Golden ratio and special angles

The golden ratio comes up in unexpected places. This post will show how the sines and cosines of some special angles can be expressed in terms of the golden ratio and its complement.

Recall the golden ratio is

φ = (1 + √ 5)/2

and the complementary golden ratio is

φ’ = (1 − √ 5)/2.

The derivation begins by solving the trigonometric equation

sin 2θ = cos 3θ

in two different ways. To make the solution unique, we look for the smallest positive solution.

First, note that the sine of an angle is the cosine of its complement, i.e.

sin(x) = cos(π/2 − x).

So our equation can be written as

cos(π/2 − 2θ) = cos 3θ.

The smallest positive solution satisfies π/2 − 2θ = 3θ, and so θ = π/10 or 18°.

Now let’s solve the same equation another way. First, we use the double and triple angle identities.

sin 2θ = 2 sin θ cos θ
cos 3θ = 4 cos3 θ − 3 cos θ

Set the two equations above equal to each other and divide by cos θ. Then we have

4 cos2 θ − 3 = 2 sin θ.

Substitute 1 – sin2 θ for cos2 θ and the result is a quadratic equation in sin θ:

4 sin2 θ + 2 sin θ − 1 = 0.

From the quadratic equation, the solutions are sin θ = (−1 ± √ 5)/4. The positive solution is

sin θ = (−1 + √ 5)/2 = -φ’/2.

Setting the solutions obtained from both methods equal to each other,

sin π/10 = sin 18°= -φ’/2.

We can now use common trig identities and the above result to express the sines and cosines of other angles in terms of  φ. Switching to degrees will make the following a little easier to read.

We know sin 18° =−φ’/2, and so cos 72° = −φ’/2. We can use the sum angle identities to express the sine and cosine of every multiple of 18° in terms of φ. Also, we could apply the half angle identities to express the sine of cosine of 9° in terms of φ, and then again by addition formula we could extend this to all multiples of 9°.

This post was an expanded form of a derivation given in The Divine Proportion.

2 thoughts on “Golden ratio and special angles

  1. And since you can do 15 degrees as half of 30 degrees, the difference formula will give you 3 degrees and from there you can get all multiplies of 3 degrees.

  2. I first learned this as a geometric argument.

    Let ABC be an isosceles triangle with A=72 degrees, B=36 degrees, C=72 degrees. Let AB = BC = 1 and CA = x. We will solve for x with a simple geometric argument.

    Draw the angle bisector from A intersecting BC at P. Observe that angle CAP is 36 degrees, so angle APC is 72 degrees, so triangle CAP is another 72-36-72 isosceles triangle. Therefore AP = x.

    Observe that angle PAB is 36 degrees, so triangle BPA is isosceles, so BP also = x and PC = 1-x.

    Since triangle ABC is similar to triangle CAP (both being 36-72-36 isosceles), we have PC/CA = CA/AB, or (1-x)/x = x/1. This is the formula for the golden ratio; “the smaller is to the larger as the larger is to the whole”… We wind up with x = 1/φ.

    Then just apply the Law of Cosines to triangle ABC to get the cosine of 36 degrees in terms of φ. It’s φ/2 if I recall correctly.

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