An unusual trigonometric identity

[quickfur]

Recently, while studying the dodecahedron, I came across this curious identity:

2*arctan(phi) = pi - arctan(2)

where arctan is the arctangent function, phi is the golden ratio, and pi is ... well, pi. :-)

As far as I can tell from brute-force calculation, the two sides of the equation are equal; however, I cannot find any algebraic justification for it. Can someone help me prove this algebraically?

(For the curious, the value of each side is the dihedral angle of the dodecahedron.)

[Keiji]

Hmm, interesting!

Not really sure where to go with this one, though, don't look at me for the answers

[wendy]

One should not be surprised. Any pentagonal/decagonal number that divides 5 as the smallest integer has a power that 5 divides.

(f+i)^2 = f(1+2i) a 1:2:r5 triangle.

(2+i) = 3+4i a 3:4:5 triangle

The rectangle f:1 occurs in the icosahedron, and hence the dodecahedron margin-angle.

A different rectangle is 1:f² : 3f, which occurs in the dodecahedron, and hence the icosahedral angle.

[quickfur]

One should not be surprised. Any pentagonal/decagonal number that divides 5 as the smallest integer has a power that 5 divides.

(f+i)^2 = f(1+2i) a 1:2:r5 triangle.

(2+i) = 3+4i a 3:4:5 triangle

The rectangle f:1 occurs in the icosahedron, and hence the dodecahedron margin-angle.

A different rectangle is 1:f² : 3f, which occurs in the dodecahedron, and hence the icosahedral angle.

Call me stupid, but I don't see what's the connection between this and the values of arctan.

[Keiji]

wendy

(is the connection)

[PWrong]

That's a really cool identity. I found a proof using the half angle formula for tan. I'll let you figure it out, all you need is
θ = π - arctan 2,
The right triangle 1:2:√5, and
tan (θ/2) = (1 - cos θ)/sin θ.

[quickfur]

That's a really cool identity. I found a proof using the half angle formula for tan. I'll let you figure it out, all you need is
θ = π - arctan 2,
The right triangle 1:2:√5, and
tan (θ/2) = (1 - cos θ)/sin θ.

Cool, I see the connection now! Thanks!!