Use a matrix or equations to find the value of $\sin(\pi/3)$ I was asked to use $\sin(0)=0$, $\sin(\pi/2)=1$, and $\sin(\pi)=0$ to calculate the value of $\sin(\pi/3)$ using  matrices or equations. I honestly have no idea how to solve this.
 A: Let denote $a=\cos(\frac{\pi}{3})$ and $b=\sin(\frac{\pi}{3})$ and note that $a$ and $b$ are positive. So
$$(a+ib)^3=e^{i\pi}=-1.$$
Now, we expand $(a+ib)^3=a^3+3a^2ib-3ab^2-ib^3=-1$, then we take out the real and imaginary part and we find
$$\left\{\begin{array}{llr}
a^3-3ab^2&=&-1\\
3a^2b-b^3&=&0
\end{array}\right.,$$
Hence, we find from the second equation $b^2=3a^2$ and then first equation give $8a^3=1$.
Finally, we conclude that $a=\frac{1}{2}$ and $b=\frac{\sqrt{3}}{2}$.
A: So I looked at the answer and apparently they used curve fitting in order to find an approximation of $sin(\frac \pi3)$, and somehow they got to this:
$$P(x)=-\frac {4x^2}{\pi^2}+\frac {4x}{\pi}$$
So I set a table with the values of $x$ as the angle, and the answers as $sin(x)$, and then got a $2x3$ matrix 
$$\begin{matrix}
         \frac {\pi^2}4&\frac \pi2&1 \\
         \pi^2&\pi&0\\
          \end{matrix}$$
And finally got to to the equation mentioned above, it's not the exact value of $sin(\frac \pi3)$ though, but it's close to it.
Thank you everyone anyway.
