# How do you find the exact value of cos54?

Jun 24, 2016

$\cos {54}^{o} = \frac{\sqrt{10 - 2 \sqrt{5}}}{4}$

#### Explanation:

$\cos {54}^{o} = \cos \left({90}^{o} - {36}^{o}\right) = \sin {36}^{o}$

Let $A = {36}^{o}$ hence $5 A = {180}^{o}$

or $3 A = {180}^{o} - 2 A$ and hence

$\sin 3 A = \sin \left({180}^{o} - 2 A\right) = \sin 2 A$ and expanding it we get

$3 \sin A - 4 {\sin}^{3} A = 2 \sin A \cos A$

as $\sin A \ne 0$, dividing by $\sin A$, we get

$3 - 4 {\sin}^{2} A = 2 \cos A$ or

$3 - 4 \left(1 - {\cos}^{2} A\right) - 2 \cos A = 0$ or

$4 {\cos}^{2} A - 2 \cos A - 1 = 0$

and $\cos A = \frac{2 \pm \sqrt{{2}^{2} - 4 \cdot 4 \cdot \left(- 1\right)}}{2 \cdot 4} = \frac{2 \pm \sqrt{20}}{8} = \frac{1 \pm \sqrt{5}}{4}$

As $\cos A$ cannot be negative, $\cos A = \frac{1 + \sqrt{5}}{4}$ and

$\sin A = \sqrt{1 - {\left(1 + \sqrt{5}\right)}^{2} / 16} = \sqrt{\frac{16 - \left(6 + 2 \sqrt{5}\right)}{16}}$

= $\sqrt{\frac{10 - 2 \sqrt{5}}{16}} = \frac{\sqrt{10 - 2 \sqrt{5}}}{4}$

Hence $\cos {54}^{o} = \frac{\sqrt{10 - 2 \sqrt{5}}}{4}$