# How do I evaluate cos(pi/5) without using a calculator?

Sep 8, 2015

Cos ($\pi$ /5) = cos 36° = (sqrt5 + 1)/4.

#### Explanation:

If $\theta$ = $\pi$/10, then 5$\theta$ = $\pi$/2 $\implies$ cos3$\theta$ = sin2$\theta$.[ cos ($\pi$ /2 - $\alpha$) = sin$\alpha$}.
$\implies$ 4${\cos}^{3}$ $\theta$ - 3cos$\theta$ = 2sin$\theta$cos$\theta$$\implies$ 4 ${\cos}^{2}$$\theta$ - 3 =2 sin $\theta$.
$\implies$ 4 ( 1 - ${\sin}^{2}$ $\theta$) - 3 = 2 sin$\theta$. $\implies$ 4${\sin}^{2}$ $\theta$+2sin$\theta$ - 1 = 0$\implies$
sin$\theta$ =( sqrt 5 - 1 ) /4.
Now cos 2$\theta$ = cos $\pi$/5 = 1 - 2${\sin}^{2}$ $\theta$, gives the result.

Feb 13, 2016

Cos (pi/5) = (sqrt (5)+1)/4.

#### Explanation:

Let $a = \cos \left(\frac{\pi}{5}\right)$, $b = \cos \left(2 \cdot \frac{\pi}{5}\right)$. Thus $\cos \left(4 \cdot \frac{\pi}{5}\right) = - a$. From the double angle formulas:

$b = 2 {a}^{2} - 1$
$- a = 2 {b}^{2} - 1$

Subtracting,

a+b = 2(a^2-b^2) = 2(a+b)(a-b)

$a + b$ is not zero, as both terms are positive, so $a - b$ must be $\frac{1}{2}$. Then

$a - \frac{1}{2} = 2 {a}^{2} - 1$
$4 {a}^{2} - 2 a - 1 = 0$

and the only positive root is

$a = \cos \left(\frac{\pi}{5}\right) = \frac{\sqrt{5} + 1}{4}$.

And b = cos (2*pi/5) = a-1/2 = (sqrt(5)-1)/4.