Is there an easy way of showing $8\cos^3(12^o)-6\cos(12^o)=\phi$?

Question

Is there an easy way of showing (1)

(1)

$$8\cos^3(12^o)-6\cos(12^o)=\phi$$

with out substituting into the equation?

$8\left(\frac{1}{8}\left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]\right)^3-6\cdot\frac{1}{8}\left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]=\phi$

Trigonometric constant expressed in real radicals-Wikipedia.

$\cos(12^o)=\frac{1}{8}\left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]$

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

http://math.stackexchange.com/questions/827540/proving-trigonometric-equation-cos36-circ-cos72-circ-1-2 can be used to prove $$\phi=1+\dfrac1\phi$$ where $$\phi=2\cos36^\circ$$ — lab bhattacharjee, May 21 '16 at 17:06

score 4 · Accepted Answer · answered May 21 '16 at 12:27

4

Recall that $$\cos 3x=4\cos^3 x- 3\cos x$$ $$\implies 2(4\cos^3 12^{\circ}-3\cos 12^{\circ})=2\cos36^\circ$$ Using $$\cos 36^\circ=\frac{1+\sqrt{5}}4$$ We get the that $$8\cos^3 12^{\circ}-6\cos 12^{\circ}=\frac{1+\sqrt{5}}2=\phi$$

answered May 21 '16 at 12:27

Nikunj

6,160

1

For the value of $\cos 36^\circ$, see https://en.wikipedia.org/wiki/Pentagon#.E2.80.A0_Proof_that_cos_36.C2.B0_.3D. – lhf May 21 '16 at 12:48

Is there an easy way of showing $8\cos^3(12^o)-6\cos(12^o)=\phi$?

1 Answers1