Can the following expression be represented in terms of hypergeometric function
$$\sqrt{3}\sin(\arcsin(7/25)/3)-\cos(\arcsin(7/25)/3)$$
It looks similar to the one presented on [this site][1]
[1] :http://www.statemaster.com/encyclopedia/Cubic-equation#Chebyshev_radicals
Let $\theta = \sin^{-1}\frac{7}{25}$, the expression at hand equals to
$$\sqrt{3}\sin\frac{\theta}{3} - \cos\frac{\theta}{3} = -2\left(\cos\frac{\theta}{3}\cos\frac{\pi}{3} - \sin\frac{\theta}{3}\sin\frac{\pi}{3}\right) = -2\cos\frac{\theta+\pi}{3} $$ For $x \in [0,\pi]$, we have
$$\begin{align} & {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{1-\cos x}{2}\right) = \cos\frac{x}{3}\\ \implies & {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{1+\cos x}{2}\right) = \cos\frac{\pi-x}{3} = -\cos\frac{x+2\pi}{3} \end{align} $$ Since $\cos\frac{x}{3} + \cos\frac{x+2\pi}{3} + \cos\frac{x+4\pi}{3} = 0$, we have
$$ \begin{align}\cos\frac{x+\pi}{3} & = -\cos\frac{x+4\pi}{3} = \cos\frac{x}{3} + \cos\frac{x+2\pi}{3}\\ & = {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{1-\cos x}{2}\right) - {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{1+\cos x}{2}\right) \end{align} $$ Now $\sin\theta = \frac{7}{25} \implies \cos\theta = \frac{24}{25}$, the expression we have can be rewritten as a difference of two hypergeometric functions.
$$\sqrt{3}\sin\frac{\theta}{3} - \cos\frac{\theta}{3} = -2\left[ {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{1}{50}\right) - {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{49}{50}\right) \right]$$
