Find and proof the exact values to the equations $\sin(\pi/3), \cos(\pi/3), \sin(\pi/6), \cos(\pi/6)$ using $\sin(3x)=3\sin(x)-4\sin^3(x)$

Question

Find and proof the exact values to the equations $\sin(\pi/3), \cos(\pi/3), \sin(\pi/6), \cos(\pi/6)$ using the proven equality $\sin(3x)=3\sin(x)-4\sin^3(x)$.

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I figured out $\sin(\pi/3)$ using this quickly. Observe that $\sin(3\cdot\pi/3)=0$, then \begin{align*} &0 = 3\sin(\pi/3)-4\sin^3(\pi/3)\\ \Leftrightarrow\,\,\,\, &3\sin(\pi/3)=4\sin^3(\pi/3)\\ \Leftrightarrow\,\,\,\, &3=4\sin^2(\pi/3)\\ \Leftrightarrow\,\,\,\, &\sqrt{3/4}=\sqrt3/2=\sin(\pi/3). \end{align*}

However I do have some trouble with the rest. Can somebody help me out?

Answer 1

use that $$\sin\left(\frac{\pi}{3}\right)^2+\cos\left(\frac{\pi}{3}\right)^2=1$$ construct a equalleteral triangle with sidelength $a$, then the higt is given by $$4h=\frac{\sqrt{3}}{2}a$$ and $$\cos\left(\frac{\pi}{6}\right)=\frac{\frac{\sqrt{3}}{2}a}{a}=\frac{\sqrt{3}}{2}$$

Answer 2

If other identities are allowed then \begin{align} \sin\left(\frac π3\right) & = \cos\left(\frac π2 - \frac π3\right) \\ & = \cos\left(\frac π6\right) \\ & = {\sqrt 3\over 2} \\ \end{align} For $\sin\left(\frac π6\right)$we can do $$\sin\left(3*\frac π6\right) =1$$ $$3\sin\left(\frac π6\right) - 4\sin^3\left(\frac π6\right) = 1$$ $$4\sin^3\left(\frac π6\right) - 3\sin\left(\frac π6\right) + 1 = 0$$ $$\left(\sin\frac π6 + 1\right)\left(2\sin\frac π6 - 1\right)\left(2\sin\frac π6 - 1\right) = 0$$

Since, we know that $$\sin\left(\frac π6\right)\neq -1$$ $$\implies\sin\left(\frac π6\right) = \frac 12$$