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Solve for $0\leq\alpha\leq\pi$, $$\left(\frac{\left(r-t\right)\cos\left(\alpha\right)+\left(r-t\right)\sin^{2}\left(\alpha\right)+t}{r}\right)^{2}+\left(\frac{4\left(t-r\right)\sin\left(\frac{\alpha}{2}\right)\cos^3\left(\frac{\alpha}{2}\right)}{t}\right)^{2}=1.$$ I've been playing with this for a while, and apparently the solution can be described simply by \begin{align} \alpha=2\arccos\left(\sqrt{\frac{t}{t+r}}\right) \end{align} and graphing the function with $\alpha$ on the $y$-axis, $t$ on the $x$-axis, and $r$ being an arbitrary constant, it does appear true. I'm not sure if this value of $\alpha$ holds true for all $r$ and $t$, though. Graph of the equation with <span class=$\alpha$ on the $y$-axis and $t$ on the $x$-axis." />

In the above graph, the red curve is $\alpha=2\arccos\left(\sqrt{\frac{t}{t+r}}\right)$ and the purple is the original equation.

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    $\begingroup$ Does $\cos\left(\frac\alpha2\right)^3$ mean $\cos^3\left(\frac\alpha2\right)$ or does it mean $\cos\left(\frac{\alpha^3}8\right)$? $\endgroup$
    – saulspatz
    Commented Sep 2, 2021 at 19:19
  • $\begingroup$ Have you tried the Weierstrass substitution $\beta = \tan\frac\alpha2$? $\endgroup$
    – saulspatz
    Commented Sep 2, 2021 at 19:24
  • $\begingroup$ @saulspatz my bad it's $ \cos^3(\alpha/2)$. I have changed it. $\endgroup$ Commented Sep 2, 2021 at 19:26
  • $\begingroup$ @saulspatz no I haven't tried it yet. $\endgroup$ Commented Sep 2, 2021 at 19:29
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    $\begingroup$ For a start substitute $1-\cos^2 \alpha$ for $\sin^2\alpha$......And the 2nd expression, expanded, includes the term $\sin^2(\alpha /2)\cdot (\cos^2(\alpha/2))^3=$ $((1-\cos \alpha)/2)\cdot ((1+\cos \alpha)/2)^3$.... So you have a 4th degree polynomial equation $P(x)=0$ with $x=\cos \alpha.$ $\endgroup$ Commented Sep 2, 2021 at 20:05

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Let $$x := \left(\frac{\left(r-t\right)\cos\left(\alpha\right)+\left(r-t\right)\sin^{2}\left(\alpha\right)+t}{r}\right)^{2}+\\ \left(\frac{4\left(t-r\right)\sin\left(\frac{\alpha}{2}\right)\cos^3\left(\frac{\alpha}{2}\right)}{t}\right)^{2}-1. \tag{1}$$ Wanted: find the values of $\,\alpha\,$ when $\,x=0.$ There are a few ways to solve this problem. One way with computer technology is to use the substitution $$\alpha=\frac{\log(A)}i,\;\; \sin(\alpha)=\frac{A-\frac1A}{2i},\;\; \cos(\alpha)=\frac{A+\frac1A}2 \tag{2}$$ in equation $(1)$ and factor the expression using a CAS (Computer Algebra System) to get $$ x = (1 - A)^2 (t - r) u^2 v/(4 A^2 r\, t)^2 \tag{3}$$ where $$ u = r + 2 A r + A^2 r - t - A^2 t,\tag{4}$$ $$ v = r + 2 A r + A^2 r + t - 2 A t + A^2 t. \tag{5}$$ The factorization in equation $(3)$ implies that there are three solutions for $\alpha.\,$ The first is $\,\alpha=0\,$ when $\,A=1.\,$ The other two are for $\,u=0\,$ and $\,v=0.\,$

Note that $$ \frac{2r}t- \frac{\cos(\alpha)}{\cos(\alpha/2)^2} = \frac{2u}{t(1+A)^2} \tag{6}$$ and $$ \cos(\alpha/2)^2 - \frac{t}{t+r} = \frac{v}{4(r+t)A}. \tag{7}$$ Thus $\,u=0\,$ when $$ \frac{2r}t = \frac{\cos(\alpha)}{\cos(\alpha/2)^2} \;\;\text{ or }\;\; \frac{t}{2(t-r)} = \cos(\alpha/2)^2 \tag{8}$$ is one solution and $\,v=0\,$ when $$ \frac{t}{t+r} = \cos(\alpha/2)^2 \quad \text{ or } \quad\frac{r}{t} = \tan(\alpha/2)^2\tag{9}$$ is another which agrees with your solution $$ \alpha=2\arccos\left(\sqrt{\frac{t}{t+r}}\right). \tag{10}$$

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  • $\begingroup$ A humble question from the sideline: How did you find your substitution $\frac{log(A)}{i}$ for alpha? Related: Why do we get an algebraic fraction by inserting the log fraction into the sinus term? $\endgroup$ Commented Sep 2, 2021 at 21:38
  • $\begingroup$ @user7427029 As indicated in equation (2), $\sin(\alpha)$ and $\cos(\alpha)$ are rational in $A$ since $e^{i\alpha}=A$. Use Euler's formula $e^{i\alpha} = \cos(\alpha)+i\sin(\alpha)$ to solve for $\sin(\alpha)$ and $\cos(\alpha)$. $\endgroup$
    – Somos
    Commented Sep 2, 2021 at 21:43
  • $\begingroup$ How exactly did you find (6) and (7)? $\endgroup$ Commented Sep 2, 2021 at 22:00
  • $\begingroup$ @bravesheeptribe Good question! I used Mathematica to solve equation (1). It gave me the first expression in equation (8) and the second in equation (9). Then I factored the differences to get equations (6) and (7). $\endgroup$
    – Somos
    Commented Sep 2, 2021 at 22:24
  • $\begingroup$ Do you think there’s a way to tackle this without using technology or no? $\endgroup$ Commented Sep 2, 2021 at 23:26

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