# Trigonometry/The sine of 18 degrees

The trigonometric functions of angles that are multiples of 3º can all be expressed as surds. This page will only deal with angles that are multiples of 18º.

Consider the equation

${\displaystyle \sin(5x)-1=0}$ .

Clearly, the only solutions are that 5x = 90º, 90º + 360º, 90º + 2x360º, 90º + 3x360º ... Thus x = 18º, 90º, 162º, 234º or 306º. (There is no need to go further, since increasing x by 360º does not affect its sine.) ${\displaystyle \sin(x)}$ is, after using various simplifications, sin(18º) twice, 1, -sin(54º) twice.

This equation can be solved in another way. Using the formula for ${\displaystyle \sin(a+b)}$ , we can expand ${\displaystyle \sin(5x)}$ to get

${\displaystyle 16\sin ^{5}(x)-20\sin ^{3}(x)+5\sin(x)-1=0}$

If we write ${\displaystyle s=\sin(x)}$ , this becomes a quintic equation in s,

${\displaystyle 16s^{5}-20s^{3}+5s-1=0}$

i.e.

${\displaystyle (s-1)(4s^{2}+2s-1)^{2}=0}$

so either ${\displaystyle s=1}$ (corresponding to x = 90º) or

${\displaystyle s={\frac {{\sqrt {5}}-1}{4}}}$

(corresponding to x = 18º or 54º). Thus

${\displaystyle \sin(18^{\circ })={\frac {{\sqrt {5}}-1}{4}}\approx 0.30902}$ .
${\displaystyle \sin(54^{\circ })={\frac {{\sqrt {5}}+1}{4}}\approx 0.80902}$.

 Exercise: More angles By considering the equation ${\displaystyle \sin(5x)=0}$ show that ${\displaystyle \sin(36^{\circ })={\frac {\sqrt {10-2{\sqrt {5}}}}{4}}\approx 0.58779}$ . ${\displaystyle \sin(72^{\circ })={\frac {\sqrt {10+2{\sqrt {5}}}}{4}}\approx 0.95106}$ .

The cosines are found immediately, e.g. cos(18º) = sin(72º).