<s>[INST] Are there formulas for $\sin\left(\frac{\theta}{3}\right)$ and $\cos\left(\frac{\theta}{3}\right)$? So far, I have seen the half angle formula (i.e.)
$$\sin^2\left(\frac{\theta}{2}\right) = \frac{1-\cos\theta}{2}$$
Is there a way to find $\sin\left(\frac{\theta}{3}\right)$ and $\cos\left(\frac{\theta}{3}\right)$ in terms of $\sin\theta$ and/or $\cos\theta$ (i.e. a closed form solution)?
 [/INST] There is no analytic formula for $\sin\frac{\theta}{3}$ & $\cos\frac{\theta}{3}$ . But the value of $\sin\frac{\theta}{3}$ & $\cos\frac{\theta}{3}$ can be derived in terms of $\sin\theta $ & $\cos\theta$ using triple angle formula (which may also give complex roots) as follows
$$\sin3x=3\sin x-4\sin^3x$$
setting $3x=\theta$ or $x=\frac{\theta}{3}$ in above trig. identity, 
$$\sin\theta=3\sin \frac{\theta}{3}-4\sin^3\frac{\theta}{3}$$$$\boxed{4\sin^3\frac{\theta}{3}-3\sin \frac{\theta}{3}+\sin\theta=0}$$
Similarly,
$$\cos3x=4\cos^3x-3\cos x$$
Setting $3x=\theta$ or $x=\frac{\theta}{3}$ in above trig. identity, 
$$\boxed{4\cos^3\frac{\theta}{3}-3\cos \frac{\theta}{3}-\cos\theta=0}$$
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