How do you express the value of `cos 2^o15'`, in mathematical exactitude?

If you know that `cos 36^@ = 1/4(sqrt(5)+1)` then you can use the half angle formula:

`cos(theta/2) = +-sqrt(1/2+1/2cos theta)`

We need to apply this `4` times to divide our known angle by `16 = 2^4`. We can take the positive square root each time as all of the angles we are working with are in Q1.

So:

`cos 2^@15' = sqrt(1/2 + 1/2cos 4^@30')`

`color(white)(cos 2^@15') = sqrt(1/2 + 1/2sqrt(1/2 + 1/2cos 9^@))`

`color(white)(cos 2^@15') = sqrt(1/2 + 1/2sqrt(1/2 + 1/2sqrt(1/2+1/2cos 18^@)))`

`color(white)(cos 2^@15') = sqrt(1/2 + 1/2sqrt(1/2 + 1/2sqrt(1/2+1/2sqrt(1/2+1/2cos 36^@))))`

`color(white)(cos 2^@15') = sqrt(1/2 + 1/2sqrt(1/2 + 1/2sqrt(1/2+1/2sqrt(1/2+1/2(1/4(sqrt(5)+1)))))`

This can be simplified a little, but has a pleasing form as is, which indicates the derivation.

`color(white)()`
Derivation of `cos 36^@`

Here's a derivation of the formula for `cos 36^@` used above...

`-1 = cos 180^@ + i sin 180^@`

`color(white)(-1) = (cos 36^@ + i sin 36^@)^5`

`color(white)(-1) = (cos^5 36^@ - 10 cos^3 36^@ sin^2 36^@ + 5 cos 36^@ sin^4 36^@) + i (5cos^4 36^@ sin 36^@ - 10cos^2 36^@ sin^3 36^@ + sin^5 36^@)`

Equating Real parts:

`-1 = cos^5 36^@ - 10 cos^3 36^@ sin^2 36^@ + 5 cos 36^@ sin^4 36^@`

`color(white)(-1) = cos^5 36^@ - 10 cos^3 36^@ (1-cos^2 36^@) + 5 cos 36^@ (1-cos^2 36^@)^2`

`color(white)(-1) = 16 cos^5 36^@ - 20 cos^3 36^@ + 5 cos 36^@`

Hence `cos 36^@` is a zero of:

`16x^5-20x^3+5x+1 = (x+1)(16x^4-16x^3-4x^2+4x+1)`

`color(white)(16x^5-20x^3+5x+1) = (x+1)(4x^2-2x-1)^2`

We can discard `(x+1)` and look for zeros of the repeated quadratic factor, finding:

`0 = 4(4x^2-2x-1)`

`color(white)(0) = 16x^2-8x+1-5`

`color(white)(0) = (4x-1)^2-sqrt(5)^2`

`color(white)(0) = ((4x-1)-sqrt(5))((4x-1)+sqrt(5))`

`color(white)(0) = (4x-1-sqrt(5))(4x-1+sqrt(5))`

Hence:

`x = 1/4(1+-sqrt(5))`

Then since `36^@` is in Q1, `cos 36^@ > 0`, so choose the positive root to find:

`cos 36^@ = 1/4(sqrt(5)+1)`