What is #cos 36^@# ?

3 Answers
Surya K.
Jan 31, 2018

#cos(36^@)=0.809#

Explanation:

#cos(36^@)=0.809#

There really is nothing to say.

George C.
Jan 31, 2018

#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#

Sridhar V.
Jan 31, 2018

#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#

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