Look again at the sharp and flat triangles of the pentagon that we saw above. If we divide each in half, we have right angled triangles with sides 1 and Phi/2 around the 36° angle in the flat triangle and sides 1/2 and Phi around the 72° angle in the sharp triangle. So:

cos (72°) = cos| |
'2 |
J = sin(18°) = sin |
! 2 20 | |

cos(36°) = cos |
(5) : |
I 2 2f |

We have sin(18°) but what about cos(18°)? This has a somewhat more awkward expression as:

Now we know the sin and cos of both 30° and 18° we can find the sin and cos of their difference using:

and get:

AAAAgh! as Snoopy might have said.

Is there a neater (that is, a simpler) expression? Perhaps you can find one. Let me know if you do and it will be added here with your name!

This form of cos(12°) is derived from the expression on page 42 of

^^^ Roots of (H-L)/15 Recurrence Equations in Generalized Pascal Triangles by C Smith and V E Hoggatt Jr. in The Fibonacci Quarterly vol 18 (1980) pages 36-42.

What about other angles? From an equilateral triangle cut in half we can easily show that:

and from a 45-45-90 degree triangle we can derive:

1 V2

and not forgetting, of course:

Can you find any more angles that have an exact expression (not necessarily involving Phi or phi)? Let me know what you find and let's get a list of them here.

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