How can you solve this trigonometric equation algebraically?

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. $\alpha$ 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.

• Does $\cos\left(\frac\alpha2\right)^3$ mean $\cos^3\left(\frac\alpha2\right)$ or does it mean $\cos\left(\frac{\alpha^3}8\right)$? Commented Sep 2, 2021 at 19:19
• Have you tried the Weierstrass substitution $\beta = \tan\frac\alpha2$? Commented Sep 2, 2021 at 19:24
• @saulspatz my bad it's $\cos^3(\alpha/2)$. I have changed it. Commented Sep 2, 2021 at 19:26
• @saulspatz no I haven't tried it yet. Commented Sep 2, 2021 at 19:29
• 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 $x=\cos \alpha.$ Commented Sep 2, 2021 at 20:05

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 $$\alpha=2\arccos\left(\sqrt{\frac{t}{t+r}}\right). \tag{10}$$
• 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? Commented Sep 2, 2021 at 21:38
• @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 $\cos(\alpha)$. Commented Sep 2, 2021 at 21:43