What's the exact value of sin36°?

Question

I know cos36° is (1+root 5)/4; can sin36° be expressed without any square root inside a square root?

The answer CANNOT be rounded! And please read the requirements carefully!

Answer 1

If cos 36o = (1+sqrt(5))/4, then
cos2 36o = (6+2*sqrt(5))/16.
Therefore, sin2 36o = (10-2*sqrt(5))/16.
Or sin 36o = sqrt(10-2*sqrt(5))/4.

Now, let 10-2*sqrt(5) = (sqrt(a)-sqrt(b))2
Considering that the rational and irrational parts are equal, we get:
a+b = 10 ..(1)
and 2*sqrt(a*b)=2*sqrt(5) or
a*b = 5. ..(2)
Therefore, a*(10-a) = 5.
or a^2-10*a+5 = 0
or a = (10±sqrt(80))/2.
Since sqrt(80) is irrational number, it cannot be further factorized.

Answer 2

The answer is
sqrt[10-2sqrt(5)]/4.

This cannot be written as a quadratic radical. It is also not possible to write it as a sum of two sqare roots.

Answer 3

if cos36 = (1+root5)/4 than;
we know that sin^2a + cos^2a = 1 (^2 means to the power of 2 and a=36 degrees)
so
sin^2a = 1-cos^2a than sin^2a = 1-((1+root5)/5)^2
after you do what you have to do you get:

sin36 = (root(29+2root5))/5

of course i could have made a mistake

Answer 4

sin 36 deg.= 0.58778525229247312916870595463907


if u want it be expressed in terms of cos 36, u can directly go in for [square root of (1-cos^2 36deg)]

enjoyyyyyyyyyy

Answer 5

s^2 + c^2 = 1
s = + root{1 - c^2} =
V{1 - (3 + V5)/8} =
V{(5 -V5)/8}
You never loose V5.

Th

Answer 6

sin 36 deg.= 0.5877852522924731291687059546..