Using the properties of a regular pentagon, how can I show the following? $$\cos36^\circ=\frac{\sqrt{5}+1}{4}$$
How can I use the sine law or the cosine law?
I can show the result using $\cos(2\theta)=\cos(3\theta)$, but how can I use the properties of a regular pentagon?
Best Answer
If $ABCDE$ is a regular pentagon, then $\triangle ABC$ and $\triangle ACD$ are both isosceles triangles with the angle at $A$ equal to $36^\circ$. If we let $s$ be the side length of the pentagon, and $t$ be the length of the diagonal of the pentagon, then the law of cosines on these two triangles gives $$ s^2 = s^2+t^2-2st\cos 36^\circ\\ s^2 = t^2+t^2-2t^2\cos 36^\circ $$ Simplifying the first equation (and assuming $t \neq 0$) gives $$ t = 2s\cos 36^\circ $$ Inserting this into the second equation (and assuming that $s\neq 0$) gives $$ s^2 = 8s^2\cos^236^\circ - 8s^2\cos^236^\circ\\ 1 = 8\cos^236^\circ - 8\cos^336^\circ $$ Letting $x = 2\cos36^\circ$, we can rewrite this as $$ 1 = 2x^2-x^3\\ x^3-2x^2+1 = 0 $$ It's not difficult to see that $1$ is a solution (the rational root theorem tells us that $\pm 1$ are the only possible rational solutions, so checking them is a reasonable thing to do). But we know that $\cos36^\circ \neq \frac12$. (That's $\cos 60^\circ$, the only other case where two isosceles triangles with side lengths $a, a, b$ and $a, b, b$ have an angle in common: $a = b$. This is also a way to spot that $x = 1$ is a solution to the cubic equation.) So we factor out that solution, and what remains is $$ x^2-x-1 = 0 $$ which has one positive solution: $x = \frac{1+\sqrt5}2$.