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Find the value of $x$ such that $$\frac{5}{8} \cot36^\circ = \cos^3x$$ The answer is $x=18^\circ$.

It's really messy to plug in the standard values of $\cos36^\circ$, $\sin36^\circ$ and miraculously guess a suitable value of $x$ and prove that our guess is correct.

Is there any nice way to find the value of $x$?

Michael Rozenberg
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user600016
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3 Answers3

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$$\dfrac58\cdot\cot36^\circ=\dfrac{5\cos36^\circ}{8\sin36^\circ}=\dfrac{5\cos^236^\circ}{4\cos18^\circ}$$

Using Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$

$$\cos36^\circ-(2\cos^236^\circ-1)=\frac12\iff5\cos^2 36^\circ=(1+\cos36^\circ)^2=(2\cos^218^\circ)^2$$ Can you take it from here?

lab bhattacharjee
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Since $$\cos^3x=\frac{3\cos{x}+\cos{3x}}{4},$$ we need to prove that: $$\frac{5\cos36^{\circ}}{8\sin36^{\circ}}=\frac{3\cos18^{\circ}+\cos54^{\circ}}{4}$$ or $$5\cos36^{\circ}=3\sin54^{\circ}+3\sin18^{\circ}+\sin90^{\circ}-\sin18^{\circ}$$ or $$2\cos36^{\circ}+2\cos108^{\circ}=1,$$ which is true because $$2\cos36^{\circ}+2\cos108^{\circ}=\frac{2\sin36^{\circ}\cos36^{\circ}+2\sin36^{\circ}\cos108^{\circ}}{\sin36^{\circ}}=$$ $$=\frac{\sin72^{\circ}+\sin144^{\circ}-\sin72^{\circ}}{\sin36^{\circ}}=1.$$

Michael Rozenberg
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    Very nice solution +1. It's really tough to decide who to choose as the accepted answer. I am going with Lab bhattacarjee's answer as there was no "reverse solution" and it appeared a bit more "one way" although your solution is more natural – user600016 Feb 26 '20 at 13:24
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    And pls don't reverse the steps of the solution for that, it would spoil its essence :) – user600016 Feb 26 '20 at 13:30
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Yet an other way to proceed, using algebraic only manipulations after a quick reshape...

We want to show $$ \frac 58\cdot \frac{\cos 36^\circ}{\sin 36^\circ} \overset{!}=\sin^3 72^\circ \ . $$ Let us denote by $c,s$ respectively the values for $\cos 36^\circ$ and $\sin 36^\circ$. Then we want: $$ \frac 58\cdot\frac cs \overset{!}= 8s^3c^3\ ,\text{ or equivalently } 5 - 64s^4c^2=0\ . $$ We start with $(s+ic)^5=\cos (5\cdot 36^\circ)+i\sin(5\cdot 36^\circ)=-1$, and consider only the imaginary part, thus getting $$ \begin{aligned} 0 &= 5sc^4-10s^3c^2+s^5\ ,\text{ and since $s\ne 0$ we get}\\ 0 &= 5c^4-10s^2c^2+s^4\\ &=5(1-2s^2+s^4)-10s^2(1-s^2)+s^4\\ &=5-20s^2+16s^4\ . \qquad\text{From here:} \\[3mm] 5-64s^4c^2 &=5 - 64s^4(1-s^2) \\&= 5-64s^4+64s^6\\ &=\underbrace{(5-20s^2+16s^4)}_{=0}(1+4s^2)=0\ . \end{aligned} $$ $\square$

dan_fulea
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