Question says if...
$\displaystyle cos 72 = \frac{\sqrt5 - 1}{4}$
then find the exact value of sin 3
How would one go about doing this?
Start with $\displaystyle \displaystyle \begin{align*} \cos{ \left( 72^{\circ} \right) } = \sin{ \left( 90^{\circ} - 72^{\circ} \right) } = \sin{\left(18^{\circ}\right)} \end{align*}$, so $\displaystyle \displaystyle \begin{align*} \sin{\left(18^{\circ}\right)} = \frac{\sqrt{5} - 1}{4} \end{align*}$. We then have
$\displaystyle \displaystyle \begin{align*} \cos{\left(18^{\circ}\right)} &= \sqrt{ 1 - \left[ \sin{ \left( 18^{\circ} \right) } \right]^2} \\ &= \sqrt{1 - \left(\frac{\sqrt{5} - 1}{4}\right)^2} \\ &= \sqrt{ 1 - \frac{5 - 2\sqrt{5} + 1}{16} } \\ &= \sqrt{ 1 - \frac{6 - 2\sqrt{5}}{16} } \\ &= \sqrt{ \frac{16 - \left(6-2\sqrt{5}\right)}{16} } \\ &= \frac{\sqrt{10 + 2\sqrt{5}}}{4} \end{align*}$.
Now note that
$\displaystyle \displaystyle \begin{align*} \sin{\left(3^{\circ}\right)} &= \sin{\left(18^{\circ} - 15^{\circ}\right)} \\ &= \sin{\left(18^{\circ}\right)}\cos{\left(15^{\circ} \right)} - \cos{\left(18^{\circ}\right)}\sin{\left(15^{\circ} \right)} \end{align*}$
To evaluate $\displaystyle \displaystyle \begin{align*} \sin{\left(15^{\circ}\right)} \end{align*}$ and $\displaystyle \displaystyle \begin{align*} \cos{\left(15^{\circ}\right)} \end{align*}$, apply the half angle formula to $\displaystyle \displaystyle \begin{align*} \sin{\left(30^{\circ}\right)} \end{align*}$ and $\displaystyle \displaystyle \begin{align*} \cos{\left(30^{\circ}\right)} \end{align*}$. Good luck
Here it is from latex.codecogs:
[IMG]http://latex.codecogs.com/png.latex?\displaystyle%20\begin{align*}\frac{\sqr t{3%20-%20\sqrt{%205%20}%20+%20\frac{\left%28%203%20sqrt{ 3}%20-%20sqrt{15}%20\right%29}{2}}%20-%20\sqrt{5%20+%20\sqrt{%205%20}%20-%20\frac{\left%28%205%20sqrt{3}%20+%20sqrt{15}%20\ right%29}{2}}}{4}\\%20\end{align*}[/IMG]
Write 72=45+(30-3) and apply the formulas for the cosine and the sine of a sum of two angles. This will give you an equation involving sin(3) and cos(3), to which you apply the Pythagorean theorem to express cos(3) in terms of sin(3); this will result in a second-order equation in sin(3), whose positive solution you seek.