# How do you express the value of cos 2^o15' , in mathematical exactitude?

Jan 3, 2017

cos 2^@15' = sqrt(1/2 + 1/2sqrt(1/2 + 1/2sqrt(1/2+1/2sqrt(1/2+1/2(1/4(sqrt(5)+1)))))

#### Explanation:

If you know that $\cos {36}^{\circ} = \frac{1}{4} \left(\sqrt{5} + 1\right)$ then you can use the half angle formula:

$\cos \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1}{2} + \frac{1}{2} \cos \theta}$

We need to apply this $4$ times to divide our known angle by $16 = {2}^{4}$. We can take the positive square root each time as all of the angles we are working with are in Q1.

So:

$\cos {2}^{\circ} 15 ' = \sqrt{\frac{1}{2} + \frac{1}{2} \cos {4}^{\circ} 30 '}$

$\textcolor{w h i t e}{\cos {2}^{\circ} 15 '} = \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2} + \frac{1}{2} \cos {9}^{\circ}}}$

$\textcolor{w h i t e}{\cos {2}^{\circ} 15 '} = \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2} + \frac{1}{2} \cos {18}^{\circ}}}}$

$\textcolor{w h i t e}{\cos {2}^{\circ} 15 '} = \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2} + \frac{1}{2} \cos {36}^{\circ}}}}}$

color(white)(cos 2^@15') = sqrt(1/2 + 1/2sqrt(1/2 + 1/2sqrt(1/2+1/2sqrt(1/2+1/2(1/4(sqrt(5)+1)))))

This can be simplified a little, but has a pleasing form as is, which indicates the derivation.

$\textcolor{w h i t e}{}$
Derivation of $\cos {36}^{\circ}$

Here's a derivation of the formula for $\cos {36}^{\circ}$ used above...

$- 1 = \cos {180}^{\circ} + i \sin {180}^{\circ}$

$\textcolor{w h i t e}{- 1} = {\left(\cos {36}^{\circ} + i \sin {36}^{\circ}\right)}^{5}$

$\textcolor{w h i t e}{- 1} = \left({\cos}^{5} {36}^{\circ} - 10 {\cos}^{3} {36}^{\circ} {\sin}^{2} {36}^{\circ} + 5 \cos {36}^{\circ} {\sin}^{4} {36}^{\circ}\right) + i \left(5 {\cos}^{4} {36}^{\circ} \sin {36}^{\circ} - 10 {\cos}^{2} {36}^{\circ} {\sin}^{3} {36}^{\circ} + {\sin}^{5} {36}^{\circ}\right)$

Equating Real parts:

$- 1 = {\cos}^{5} {36}^{\circ} - 10 {\cos}^{3} {36}^{\circ} {\sin}^{2} {36}^{\circ} + 5 \cos {36}^{\circ} {\sin}^{4} {36}^{\circ}$

$\textcolor{w h i t e}{- 1} = {\cos}^{5} {36}^{\circ} - 10 {\cos}^{3} {36}^{\circ} \left(1 - {\cos}^{2} {36}^{\circ}\right) + 5 \cos {36}^{\circ} {\left(1 - {\cos}^{2} {36}^{\circ}\right)}^{2}$

$\textcolor{w h i t e}{- 1} = 16 {\cos}^{5} {36}^{\circ} - 20 {\cos}^{3} {36}^{\circ} + 5 \cos {36}^{\circ}$

Hence $\cos {36}^{\circ}$ is a zero of:

$16 {x}^{5} - 20 {x}^{3} + 5 x + 1 = \left(x + 1\right) \left(16 {x}^{4} - 16 {x}^{3} - 4 {x}^{2} + 4 x + 1\right)$

$\textcolor{w h i t e}{16 {x}^{5} - 20 {x}^{3} + 5 x + 1} = \left(x + 1\right) {\left(4 {x}^{2} - 2 x - 1\right)}^{2}$

We can discard $\left(x + 1\right)$ and look for zeros of the repeated quadratic factor, finding:

$0 = 4 \left(4 {x}^{2} - 2 x - 1\right)$

$\textcolor{w h i t e}{0} = 16 {x}^{2} - 8 x + 1 - 5$

$\textcolor{w h i t e}{0} = {\left(4 x - 1\right)}^{2} - {\sqrt{5}}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(4 x - 1\right) - \sqrt{5}\right) \left(\left(4 x - 1\right) + \sqrt{5}\right)$

$\textcolor{w h i t e}{0} = \left(4 x - 1 - \sqrt{5}\right) \left(4 x - 1 + \sqrt{5}\right)$

Hence:

$x = \frac{1}{4} \left(1 \pm \sqrt{5}\right)$

Then since ${36}^{\circ}$ is in Q1, $\cos {36}^{\circ} > 0$, so choose the positive root to find:

$\cos {36}^{\circ} = \frac{1}{4} \left(\sqrt{5} + 1\right)$