<s>[INST] 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 $\cos(π/3)$?
 [/INST] hint: Put $x = \sin \left(\frac{\pi}{3}\right)$, then the equation $0=\sin\left(3\cdot\dfrac{\pi}{3}\right)= 3\sin\left(\frac{\pi}{3}\right) - 4\sin^3(\frac{\pi}{3})= 3x - 4x^3$ gives: $4x^3-3x = 0 \to x(4x^2-3) = 0$. Can you solve for $x$. Note that $x \neq 0$.
</s>