$\sin \pi/6 =-1$? From sum of angles formula. I was solving this problem from Apostol's calculus book and encountered the problem of getting the value of $\sin(\pi/6)$ with the aid of the equality $\sin3x=3\sin x-4\sin^3x $ (from sum of angles). 
I replaced $x$ in the equation with $\pi/6$ then proceeded by replacing $\sin(\pi/6)$ with some variable and solve for the solution of the resulting cubic equation.  I got $-1$ or $0.5$. For $0.5$ I can confirm from calculator but $-1$ is a baffling result.  What concept might have I forgotten?  
 A: Once you have established that $\sin\frac{\pi}{6}$ is one solution of $$4X^3-3X+1=0$$ as you did, it is not sufficient to solve the equation: Indeed, this equation has 2 solutions, and $\sin\frac{\pi}{6}$ is only one of the two.
This amounts to a different between "neceassry" and "sufficient." $\sin\frac{\pi}{6}$ is necessarily a solution of the equation, and therefore $\sin\frac{\pi}{6}\in\{\frac{1}{2},-1\}$; but it is not sufficient to be a solution of the equation to be equal to $\sin\frac{\pi}{6}$.
A: You equation is
$$
\sin(3x) = 3\sin(x) - 4\sin^3(x).
$$
With $x = \pi/6$, this gives
$$
\sin(\pi/2) = 3\sin(\pi/6) -4 \sin^3(\pi/6)
$$
With $\sin(\pi/6) = y$, this gives
$$
1 = \sin(\pi/2) = 3y - 4y^3.
$$
This is the equation you want to solve. The solutions are $1/2$ and $-1$. All this means is that $\sin(\pi/6)$ is $1/2$ or $-1$. Using "other informaiton" you can pick the right value.
A: I don't know what you mean. $\sin(\pi/6) = 1/2$ (which someone reading Apostol should know how to prove!) and thus
$$\sin(\pi/2) = 3\sin(\pi/6) - 4\sin^3(\pi/6) = {3 \over 2} - {4 \over 8} = 1$$
A: $-\pi/2$ is also a solution to $\sin3\theta=1$ because $\sin-3\pi/2=\sin(-3\pi/2+2\pi)=\sin\pi/2$
A: $$\sin3x=1\implies3x=2n\pi+\dfrac\pi2=\dfrac\pi2(4n+1)$$ where $n$ is any integer
$$\implies x=\dfrac\pi6(4n+1)$$ where $n\equiv-1,0,1\pmod3$
Now $\sin\dfrac\pi6[4(-1)+1]=\sin\left(-\dfrac\pi2\right)=-\sin\dfrac\pi2=-1$
$\sin\dfrac\pi6[4\cdot0+1]=\sin\dfrac\pi6=?$
$\sin\dfrac\pi6[4\cdot1+1]=\sin\left(\dfrac{5\pi}6\right)=\sin\left(\pi-\dfrac\pi6\right)=\sin\dfrac\pi6=?$
Finally, $4y^3-3y-1=(y+1)(2y-1)^2$
