Exact Sines and Cosines

Civilization has fallen. Every computer has ceased to function. All libraries are rubble. Worse, every copy of Abramowitz and Stegun has been burnt.

Your barbarian captor tells you that you have one day to calculate the sine of 12° or he will feed you to his pack of feral Vietnamese Pot-Bellied Pigs.

Watcha gonna do?

---

Way, way back in high school geometry Mr. Wye taught me to draw triangles like these:

Which makes it easy to calculate the sines and cosines of 30°, 45°, and 60°:

     sin 30° = cos 60° = ½   (1)
     sin 45° = cos 45° = 1/√2   (2)
     cos 60° = sin 30° = ½ √3   (3)

From that you can use the half-angle formulas

     sin θ = [ 1 − cos 2 θ ]^½   (4)  and
     cos θ = [ 1 + cos 2 θ ]^½ .   (5)

to give you, e.g.,

     sin 15° = (√3 − 1)/√8   (6)   and
     cos 15° = (√3 + 1)/√8    .   (7)

And then you can go on to the values for 7.5°, 3.75°, 1.875°, … . You could then use the addition formulas

     sin x+y = sin x cos y + cos x sin y   (8)   and

     cos x+y = cos cos y − sin x sin y   (9)

to compute

     sin (15° − 3.75°) = sin 11.25°

and so on, eventually getting really, really close to sin 12°.

Of course you'll never quite get there, and that means you'll be sleeping with the piggies.

But wait! We don't have to use pictures to compute sines and cosines. In addition to all of the formulas above, we we've got a large number of trigonometric relations that we can draw on:

      sin 0 = 0   ,  (10)
      cos 0 = 1   ,  (11)
      sin 90° = 1   ,  (12)
      cos 90° = 0   ,  (13)
      sin² x + cos² y = 1   ,  (14)
      sin (90° − x) = cos x   ,  (15)
      sin (−x) = − sin x  ,  (16)
      cos (−x) = cos x  ,  (17)   and
      cos (x + 360°) = cos x , sin (x + 360°) = sin x (18)
      sin x > 0 , cos x > 0 , 0 < x < 90°   .  (19)

So, for example, by using (15) with x = 45° we find

      sin 45° = cos 45°   (20)  .

Combine this with the Pythagorean Theorem (14) and we get

      2 sin² 45° =1  ,  (21)   or

      sin 45° = cos 45° = 1/√2   .  (22)

For 30/60°, repeated application of (8) and (9) gives

      cos 3 x = 4 cos³ x −3 cos x   .  (23)

if we set 3 x = 90°, then the left hand side of (23) must vanish and, using (19)

      cos 30° = ½ √3  .  (24)

and by (14)

      sin 30° = ½  .  (25)

We can get the values for 60° using (15).

That's nice, but it doesn't get us any closer to sin 12°. But wait! If (23) involves cos 3 x, what about another integer? Say

      cos 5 x = 5 cos x − 20 cos³ x + 16 cos⁵ x   .  (26)

This is a fifth-order polynomial in cos x, so if we set cos 5 x = 0, we'll get five solutions. Let's think about those a bit.

The function cos 5 x has zeros whenever 5 x = 90° + N 180°, for any integer N. If we take N = 0, 1, 2, 3, 4, we get x = 18°, 54°, 90°, 126° and 162°, respectively. Equation (26) has one zero root, which is obviously cos 90°. That leaves

      16 cos⁴ x − 20 cos² x + 5 = 0   ,  (26)

which has the solutions

     cos x = ± [ (5 ± √5)/8]^½  .  (27)

Since cos x is decreasing as x goes from 0 to 90° we can match the smaller positive value in (27) to cos 54° and the larger to cos 18°. The negative solutions are cos 126° = − cos 54° and cos 162° = − cos 18°.

So we have cos 18° and cos 54°. Then we can use (14) to give us the sines of those angles. Even better, using (15) we can get the sines and cosines of 36° and 72°. And guess what! From the values for 18° and 15° we can get sin 3° and cos 3°, and using those and the values for 15° we can get sine 12° and cos 12°. You're not going to be pig supper!!!

---

In fact, you can calculate the sine and cosine at every 3 degrees. I've attached a table of these values below. Save it for future reference, just in case your barbarian captor doesn't give you a full 24 hours to do the calculations:

Values of the trigonometric functions at special angles

θ	θ	sinθ = cosϕ	ϕ	ϕ
(Radians)	(degrees)	(θ+ ϕ = π/2)	(Radians)	(degrees)
0	0	0	π/2	90
π/60	3	(1/4) √{8−√3 (√5+1) −√{2(5−√5)}}	29π/60	87
π/30	6	(1/4) √{9−√5−√{6 (5+√5)}}	7π/15	84
π/20	9	(1/2)  √{2 − √{(5+√5)/2}}	9π/20	81
π/15	12	(1/4) √{7−√5−√{6 (5−√5)}}	13π/30	78
π/12	15	(√3−1)/(2√2)	5π/12	75
π/10	18	(1/4)  (√5 − 1)	2π/5	72
7π/60	21	(1/4) √{8 −√3(√5−1)−√{2(5+√5)}}	23π/60	69
2π/15	24	(1/4)√{7+√5−√{6(5+√5)}}	11π/30	66
3π/20	27	(1/2)  √{2 − √{(5−√5)/2}}	7π/20	63
π/6	30	1/2	π/3	60
11π/60	33	(1/4) √{8−√3 (√5+1) + √{2 (5−√5)}}	19 π/60	57
π/5	36	(1/2)  √{(5−√5)/2}	3π/10	54
13π/60	39	(1/4) √{8 + √3 (√5−1) − √{2 (5+√5)}}	17π/60	51
7π/30	42	(1/4) √{9+√5−√{6 (5−√5)}}	4π/15	48
π/4	45	1/√2	π/4	45
4π/15	48	(1/4) √{7−√5+√{6(5−√5)}}	7π/30	42
17π/60	51	(1/4) √{8−√3(√5−1) + √{2(5+√5)}}	13π/60	39
3π/10	54	(1/4)  (√5 + 1)	π/5	36
19π/60	57	(1/4) √{8+√3 (√5+1) − √{2(5−√5)}}	11π/60	33
π/3	60	(1/2)  √3	π/6	30
7π/20	63	(1/2)  √{2 + √{(5−√5)/2}}	3π/20	27
11π/30	66	(1/4) √{9 − √5 + √{6 (5 + √5)}}	2π/15	24
23π/60	69	(1/4) √{8 +√3(√5−1)+√{2(5+√5)}}	7π/60	21
2π/5	72	(1/2)  √{(5+√5)/2}	π/10	18
5π/12	75	(√3+1)/(2√2)	π/12	15
13π/30	78	(1/4)√{9+√5+√{6(5−√5)}}	π/15	12
9π/20	81	(1/2) √{2 + √{(5+√5)/2}}	π/20	9
7π/15	84	(1/4) √{7+√5+√{6 (5+√5)}}	π/30	6
29π/60	87	(1/4) √{8+√3 (√5+1) +√{2 (5−√5)}}	π/60	3
π/2	90	1	0	0

File translated from T_EX by T_TH, version 4.03.
On 2 Aug 2013, 14:50.

A few notes:

• Note that you read the sines from the top down, and the cosines from the bottom up.
• I haven't used radians here, because your captor won't know them from arc seconds, but what we've done is compute the sine and cosine for any value N π/60 radians.
• I've checked all the formulas by plugging them into a calculator and then comparing the values to the calculator's value for the sine. Several times. So I'm pretty sure they are all correct.
• Hopefully this is sufficiently readable. There isn't a good way of translating LaTeX to HTML on Blogger. T_TH is rather old, but still workable. If you'd like a PDF (or the original LaTeX source), shoot me an email and I'll send it to you.
• You might think we can use cos 9 x = 0 to compute cos 10°, which means we could compute cos 1° = cos(10°-9°), and have an analytic expression for the sine and cosine at every degree. We'll explore why that doesn't work next time.

The Logarithmic Function From Scratch

That counterpart to last week's development of the exponential function is the logarithm. As anyone who ever played with the antique known as a slide rule, the logarithmic function converts multiplication into addition:

Log(a b) = Log(a) + Log(b)  ,  (1)

but that's getting ahead of ourselves.

Assume that (1) is all we know: Consider a real function G defined on at least some part of the real numbers, and having the property

G(a b) = G(a) + G(b)  ,  (2)

for any real numbers a and b. What can we say about this function?

First, take a = 0. Then we have

G(0 b) = G(0) + G(b)  , or
G(0) = G(0) + G(b)  .  (3)

So either G(b) = 0 for all b, or G(0) is undefined. (Yes, we can say that G(0) = ∞, but I'd rather not go there right now.) Since G = 0 is uninteresting, we'll assume that G(0) is undefined. For now, at least, we'll restrict ourselves to x > 0 and so avoid the problem.

If we set a = 1 in (2) we get

G(b) = G(1) + G(b)  , which means that

G(1) = 0  ,  (4)

which implies G[a (1/a)] = G(a) + G(1/a) = G(1) = 0  , or

G(1/a) = - G(a)  .  (5)

For any integer M we can repeatedly apply (2) and get

G(a^M) = M G(a)  .  (6)

If we let b = a^M, or a = b^1/M, then by turning around (6) we get

G(b^1/M) = (1/M) G(b)  ,  (7)

and combining the two we get

G(x^M/N) = (M/N) G(x)    (8) for any integers M and N.

Furthermore, we can play the same trick as we did last week and extend this to all positive real numbers y:

G(x^y) = y G(x)  .  (9)

What about continuity? I'm going to be a little more physicist-like this week (though last week was probably not properly mathematical anyway), and go through it rather quickly. Let's take two numbers, x > 0 and y > -x. Then we can always write

x + y = x Q  ,  Q = 1+y/x > 0  ,

and we want to study the behavior of

G(x + y) = G(x Q) = G(x) + G(Q)   as y → 0 or Q → 1.

When we put it like that, the answer is obvious. Write Q = h^1/N. Then as N → ∞, Q → ∞ 1, and

lim_{N → ∞} G(Q) = lim_{N → ∞} G(h)/N → 0  .  (10)

Thus

lim_y→0 |G(x+y) - G(x)| = 0  ,  (11)

and G(x) is a continuous function for x > 0.

Using the same tricks, let's look at the derivative, if it exists:

G'(x) = lim_y→∞0 [G(x+y)-G(x)]/y  ,  (12)

With the above substitutions we can write this as

G'(x) = lim_N→∞ [G(x h^1/N) - G(x)]/[x (h^1/N-1)  ,  (13)

which reduces to G'(x) = [G(h)/x]/ [lim_N→∞ N (h^1/N - 1)]  .  (14)

Last week we showed that the limit in the denominator of (14) is finite, so it follows that G'(x) is well defined for x > 0.

To evaluate the derivative, we do the standard tricks:

d/dy G(x y) = x G'(x y) = d/dy [G(x) + G(y)] = G'(y)

and set y = 1. Then

G'(x) = G'(1)/x  , x > 0  .  (15)

Thus each and every log function (2) is uniquely defined once we define G'(1). Since G(1) = 0, we can integrate (15) to find

G(x) = G'(1) ∫₁^x dt / t  .  (16)

The logical prototype logarithm function (which we could call the, ahem, natural logarithm) is then the function with G'(1) = 1:

ln x = ∫₁^x dt / t  .  (17)

Which is exactly how my college calculus course started out.

One more thing: we want to show that ln x and exp x are inverse functions of one another. Write

H(x) = ln(exp x)  .  (18)

Obviously H(0) = ln(exp 0) = ln(1) = 0. Furthermore,

H'(x) = [d/dx exp x]/exp x = exp x/exp x = 1  ,  (19)

so H(x) = x. We could go the opposite way as well and show that

exp(ln x) = x  .  (20)

That concludes this little exploration of the log and exponential functions. Next week we'll take a look at sines and cosines.

The Exponential Function From Scratch

Welcome to Stupid Math Tricks. Over the last mumblty years or so I've found myself scratching out various proofs of things that are probably well known to anyone who graduated with a degree in math. But I'm a physicist, so what do I know?

So when I can't sleep at night, I start thinking about these little mathematical tidbits, and sometimes obsessing about them, to the point where I really can't get to sleep. The hope is that by writing them down in a permanent place I'll be able to forget them. Of course, maybe a real mathematician will come along and tell me how badly I've gotten things wrong, but that's the chance we'll have to take.

So for my first SMT, let's look at the exponential function. Back in Calculus 101, the book we used at KU started this subject by defining the natural logarithm using the formula:

ln x = ∫₁^x dt / t .    (1)

This is actually a neat way to go about it, as it answers one of the questions on the mind of every calculus student once integrals are taught. Since

∫₁^x tⁿ dt = [xⁿ⁺¹ - 1]/(n+1)  , n ≠ -1,  (2)

what happens when n in (2) is -1? The answer is, well, we have to define a new function. Once we accepted that, it was rather easy to show that ln x has the properties of a logarithm, i.e.

ln (a b) = ln a + ln b  .    (3)

Once we accepted that, the text defined a function exp(x) as the inverse of ln(x):

exp(ln x) = x   ;   ln(exp x) = x ,    (4)

and it was shown that exp(x) had the properties of an exponential:

exp(x + y) = exp(x) exp(y)  .    (5)

But you don't have to do it that way. A few years later, in a thermodynamics class, we were discussing how Maxwell derived the distribution of speeds in an ideal gas. He did this by assuming that (1) the movement of a gas molecule along the x-axis was independent of its movement along the y- or z- axis, and, indeed, independent of the choice of directions. Which meant that the distribution of velocities in an ideal gas must obey the relation

F(v_x²) F(v_y²) F(v_z²) = F(v_x² + v_y² + v_z²)    (6)

where v_x is the component of the velocity in the x direction.

Comparing (5) and (6), and adding a little physics, namely that no atom ever gets up to infinite speed, then the distribution of molecular speeds in an ideal gas must have the form

F(v) = A exp(- B v²)  .    (7)

where the values of A and B depend on temperature.

Believe it or not, the idea that the requirement F(x+y) = F(x) F(y) requires that F(x) be an exponential function was new to me, as I'd been taught to think of exponentials using (2)-(6).

Which, finally, brings us to the topic of the inaugural Stupid Math Tricks post, namely,

What are the properties of a function F(x)
if F(x+y) = F(x) F(y)?

(And try to be formal about it.)

Define a real function F(x) over the real numbers. We assume that

F(x + y) = F(x) F(y)  .    (8)

Let's see what properties we can derive:

1. If x = y = 0, then
   
   F(0) = F(0)²  ,    (9)
   
   so F(0) is either 0 or 1. F(0) = 0 would lead to a very boring function, since then
   
   F(x) = F(x+0) = F(x) F(0) = 0   ∀ x  
   
   so we'll take
   
   F(0) = 1  .    (10)
   
   
2. F(x) must be positive. To see this, first note that
   
   F(x) = F(x/2 + x/2) = F(x/2)²   ,
   
   so F(x) is either positive or 0. But if there was some point x₀ such that F(x₀) = 0, then we could always write
   
   F(x) = F(x-x₀+x₀) = F(x-x₀) F(x₀) = 0   ,
   
   and in particular
   
   F(0) = F(x₀) F(-x₀) = 0  ,
   
   which violates (10). Thus
   
   F(x) > 0   for all real x.   (11)
   
   
3. It follows from (8) and (10) that
   
   F(-x) = 1/F(x)   .   (12)
   
   
4. Since N = 1 + 1 + … + 1 (N times), we can write
   
   F(N) = F(1)^N   .   (13)
   
   
5. Conversely, since 1 = 1/N + 1/N + … + 1/N ,
   
   F(1) = F(1/N)^N   ,
   
   from which it follows that
   
   F(1/N) = F(1)^1/N   .   (14)
   
   
6. Combining (13) and (14), we get
   
   F(R) = F(1)^R   for any rational R   .   (15)
   
   
7. We can extend (14) to any real number x:
   1. For any non-rational real number x and positive integer N, define an integer function M(N;x) such that
      
      M(N;x)/N < x < [M(N;x) + 1]/N   .   (16)
      
      
   2. Then define
      
      F(x) = lim_{N → ∞} F[M(N;x)/N]
      
           = lim_{N → ∞} F(1)^M(N;x)/N   .     (17)
      
      
   3. This also proves that F(x) is a continuous function of x, since for any N,
      
      F(x + 1/N) = F(x) F(1/N) = F(x) F(1)^1/N   ,   (18)
      
      and
      
      lim_{N → ∞} F(1)^1/N = 1 for all F(1) > 0   .   (19)
      
      

OK, so F(x) is a positive, continuous function on the reals. But is it differentiable? You'd think so, but there are many continuous functions which are not everywhere differentiable. So let's see. In the following we'll take F(1) = A > 0, without any loss of generality.

7. We can write
   
   F'(x) = lim_{N → ∞} [ F(x+1/N) - F(x) ]/(1/N)    .   (20)
   
   Using F(x + 1/N) = F(x) F(1/N) = F(x) A^1/N ,
   
   F'(x) = F(x) lim_{N → ∞} N [ A^1/N – 1 ]    .   (21)
   
   
8. Note that
   
   A^1/N – 1 = [A – 1]/[ ∑_n=0^N-1 A^n/N ]    .   (22)
   
   
9. If A > 1, then the sum in the denominator of (22) must be greater than N and
   
   lim_{N → ∞} N [ A^1/N – 1 ] < A – 1    .   (23)
   
   
10. On the other hand, if A < 1, then the denominator of (22) must be greater than N A, and, keeping in mind that [ A^1/N – 1 ] is now negative,
    
    lim_{N → ∞} N [ A^1/N – 1 ] > (A – 1)/A    .   (24)
    
    
11. In either case, the limit in (21) is finite, and so the derivative F'(x) exists for all real x.

Given that, what is the derivative of F(x)?

12. Go back to (8), and differentiate both sides of the equation with respect to y:
    
    F'(x + y) = F(x) F'(y)    .   (25)
    
    
13. If we take y = 0, then
    
    F'(x) = F'(0) F(x)    .   (26)
    
    
14. Note that since F(0) = 1, this implies that the entire set of functions F(x) is defined solely by the value of F'(0). So define a special function of F(x) where F'(0) = 1. Oh, what the heck, let's call it exp(x), where
    
    exp(x+y) = exp(x) exp(y)   by (8),
    
    exp(0) = 1   by (10), and
    
    exp'(x) = exp(x)    .   (27)
    
    
15. By the chain rule
    
    d/dx exp( λ x ) = λ exp(λ x)    ,   (28)
    
    so we can define the general function
    
    F(x) = exp[ F'(0) x]    ,   (29)
    
    
16. and, by (27), for any integer N > 0, the N^th derivative of exp(x) is itself,
    
    d^N/dx^N exp(x) = exp(x)    ,   (30)
    
    
17. Leading to the most famous Taylor Series ever,
    
    exp(x) = ∑_n=0^∞ xⁿ/n!   .   (31)
    
    Which is how my complex analysis textbook started the discussion of exp(x).
    
    
18. (31), of course, lets us evaluate
    
    e = exp(1) = 2.71828182…    .   (32)
    
    
19. And, with the help of (15) and the discussion in VII, write
    
    exp(x) = e^x    .   (33)

And there you have it. From (8), we've derived all of the properties of exp(x), without ever invoking natural logarithms and their inverses. Not that I don't appreciate the other way of looking at things, but, hey, it's nice to know that Maxwell was right.

But wait! Is the exp(x) in (27) the same as the inverse of the function (1)? That remains to be seen, and will be left until next time.