2

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)?

Harish Chandra Rajpoot
  • 37,450
Superman
  • 255
  • You have gotten it wrong. It's $\sin^2 \frac{\theta}{2}=\frac{1-\cos \theta}{2}$ – AryanSonwatikar May 20 '20 at 04:29
  • Oops, sorry, a typo! Just fixed it. – Superman May 20 '20 at 04:30
  • Are you familiar with the formulae of $\sin 3A$ and $\cos 3A$ and their derivations? – AryanSonwatikar May 20 '20 at 04:32
  • @AryanSonwatikar I have tried that. Replacing $A$ by $A/3$ in the formulae for $\sin3A$ and $\cos3A$ gives a complicated mess. – impopularGuy May 20 '20 at 04:37
  • @impopularGuy, it seems like there's no closed form, is it? – Superman May 20 '20 at 04:47
  • @impopularGuy now that I realize, my approach doesn't give a closed form. You can try and solve the cubic generated though. – AryanSonwatikar May 20 '20 at 05:03
  • How about $\cos{x/3}=(e^{ix/3}+e^{-ix/3})/2$ ? Replace $e^{ix}=\cos{x}+i\sin{x}$. – impopularGuy May 20 '20 at 05:21
  • @impopularGuy then it strays into the territory of complex numbers. And taking cube root of a complex number is not that easy. – AryanSonwatikar May 20 '20 at 05:25
  • Have a look at this – impopularGuy May 20 '20 at 05:27
  • The knowledge in my link is super advanced, and it may not necessarily apply to all $\theta$... – Superman May 20 '20 at 05:42
  • 4
    If you don't accept Cardano's formula for the roots of a cubic (coming from triple angle formula) then there is no formula. This is because an approach to solving a cubic is also to use that same triple angle formula, reducing solving a cubic to find $\cos(\theta/3)$. In other words, the two problems are equivalent. Either you accept complex third roots, or accept that there is no formula. Yes, I'm aware that for some cubics you would rather use hyperbolic cosine as opposed to the trig cosine, but that's not really relevant. – Jyrki Lahtonen May 20 '20 at 06:03
  • 3
    Mind you. In the same sense the ability to calculate half-angle trig functions is equivalent to the ability to solve quadratics (or to take square roots). – Jyrki Lahtonen May 20 '20 at 06:07

1 Answers1

1

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, $$\boxed{4\cos^3\frac{\theta}{3}-3\cos \frac{\theta}{3}-\cos\theta=0}$$

Harish Chandra Rajpoot
  • 37,450