Find a formula for $\sin(3a)$ and use to calculate $\sin(π/3)$ and $\cos(π/3)$? Problem:
Find a formula for $\sin(3a)$ in terms of $\sin(a)$ and $\cos(a)$. Use this to calculate $\sin(π/3)$ and $\cos(π/3)$. 

My attempt: 
\begin{align}
\sin(3a) &= \sin(2a + a) = \sin(2a)\cos(a) + \cos(2a)\sin(a) \\
&=  \sin(a + a)\cos(a) + \cos(a + a)\sin(a) \\
&= [\sin(a)\cos(a) + \cos(a)\sin(a)]\cos(a) + [\cos(a)\cos(a) - \sin(a)\sin(a)]\sin(a).
\end{align}
It can then be simplified to 
$$2\sin(a)\cos^2(a) + \sin(a)\cos^2(a) - \sin^3(a) = 3\sin(a)\cos^2(a) - \sin^3(a).$$

My question is this: How am I supposed to use this formula to find $\sin(π/3)$ and $\cos(π/3)$?

$$\sin 3a = \sin (2a+a) = \sin(2a)\cos(a)+ \cos(2a)\sin(a)= 2\sin(a)\cos^2(a)+\sin(a)(1-2\sin^2(a))=2\sin(a)(1-\sin^2(a))+\sin(a)(1-2\sin^2(a)) =3\sin(a)-4\sin^3(a) $$
So...
$$\sin(\pi) = 0 = 3\sin(\dfrac{\pi}{3})-4\sin^3(\dfrac{\pi}{3})\implies \dfrac{3}{4} =\sin^2(\dfrac{\pi}{3})\implies \sin(\dfrac{\pi}{3})=\dfrac{\sqrt{3}}{2}$$
Using $cos^2(a)+\sin^2(a)=1$
$$\cos^2(\dfrac{\pi}{3})+\dfrac{3}{4} = 1\implies \cos^2(\dfrac{\pi}{3}) =\dfrac{1}{4}\implies \cos(\dfrac{\pi}{3}) =\dfrac{1}{2}$$
