A curious approximation to $\cos (\alpha/3)$ The following curious approximation
$\cos\left ( \frac{\alpha}{3} \right ) \approx \frac{1}{2}\sqrt{\frac{2\cos\alpha}{\sqrt{\cos\alpha+3}}+3}$
is accurate for an angle $\alpha$ between $0^\circ$ and $120^\circ$
In fact, for $\alpha = 90^\circ$, the result is exact.
How can we derive it?
 A: Let $y = \cos \alpha$ and $x = \cos(\alpha/3)$ then we know that
$$y = x(4x^{2} - 3)\tag{1}$$
Your approximation says that
$$x \approx \frac{1}{2}\sqrt{\frac{2y}{\sqrt{y + 3}} + 3}\tag{2}$$
or using $(1)$ we get $$\frac{y}{x} = 4x^{2} - 3 \approx \frac{2y}{\sqrt{y + 3}}\tag{3}$$ Canceling $y$ we get $$\frac{1}{x}\approx \frac{2}{\sqrt{y + 3}}\tag{4}$$ or $$4x^{2} \approx y + 3$$ or $$y \approx 4x^{2} - 3$$ Comparing with $(1)$ we see that the above approximation is true if $x = 1$ and is good enough if $x$ is near $1$. So the overall approximation is good enough if $\alpha$ is near $0$. The approximation is also correct if $y = 0$ because in this case the LHS and RHS of $(3)$ are equal to $0$.
A: Close but not exact even with regard to the linear term. With $x$ and $y$ as per Paramanand Singh's answer, if $x=1-\delta$, then, ignoring $o(\delta)$ throughout,
$$\begin{align}
y&=(1-\delta)[4(1-2\delta)-3]\\
&=(1-\delta)(4-8\delta-3)\\
&=(1-\delta)(1-8\delta)\\
&=1-9\delta.
\end{align}$$
By Paramanand Singh's equation (2),
$$\begin{align}
x&\approx \frac12\sqrt{\frac{2(1-9\delta)}{\sqrt{1-9\delta+3}}+3}\\
&\approx \frac12\sqrt{\frac{2(1-9\delta)}{\sqrt{4-9\delta}}+3}\\
&\approx \frac12\sqrt{\frac{1-9\delta}{\sqrt{1-9\delta/4}}+3}\\
&\approx \frac12\sqrt{\frac{1-72\delta/8}{1-9\delta/8}+3}\\
&\approx \frac12\sqrt{1-63\delta/8+3}\\
&\approx \frac12\sqrt{4-63\delta/8}\\
&\approx \sqrt{1-63\delta/32}\\
&\approx 1-63\delta/64.
\end{align}$$
