 1/3^(2n-2) ],
which is equal to (finite geometric series sum)
(3^n)(sin x/3^n) - (4x^3/3^3)[9 - 1/3^(2n-2)]/[9 - 1].
Therefore, for each positive integer n, we have
sin(x) > (3^n)(sin x/3^n) - (x^3/6)[1 - 1/3^(2n)].
Taking the limit as n --> ininity of both
sides (not really calculus because we're not
introducing a derivative?) gives
sin(x) > x - (x^3/6)[1 - 0]
sin(x) > x - (1/6)x^3.
Note that (3^n)(sin x/3^n) = (sin x/3^n)/(1/3^n)
approaches x, since this is a particular sequence
instance of the limit as t --> 0 of (sin xt)/t.
To obtain the other result that I said could be
proved without calculus methods,
1 - (1/2)x^2 < cos(x) < 1 - (1/2)x^2 + (1/24)x^4,
look back at Steps 2 and 4 above.
-----------------------------------
The references below are those I happen to have
in a folder on this topic, and they don't represent
any kind of systematic search. In particular,
I've seen the inequality sin(x) > x - (1/6)x^3
in a number of 19th century trig. texts, but I
didn't record these instances anywhere when I
came across them.
Rawson [3] begins with: "Mr. Todhunter has
given a demonstration of this property in
his Trigonometry, pp. 87 and 88, based upon
the formula (sin 2x) = 2(sin x)(cos x). The
following proof depends upon the formula
(sin 3x) = 3(sin x) - 4(sin x)^3, and may be
regarded as a little more simple than that
given by Mr. Todhunter." [I've changed some
of the math symbols (e.g. x instead of $\theta$)
in the original for readability in ASCII format.]
The copy of Todhunter's book I've seen, at the
University of Iowa library, has the derivation
that Rawson published. On p. 91 of Todhunter [1]
is the parenthetical note: "(Le Cointe's
Trigonometry, and Messenger of Mathematics,
III. 101.)"
Incidentally, it is simpler to obtain the weaker
result sin(x) > x - (1/4)x^3, and many texts
(e.g. Todhunter [1], Section 120, p. 85) prove
this result first. Indeed, Todhunter outlines
how trig. tables can be calculated quite accurately
by using this weaker result on pp. 85-89. [The
expression works well for angles near zero, and
various methods involving trig. identities and
exact radical values (for angles that are multiples
of 3 degrees) are used to transfer approximations
for angles near zero to other angles.] Todhunter
gives an algebraic proof of this weaker result.
A very short geometrical proof of this weaker
result is given in [4] (p. 117). In [4] (pp. 157-158)
there is a short argument (much shorter than what
I gave above, which is what [1], [2], [3] give)
that shows how the weaker result can be used to
prove sin(x) > x - (1/6)x^3.
[1] Isaac Todhunter, "Plane Trigonometry for the use
of Colleges and Schools, with Numerous Examples",
MacMillan and Company, 1859/1895, vii + 341 pages.
[2] Ernest W. Hobson, "A Treatise on Plane and Advanced
Trigonometry", 7'th edition, Dover Publications,
1928/1957, xvi + 383 pages.
[3] R. Rawson, "Proof of the trigonometric formula
sin % > % - (1/6)%^3", Messenger of Mathematics
(old series) 3 (1866), 101-104. [% = $\theta$]
[4] Various notes in the journal Mathesis Recueil
Mathematique:
(Series 1) 10 (1890), pp. 58-60
(Series 1) 10 (1890), pp. 112-113
(Series 1) 10 (1890), pp. 157-158
(Series 2) 4 (1894), pp. 68-69
(Series 2) 4 (1894), pp. 73-76
(Series 2) 5 (1895), p. 117
(Series 3) 8 (1908), pp. 59-61
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Dave L. Renfro