Constructions for the Golden Ratio

Other angles related to Phi

Half of 1,1,Phi is 54-90-36; half of Phi,Phi,1 is 18-72-90 Look again at the sharp and flat triangles of the pentagon that we saw above. If we divide each in half, we have right angled triangles with sides 1 and Phi/2 around the 36° angle in the flat triangle and sides 1/2 and Phi around the 72° angle in the sharp triangle. So:


cos(72°) = cos ( 2 π ) = sin(18°) = sin ( π ) = φ = 1

5

10

2

2 Φ

cos(36°) = cos ( π ) = sin(54°) = sin ( 3 π ) = Φ = 1

5

10

2

2 φ


We have sin(18°) but what about cos(18°)? This has a somewhat more awkward expression as:

cos(18°) =
sqrt
Φ √5

2

Now we know the sin and cos of both 30° and 18° we can find the sin and cos of the angle differences using:

cos(A – B) = cos(A) cos(B) + sin(A) sin(B)

and get:

cos(12°) =
φ +
sqrt
3 Φ √5

4

AAAAGH! as Snoopy might have said.
Is there a neater (that is, a simpler) expression? Perhaps you can find one. Let me know if you do and it will be added here with your name!

There are several angles whose cosine is similar to this one:

cos(24°) =
Φ +
sqrt
3 φ √5

4
cos(48°) =
–φ +
sqrt
3 Φ √5

4
cos(84°) =
–Φ +
sqrt
3 φ √5

4

What about other angles? From an equilateral triangle cut in half we can easily show that:

cos(60°) =
sin(30°) = 1

2
cos(30°) =
sin(60°) = √3

2

and from a 45-45-90 degree triangle we can derive:

cos(45°) = sin(45°) = 1 = √2

√2

2

and not forgetting, of course:

sin(0°) = cos(90°) = 0
sin(90°) = cos(0°) = 1

Article: The form of cos(12°) above is derived from the expression on page 42 of
Roots of (H-L)/15 Recurrence Equations in Generalized Pascal Triangles by C Smith and V E Hoggatt Jr. in The Fibonacci Quarterly vol 18 (1980) pages 36-42.

Can you find any more angles that have an exact expression (not necessarily involving Phi or phi)?

Let me know what you find and let's get a list of them here.
Here is my initial attempt - can you add any more angles?


Binet's Formula for Fib(n) in trig. terms

These two alternative forms of Binet's formula:

Fib(n) = Phin – (–Phi)–n = Phin – (–phi)n

√5

√5

use sine and cosine functions.

A Simple Trig formula for Fib(n)

Above we saw that cos(π/5) = cos(36°) = Φ/2 and cos(2 π/5) = cos(72°) = φ/2.
Since cos(180° – A) = –cos(A), we have cos(180° – 72) = – φ/2. So

Φ = 2 cos( π/5 ), –φ = 2 cos( 3π/5 )

We can use these in Binet's formula to rewrite it as

Fib(n) = 2n [ cosn(π/5) – cosn(3π/5) ]
--
√5

Article: The Bee and the Regular Pentagon W Hope-Jones, Mathematical Gazette vol 55, 1971, pg 220 ff which is a reprint of the original 1921 version (vol 10).

Binet's Formula solely in Trig terms

This time we even replace the √5's by trig forms:

sin ( π ) sin ( ) = √5 and    sin ( ) sin ( ) = – √5

5

5

4

5

5

4

and so Binet's formula becomes this time:

Fib(n) = 2n+2 ( cosn ( π ) sin ( π ) sin ( ) +  cosn ( ) sin ( ) sin ( ) )

5

5

5

5

5

5

5

or, if you prefer degrees rather than radians:

Fib(n) = 2n+2 ( cosn(36°) sin( 36°) sin(108°)  +  cosn(108°) sin(108°) sin(–36°) )

5

Article: See Fibonacci in Trigonometric Form Problem B-374 proposed and solved by F Stern in The Fibonacci Quarterly vol 17 (1979) page 93 where another form is also given.

Phi and Powers of Pi

There is a simple (infinite) series for calculating the cosine and the sine of an angle where the angle is expressed in radians. See Radian Measure (the link opens in a new window - close it to return here) for a fuller explanation.
Basically, instead of 360 degrees in a full turn there are 2π radians. The radian measure makes many trigonometric equations simpler and so it is the preferred unit of measuring angles in mathematics.

If angle x is measured in radians then

cos( x ) = 1 – x2 + x4 x6 + ...

2!

4!

6!

sin( x ) = x – x3 + x5 x7 + ...

3!

5!

7!

Here, n! means the factorial of n which means the product of all the whole numbers from 1 to n.
For example, 4! = 1 × 2 × 3 × 4 = 24

.

So, using the particular angles above in sin(π/10) and cos(π/5) we have formulae for φ and Φ in terms of powers of π :-

φ =
2 sin ( π )

10

=
2 ( π
π3
+ π5
π7
+ ... )

10

1033!

1055!

1077!

=
π π3
+ π5
π7
+ ...

5

3,000

6,000,000

25,200,000,000

Φ =
2 cos ( π
)

5

=
2 ( 1 – π2
+ π4
π6
+ π8
–  ... )

522!

544!

566!

588!

=
2 – π2
+ π4
π6
+ π8
–  ...

25

7,500

5,625,000

7,875,000,000

In the upper formula, going to up to the π9 term only will give φ to 9 decimal places whereas stopping at the π8 term in the lower formula will give Φ to 7 decimal places.

These two formula easily lend themselves as an iterative method for a computer program (i.e. using a loop) to compute Φ and φ. To compute the next term from the previous one, multiply it by (π/5)2 or by (π/10)2 for φ and divide by two integers to update the factorial on the bottom, remembering to add the next term if the previous one was subtracted and vice versa. Finally multiply your number by 2.

You will need and an accurate value of π. Here is π to 102 decimal places:

     3. 14159 26535 89793 23846 26433 83279 50288 41971 69399 37510
58209 74944 59230 78164 06286 20899 86280 34825 34211 70679
82..

With thanks to John R Goering for suggesting this connection between Phi and π.