The title says it. I had the idea to use the $2\theta$ and $5\theta$ formulae but they are not geometric... The question asks for an algebraic solution as well.

Any ideas?

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    $\begingroup$ Proposition 10 of Book XIII of Euclid's Elements states that if an equilateral pentagon is inscribed in a circle, then the area of the square on the side of the pentagon equals the sum of the areas of the squares on the sides of the hexagon and the decagon inscribed in the same circle. The part about the decagon says something about $\pi/5,$ so it is perhaps not unrelated to this question. $\qquad$ $\endgroup$ – Michael Hardy Mar 29 '17 at 17:17

Answer does not explicitly use the formulae for sin or cos of $2\theta$ or $5\theta$ (cosine rule is used), but also uses some algebra - mainly quadratic equations


Consider a regular pentagon $ABCDE$, connect $BE$ and $AC$ so that they intersect at $F$, let the sides of this pentagon be $x$ and $AF$ be $y$, then $$AE=AB=EF=x$$ $$AF=FB=y$$ In $\triangle AEF$ by cosine rule, $$x^2 = x^2 + y^2 - 2xy\cos72^{\circ} \implies y = 2x\cos72^{\circ} --- (1)$$

In $\triangle ABE$ by cosine rule, $$(x+y)^2 = x^2 + x^2 - 2x^2 \cos108^{\circ} --- (2)$$

Substituting the value of $y$ from $(1)$ in $(2)$ and simplifying (cancel off $x^2$), we get

$$4\cos^{2}72^{\circ} + 2\cos72^{\circ} - 1 = 0$$

Solve the quadratic and get the positive answer as

$$\cos72^{\circ} = \sin18^{\circ} = \frac{\sqrt{5} - 1}{4}$$

  • $\begingroup$ I want to accept this... but if a more geometric solution like the hint mentioned in Michael hardy's comment shows up,I want to save it for that. $\endgroup$ – Logan Luther Mar 29 '17 at 18:12

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