Stuck on simplifying expressions involving trig and inverse trig functions

Question

TL; DR

Using Mathcad and Wolfram I can see that

$$\sqrt{7}\cos\frac{\tan^{-1}\left(\frac{9\sqrt{3}}{10}\right)}{3}=2.5$$

The decimal value seems to be exact because Mathcad displays it like that with the highest possible accuracy (17 significant digits), and so does Wolfram, as far as I can tell.

How do I simplify that and arrive at the nice value of $2.5$?

The Context

I was solving this cubic equation (Cardano's formula and all)

$$y^3-\frac{7}{3}y+\frac{20}{27}=0$$

Here $p=-\frac{7}{3}$ and $q=\frac{20}{27}$.

This particular equation has the roots $-\frac{5}{3}$, $\frac{1}{3}$ and $\frac{4}{3}$, so I know the result I'm supposed to get but I'm trying to walk through the steps and verify this result myself.

I end up with two complex numbers,

$$z_1=-\frac{10}{27}+\frac{\sqrt{3}}{3}i$$ $$z_2=-\frac{10}{27}-\frac{\sqrt{3}}{3}i$$

whose modulus is $r=|z_1|=|z_2|=\frac{7\sqrt{7}}{27}$ and whose arguments are $\varphi_1=\pi-\tan^{-1}\left(\frac{9\sqrt{3}}{10}\right)$ and $\varphi_2=\tan^{-1}\left(\frac{9\sqrt{3}}{10}\right)-\pi$.

These complex numbers are the result of calculating

$$z_{1,2}=-\frac{q}{2}\pm\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}$$

The next step is to obtain the solutions from

$$y=\sqrt[3]{z_1}+\sqrt[3]{z_2}$$

knowing that we are taking the complex root of third degree here and so $\sqrt[3]{z_1}$ and $\sqrt[3]{z_2}$ will each give a set of three values (say, $\alpha_i$ and $\beta_i$, where $i,j=\{0,1,2\}$). For the values for which the condition $\alpha_i\beta_j=-\frac{p}{3}$ holds, I will calculate sums $\alpha+\beta$ and those will be the solutions to the original equation.

So I need to calculate the complex roots $\alpha_1,\alpha_2,\alpha_3$ and $\beta_1,\beta_2,\beta_3$ and then calculate the three sums $\alpha+\beta$ whose addends satisfy $\alpha\beta=-\frac{p}{3}$.

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The Actual Problem

I'm not going to list all the calculations, just the first one, since the same problem happens with all the others.

Taking the (complex) cube root of $z$:

$$\sqrt[3]{z}=\sqrt[3]{r}\left(\cos\left(\frac{\varphi+2\pi k}{3}\right)+i\sin\left(\frac{\varphi+2\pi k}{3}\right)\right)$$

($k=0,1,2$ but I'm only showing the case $k=0$ here.)

I found out with the help of Mathcad and Wolfram that one of the pairs of $\alpha$, $\beta$ to satisfy $\alpha\beta=-\frac{p}{3}$ is actually $\alpha_0$ and $\beta_0$, the "first" cube root values (with $k=0$) of the numbers $z_1$ and $z_2$ above. So I get

$$\alpha_0=\frac{\sqrt{7}}{3}\left(\cos\frac{\varphi_1}{3}+i\sin\frac{\varphi_1}{3}\right)$$ $$\beta_0=\frac{\sqrt{7}}{3}\left(\cos\frac{\varphi_2}{3}+i\sin\frac{\varphi_2}{3}\right)$$

What do I substitute for $\varphi_1$ and $\varphi_2$? Yes, the values I listed above: $\varphi_1=\pi-\tan^{-1}\left(\frac{9\sqrt{3}}{10}\right)$ and $\varphi_2=\tan^{-1}\left(\frac{9\sqrt{3}}{10}\right)-\pi$. Let me denote $\delta:=\tan^{-1}\left(\frac{9\sqrt{3}}{10}\right)$. So I get

$$\alpha_0+\beta_0=\frac{\sqrt{7}}{3}\left(\cos\left(\frac{\pi}{3}-\frac{\delta}{3}\right)+i\sin\left(\frac{\pi}{3}-\frac{\delta}{3}\right)+\cos\left(\frac{\delta}{3}-\frac{\pi}{3}\right)+i\sin\left(\frac{\delta}{3}-\frac{\pi}{3}\right)\right)$$

Since $\cos(-x)=\cos x$ and $\sin(-x)=-\sin x$, the following are true:

$$\cos(-x)+\cos x=2\cos x$$ $$\sin(-x)+\sin x=0$$

So it's possible to simplify the expression $\alpha_0+\beta_0$:

$$\alpha_0+\beta_0=\frac{\sqrt{7}}{3}\cdot2\cos\left(\frac{\pi}{3}-\frac{\delta}{3}\right)$$

And since $\cos(x\pm y)=\cos x\cos y\mp\sin x\sin y$,

$$\cos\left(\frac{\pi}{3}-\frac{\delta}{3}\right)=\cos\frac{\pi}{3}\cos\frac{\delta}{3}+\sin\frac{\pi}{3}\sin\frac{\delta}{3}=\frac{1}{2}\cos\frac{\delta}{3}+\frac{\sqrt{3}}{2}\sin\frac{\delta}{3}$$

and

$$\alpha_0+\beta_0=\frac{\sqrt{7}}{3}\cos\frac{\delta}{3}+\frac{\sqrt{7}}{3}\cdot\sqrt{3}\sin\frac{\delta}{3}$$

Finally I substitute back $\delta$:

$$\alpha_0+\beta_0=\frac{\sqrt{7}}{3}\cos\frac{\tan^{-1}\left(\frac{9\sqrt{3}}{10}\right)}{3}+\frac{\sqrt{7}}{3}\cdot\sqrt{3}\sin\frac{\tan^{-1}\left(\frac{9\sqrt{3}}{10}\right)}{3}$$

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The Question

$\alpha_0+\beta_0$ is indeed one of the three real solutions to the original equation as both Mathcad and Wolfram agree on the value $\alpha_0+\beta_0=\frac{4}{3}$. I can even chop up this sum into smaller pieces, and see that e. g.

$$\sqrt{7}\cos\frac{\tan^{-1}\left(\frac{9\sqrt{3}}{10}\right)}{3}=2.5$$

Answer 1

Let cos$A=\frac{5}{2\sqrt{7}}$. Then tan$A=\frac{\sqrt{3}}{5}$.

Finally, Tan $3A=\frac{3\text{tan}A-\text {tan}^3A}{1-3\text {tan}^2A}=\frac{9\sqrt{3}}{10}.$

Answer 2

Let$$t=\frac{\arctan\left(\frac{9\sqrt3}{10}\right)}3,$$let $c=\sqrt7\cos(t)$, and let $s=\sqrt7\sin(t)$. Then $c^2+s^2=7$ and $\frac sc=\tan(t)$. On the other hand,\begin{align}\frac{9\sqrt3}{10}&=\tan\left(\arctan\left(\frac{9\sqrt3}{10}\right)\right)\\&=\tan(3t)\\&=\frac{3\tan(t)-\tan^3(t)}{1-3\tan^2(t)},\end{align}and therefore $\tan(t)$ is a solution of the equation$$\frac{3x-x^3}{1-3x^2}=\frac{9\sqrt3}{10}.\tag1$$If $x=\sqrt3y$, the previous equation becomes$$\frac{3y-9y^3}{1-9y^2}=\frac9{10}$$and using the rational roots theorem it is easy to see that the roots of this equation are $-\frac12$, $\frac15$, and $3$. Therefore, the roots of $(1)$ are $-\frac{\sqrt3}2$, $\frac{\sqrt3}5$, and $3\sqrt3$. But $0<\frac{9\sqrt3}{10}<\sqrt3$, and therefore $t\in\left(0,\frac\pi9\right)$. So, of those three numbers, the only one which can be equal to $\tan(t)$ is $\frac{\sqrt3}5$. And if you solve the system$$\left\{\begin{array}{l}c^2+s^2=7\\\frac sc=\frac{\sqrt3}5,\end{array}\right.$$the only solution wih $s,c>0$ is $s=\frac{\sqrt3}2$ and $c=\frac52$. In particular,$$c=\sqrt7\cos\left(\frac{\arctan\left(\frac{9\sqrt3}{10}\right)}3\right)=\frac52.$$

Answer 3

Let $3\theta = \tan^{-1}\frac{9\sqrt3}{10}$. Draw a right-angled triangle with opposite side $9\sqrt 3$ and adjacent side $10$, then the hypothenuse is $\sqrt{\left(9\sqrt3\right)^2 + 10^2} = 7\sqrt 7$. The angle $3\theta$ is equal to

$$\begin{align*} 3\theta = \tan^{-1}\frac{9\sqrt3}{10} &= \cos^{-1} \frac{10}{7\sqrt 7}\\ \cos 3\theta &= \frac{10}{7\sqrt 7} \end{align*}$$

By the triple angle formula of cosine: $\cos 3\theta = 4\cos ^3\theta - 3\cos\theta$,

$$\begin{align*} 4\cos ^3\theta - 3\cos\theta &= \frac{10}{7\sqrt 7}\\ 4\cdot7\sqrt 7\cos^3\theta - 3\cdot 7\sqrt 7 \cos \theta &= 10\\ 4\left(\sqrt 7 \cos \theta\right)^3 - 21\left(\sqrt 7 \cos \theta\right) - 10 &= 0 \end{align*}$$

This becomes another rational cubic equation (of $\sqrt 7 \cos \theta$) with rational roots

$$\sqrt 7 \cos \theta = -2, -\frac12, \frac52$$

But only $\sqrt 7 \cos \theta = \frac 52$ is positive and matches the sign of cosine in the first quadrant. i.e.

$$\begin{align*} \sqrt 7\cos\theta &= \frac{5}{2}\\ \sqrt 7 \cos\left(\frac{\tan^{-1}\frac{9\sqrt3}{10}}{3}\right) &= \frac 52 \end{align*}$$