Section 1.2.8: Fibonacci Numbers

Exercise 19. [M27] What is $\cos 36^\circ$?

Verified: yes
Solve time: 44s

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Setup

Determine the exact value of $\cos 36^\circ$.

Equation (4) states that

$$ F_{n+1}F_{n-1}-F_n^2=(-1)^n. \tag{4} $$

Taking $n=2$ gives

$$ F_3F_1-F_2^2=1, $$

which is $2\cdot1-1^2=1$.

Exercise 17 generalizes (4) to

F_{n+k}F_{m-k}-F_nF_m = (-1)^nF_{m-n-k}F_k. ] Taking $n=2$, $m=5$, $k=1$ yields

F_3F_4-F_2F_5

F_2F_1,

$$ hence $$

2\cdot3-1\cdot5=1.

$$ Therefore $$

F_5=F_4+F_3=3+2=5.

The numbers $1,2,3,5$ suggest the geometry of a regular pentagon, whose diagonal-to-side ratio is the golden ratio $\phi$ defined by equation (3),

\phi=\frac{1+\sqrt5}{2}.

\tag{3}

We will compute $\cos36^\circ$ from this ratio. ## Solution Let a regular pentagon have side length $1$ and diagonal length $x$. The diagonals form a smaller regular pentagon. Similarity of the large and small pentagons gives

\frac{x}{1}

\frac{1}{x-1}.

$$ Hence $$

x(x-1)=1,

$$ so $$

x^2-x-1=0.

Since $x>0$,

x=\frac{1+\sqrt5}{2}=\phi.

Now consider an isosceles triangle formed by two diagonals and one side of the pentagon. Its sides are $x,x,1$, and its vertex angle is $36^\circ$. Applying the law of cosines,

1^2=x^2+x^2-2x^2\cos36^\circ.

$$ Therefore $$

1=2x^2(1-\cos36^\circ),

$$ and $$

\cos36^\circ

1-\frac{1}{2x^2}.

Since $x=\phi$ and $\phi^2=\phi+1$,

\cos36^\circ

1-\frac{1}{2(\phi+1)}

\frac{2\phi+1}{2(\phi+1)}.

Using $\phi^2=\phi+1$,

\frac{2\phi+1}{2(\phi+1)}

\frac{2\phi+1}{2\phi^2}

\frac{\phi}{2}.

Substituting $\phi=(1+\sqrt5)/2$,

\cos36^\circ

\frac{1+\sqrt5}{4}.

$$ Thus $$

\boxed{\cos36^\circ=\frac{1+\sqrt5}{4}}.

Verification Let

c=\frac{1+\sqrt5}{4}.

$$ Then $$

2c=\frac{1+\sqrt5}{2}=\phi.

$$ Hence $$

4c^2=\phi^2=\phi+1

=\frac{3+\sqrt5}{2},

$$ so $$

c^2=\frac{3+\sqrt5}{8}.

$$ Using the double-angle identity, $$

\cos72^\circ

2c^2-1

\frac{3+\sqrt5}{4}-1

\frac{\sqrt5-1}{4},

which is the standard exact value of $\cos72^\circ$. The result is therefore consistent. ## Notes A purely algebraic derivation can be obtained from the identity

\cos 5\theta

16\cos^5\theta

20\cos^3\theta

5\cos\theta.

Setting $\theta=36^\circ$ gives $\cos5\theta=\cos180^\circ=-1$, and the positive root of the resulting polynomial equation is again

\cos36^\circ=\frac{1+\sqrt5}{4}.

$$$$