Math Is Fun Forum

henkverhelle

Member

Registered: 2008-12-28

Posts: 1

Calculating exact values for cos(x)

It's possible to calculate certain values
for cos(x) analytically
like cos(45°),cos(30°)
but also
cos(12°) and cos(18°)
My questions is how is it done e.g. cos(18°) ?

Kurre

Member

Registered: 2006-07-18

Posts: 280

Re: Calculating exact values for cos(x)

36°=pi/5 radians. For example you could proceed like this:
Sin(pi/5)=Sin(pi-pi/5)=Sin(4pi/5)=2Sin(2pi/5)Cos(2Pi/5)=4Cos(pi/5)Sin(pi/5)(2Cos(Pi/5)-1)
-->
8x^3-4x-1=0, where x=cos(pi/5)
We can guess one root to be -1/2 so Factoring yields:
8x^3-4x-1=2(x+1/2)(4x^2-2x-1)=0
but obviously cos(pi/5)≠-1/2, so Cos(pi/5) is a solution to
4x^2-2x-1, which has the solutions:
x=(1±√5)/4, but Cos(pi/5)>0, so:
Cos(pi/5)=(1+√5)/4
*I misread and thought Cos(pi/5) was what we were looking for, but we can now for example use half angle formulas and such to get Cos(18°)=Cos(pi/10). there may be quicker and more direct methods though

Identity

Member

Registered: 2007-04-18

Posts: 934

Re: Calculating exact values for cos(x)

A longer but geometrical (yay) proof

Let points BCE be collinear and let point A be such that triangle ABE is isosceles with AB = BE, and triangle ACE is isosceles with AC = AE. Let D be the point where the angle bisector of <CAE touches CE. AE = 1 and cos(<ABE) = pi/5

Then,
Since <AEC = 2pi/5 , then <ACE = 2pi/5, <BCA = pi - 2pi/5 = 3pi/5, <BAC = pi/5.
∴ ABC is isosceles. So BC = AC = AE = 1.
Now, <CAD = pi - pi/2 - 2pi/5 - pi/10, so sin(pi/10) = CD/AC = CD
∴ BD = BC + CD = 1 + sin(pi/10)

Since CD = CE (bisector), BE = AB = 1+2sin(pi/10),

So, from triangle ABD,

And solve for sin(pi/10)

You can then use formula to find cos(pi/10)

Last edited by Identity (2008-12-28 18:13:12)