Challenge Problem #6: Prove tan 18°=√(1-2/√5)

In summary, the conversation is about proving that $\tan18^\circ = \sqrt{1-\dfrac2{\sqrt5}}$, without using any cheating tools. The speaker, Dan, claims that the proof is intuitively obvious and comes from the geometry of a regular pentagon. The diagram is shown and the diagonals of the pentagon are found to be the golden ratio, which leads to the length of $PA$ being $\dfrac14(\sqrt5-1)$. Using Pythagoras, it is proved that $\tan 18^\circ = \sqrt{1 - \dfrac2{\sqrt5}}$. Another participant, Opalg, shares a geometric solution while the speaker has an analytic solution.
  • #1
Olinguito
239
0
Prove that
$$\tan18^\circ\ =\ \sqrt{1-\dfrac2{\sqrt5}}.$$
No calculator, computer program, Excel, Google, or any other kind of cheating tool allowed. (Smirk)

Have fun!
 
Last edited:
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  • #2
It's true. Proof: It's intuitively obvious.

-Dan
 
  • #3
Olinguito said:
Prove that
$$\tan18^\circ\ =\ \sqrt{1-\dfrac2{\sqrt5}}.$$
No calculator, computer program, Excel, Google, or any other kind of cheating tool allowed. (Smirk)
[sp]This comes from the geometry of the pentagon.
[TIKZ][scale=0.75]
\coordinate [label=above: $C$] (C) at (90:5cm) ;
\coordinate [label=above right: $D$] (D) at (18:5cm) ;
\coordinate [label=above left: $B$] (B) at (162:5cm) ;
\coordinate [label=below: $A$] (A) at (234:5cm) ;
\coordinate [label=below: $E$] (E) at (306:5cm) ;
\coordinate [label=below: $P$] (P) at (-4.755,-4.045) ;
\coordinate [label=below: $R$] (R) at (4.755,-4.045) ;
\coordinate [label=above right: $Q$] (Q) at (0.45,0.6) ;
\draw (A) -- node
{$1$} (B) -- node[above left] {$1$} (C) -- node[above right] {$1$} (D) -- node
{$1$} (E) -- node[below] {$1$} (A) -- (P) -- (B) -- (D) -- (R) -- (E) ;
\draw (C) -- (E) -- node[above right] {$d$}(B) ;
\draw (-3.4,-3.75) node {$72^\circ$} ;[/TIKZ]
If $ABCDE$ is a regular pentagon with side $1$ then it is a standard result that the diagonals have length $d = \frac12(\sqrt5+1)$, the golden ratio. A quick way to see that is to notice that in the above diagram the triangles $CQD$ and $BQE$ are similar (with angles $36^\circ$ and $108^\circ$), from which $QD = \frac1d$. Then $BD = BQ+QD$, so that $d = 1 + \frac1d$, a quadratic equation whose positive root is the golden ratio.

If $BP$ and $DR$ are the perpendiculars from $B$ and $D$ to $AE$ then $d = PR = PA + AE + ER = 1+2PA$, from which $PA = \frac12(d-1) = \frac14(\sqrt5-1)$.

By Pythagoras in the triangle $BPA$, $BP^2 = 1 - \frac1{16}(\sqrt5-1)^2 = \frac18(5+\sqrt5)$. Therefore $$\frac{PA^2}{BP^2} = \frac{\frac18(3-\sqrt5)}{\frac18(5+\sqrt5)} = \frac{(3-\sqrt5)(5-\sqrt5)}{(5+\sqrt5)(5-\sqrt5)} = \frac{20-8\sqrt5}{20} = 1 - \frac2{\sqrt5}.$$ But $\frac{PA}{BP} = \tan 18^\circ$, and therefore $\tan 18^\circ = \sqrt{1 - \frac2{\sqrt5}}.$

[/sp]​
 
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  • #4
Thanks, Opalg! It’s a neat geometric solution. (Clapping)

My own solution is entirely analytic. (Nerd)

Consider $\tan54^\circ$. On the one hand:
$$\tan54^\circ\ =\ \frac1{\tan36^\circ}\ =\ \frac{1-\tan^218^\circ}{2\tan18^\circ}.$$
On the other hand:
$$\tan54^\circ\ =\ \tan3(18)^\circ\ =\ \frac{3\tan18^\circ-\tan^318^\circ}{1-3\tan^218^\circ}$$
using the triple-angle formula for tangent. Hence:
$$\frac{1-\tan^218^\circ}{2\tan18^\circ}\ =\ \frac{3\tan18^\circ-\tan^318^\circ}{1-3\tan^218^\circ}$$
$\implies\ 1-4\tan^218^\circ+3\tan^418^\circ\ =\ 6\tan^218^\circ-2\tan^418^\circ$

$\implies\ 5\tan^418^\circ-10\tan^218^\circ+1\ =\ 0$

$\implies$ $\tan^218^\circ$ is a root of the quadratic equation $f(x)=5x^2-10x+1=0$

$\implies\ \tan^218^\circ=\dfrac{10\pm\sqrt{80}}{10}=1\pm\dfrac2{\sqrt5}.$

But $f(1)=-4<0$ and $f(2)=1>0$ showing that there is a root between $1$ and $2$. This root can’t be $\tan^218^\circ$ as $0<\tan18^\circ<1$. It follows that $\tan^218^\circ$ is the lesser of the two possible roots, i.e.
$$\tan^218^\circ\ =\ 1-\frac2{\sqrt5}$$
and we are done. (Cool) (The other root turns out to be $\tan^254^\circ$.)
 

FAQ: Challenge Problem #6: Prove tan 18°=√(1-2/√5)

1. How do you prove that tan 18° is equal to √(1-2/√5)?

To prove this, we can use the trigonometric identity tan 2θ = (2tanθ)/(1-tan²θ). By substituting θ = 9°, we get tan 18° = (2tan9°)/(1-tan²9°). We can then use the half-angle formula for tangent, tan θ/2 = (1-cosθ)/sinθ, to further simplify the expression. After some algebraic manipulation, we arrive at tan 18° = √(1-2/√5).

2. What is the significance of proving tan 18°=√(1-2/√5)?

This proof helps us understand the relationship between trigonometric functions and their corresponding angles. It also demonstrates the usefulness of trigonometric identities in solving problems involving angles and their functions.

3. Can this proof be used for other angle values?

Yes, this proof can be extended to other angle values by substituting the corresponding angle in place of 18°. However, the resulting expression may not be as simple as the one obtained for 18°.

4. How does this proof relate to the golden ratio?

The golden ratio, also known as phi (φ), is equal to (1+√5)/2. By substituting this value in place of √5 in the expression √(1-2/√5), we get the value of tan 18° as φ-1. This shows the connection between the golden ratio and the tangent of 18°.

5. Can this proof be used to find the value of tan 18° without a calculator?

Yes, this proof provides a way to calculate the value of tan 18° without using a calculator. By using the trigonometric identities and formulas, we can obtain the exact value of tan 18° as √(1-2/√5).

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