Question

What is `cos 36^@` ?

Answer 1

`cos(36^@)=0.809`

Explanation:

`cos(36^@)=0.809`

There really is nothing to say.

Answer 2

`cos 36^@ = 1/4(1+sqrt(5)) ~~ 1597/1974 ~~ 0.80917`

Explanation:

For brevity, write:

`{ (c = cos 36^@), (s = sin 36^@) :}`

Note that:

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

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

`color(white)(-1) = (c + i s)^5`

`color(white)(-1) = (c^5-10c^3s^2+5cs^4) + i(5c^4s-10c^2s^3+s^5)`

Looking at the imaginary part and dividing by `s`, we find:

`0 = 5c^4-10c^2s^2+s^4`

`color(white)(0) = 5c^4-10c^2(1-c^2)+(1-c^2)^2`

`color(white)(0) = 5c^4-10c^2+10c^4+1-2c^2+c^4`

`color(white)(0) = 16c^4-12c^2+1`

`color(white)(0) = 16c^4-8c^2+1-4c^2`

`color(white)(0) = (4c^2-1)^2-(2c)^2`

`color(white)(0) = ((4c^2-1)-2c)((4c^2-1)+2c)`

`color(white)(0) = (4c^2-2c-1)(4c^2+2c-1)`

So using the quadratic formula, we find:

`c = (+-color(blue)(2)+-sqrt(color(blue)(2)^2-4(color(blue)(4))(color(blue)(-1))))/(2(color(blue)(4)))`

`color(white)(c) = (+-2+-2sqrt(5))/8`

`color(white)(c) = 1/4(+-1+-sqrt(5))`

Which combinations of signs do we need?

Note that `cos 36^@ > 0` since `36^@` is in Q1.

So:

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

Then `cos 36^@ ~~ cos 30^@ = sqrt(3)/2 ~~ 0.866`. Hence we need `1/4(1+sqrt(5)) ~~ 0.8` rather than `1/4(-1+sqrt(5)) ~~ 0.3`

So:

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

Notice that this is `1/2` of the golden ratio `1/2(1+sqrt(5)) ~~ 1.618`, which is the limiting ratio of the Fibonacci sequence.

So to get a good rational approximation for `cos 36^@` we can calculate some terms of the Fibonacci sequence, divide the last two terms by one another and by `2`...

`1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597`

So:

`cos 36^@ ~~ 1597/(2*987) = 1597/1974 ~~ 0.80917`

Answer 3

`color(blue)(cos 36 ^{\circ} ~~ (sqrt(5)+1)/4`

Explanation:

Given:

`color(blue)(cos 36 ^{\circ}`

We will write `color(green)(theta = 36^{\circ}`

`color(green)(rArr theta = 2*18^{\circ}`

Let us say `color(red)(A = 18^{\circ}`

`:. color(blue)(cos 36 ^{\circ} = cos(2*18^{\circ}) = cos(2A)`

Note that:

`color(green)(cos(2A) = 1-2sin^2A`

Let us consider `color(blue)(cos(2*18^{\circ})`

`rArr 1-2 sin^2 18^{\circ}`

Note that:

`color(green)(sin(18^{\circ}) = (sqrt(5) - 1)/4`

`:. 1-2 sin^2 18^{\circ} = 1-2 [(sqrt(5)-1)/4]^2`

Remember:

`color(brown)((a-b)^2 -= (a^2 - 2ab + b^2)`

We use this identity to simplify

`1-2 [(sqrt(5)-1)/4]^2`

`rArr 1-2 [((sqrt(5))^2-2*sqrt(5)*1+1^2)/16]`

`rArr 1-2 [[5-2*sqrt(5)+1)/16]`

`rArr 1- [[6-2*sqrt(5))/8]`

`rArr (8-6+2*sqrt(5))/8`

`rArr (2+2*sqrt(5))/8`

`rArr [2(1+sqrt(5))]/8`

`rArr (1+sqrt(5))/4`

Rearranging the terms, we get

`(sqrt(5)+1)/4`

Hence,

`color(blue)(cos 36 ^{\circ} ~~ (sqrt(5)+1)/4`