How prove this $\alpha+\beta+\gamma=n\pi$

Question

let $\theta\in R$,and $\alpha\neq\beta\neq\gamma$ and such $$\dfrac{\cos{(\alpha+\theta)}}{\sin^3{\alpha}}=\dfrac{\cos{(\beta+\theta)}}{\sin^3{\beta}}=\dfrac{\cos{(\gamma+\theta)}}{\sin^3{\gamma}}$$

prove $$\alpha+\beta+\gamma=n\pi$$

My try: let $$\dfrac{\cos{(\alpha+\theta)}}{\sin^3{\alpha}}=\dfrac{\cos{(\beta+\theta)}}{\sin^3{\beta}}=\dfrac{\cos{(\gamma+\theta)}}{\sin^3{\gamma}}=k$$ then $$\cos{(\alpha+\theta)}=k\sin^3{\alpha},\quad \cos{(\beta+\theta)}=k\sin^3{\beta},\quad\cos{(\gamma+\theta)}=k\sin^3{\gamma}$$ so $$\cos{(\alpha-\beta)}=\cos{[(\alpha+\theta)-(\beta+\theta)]}=\cos{(\alpha+\theta)}\cos{(\beta+\theta)}+\sin{(\alpha+\theta)}\sin{(\beta+\theta)}$$ and follow maybe can't work.Thank you

Answer 1

Let the ratios are $=k$

$$\implies\cos(\alpha+\theta)=k\sin^3\alpha$$

$$\implies\cos\alpha\cos\theta-\sin\alpha\sin\theta=k\sin^3\alpha$$

Dividing either sides by $\cos\alpha,$

$$\cos\theta-\tan\alpha\sin\theta=k\tan\alpha\sin^2\alpha=\frac{k\tan\alpha}{\csc^2\alpha}=\frac{k\tan\alpha}{1+\cot^2\alpha}$$

$$\implies \cos\theta-\tan\alpha\sin\theta=\frac{k\tan^3\alpha}{1+\tan^2\alpha} $$

Rearrange to form a cubic equation in $\tan\alpha,$

$$\displaystyle(k+\sin\theta)\tan^3\alpha-\cos\theta\tan^2\alpha+\sin\theta \tan\alpha-\cos\theta=0$$

Observe that $\tan\beta,\tan\gamma$ also satisfy the cubic equation

$\displaystyle\implies \tan\alpha,\tan\beta,\tan\gamma$ are the roots of $$\displaystyle(k+\sin\theta)t^3-t^2\cos\theta+t\sin\theta-\cos\theta=0$$

Using Vieta's formulas,

$\displaystyle\tan\alpha+\tan\beta+\tan\gamma=\frac{\cos\theta}{k+\sin\theta}$ and $\displaystyle\tan\alpha\tan\beta\tan\gamma=\frac{\cos\theta}{k+\sin\theta}$

$\displaystyle\implies\sum\tan\alpha=\prod\tan\alpha$

Now we know this

Answer 2

The question as written allows the solution $\beta = \alpha + m\pi$ and $\gamma = \alpha + n\pi$ for non-zero integers $m\neq n$ and any $\alpha \neq k\pi$. In the following, I'll rule-out this solution.

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Write $$a := e^{2i\alpha} \qquad b := e^{2i\beta} \qquad c := e^{2i\gamma} \qquad t := e^{2i\theta}$$ Presumably, the sines of the angles $\alpha$, $\beta$, $\gamma$ are non-zero, so we have $a \neq 1$ (likewise for $b$ and $c$). We'll also assume $a$, $b$, $c$ to be distinct (in order to avoid the solution in the preamble).

Then invoking the formulas $\cos x = \frac{1}{2}\left( e^{ix} + e^{-ix}\right)$ and $\sin x = \frac{1}{2i}\left( e^{ix} - e^{-ix} \right)$, the initial equation becomes (after some cancellation) $$\frac{a \left( a t + 1 \right)}{\left( a - 1 \right)^3} = \frac{b \left( b t + 1 \right)}{\left( b - 1 \right)^3} = \frac{c \left( c t + 1 \right)}{\left( c - 1 \right)^3}$$

The first equality (with $a\neq b$) and second equality (with $b\neq c$) imply, $$\frac{- 1 + 3 a b - a^2 b - a b^2}{a + b - 3 a b + a^2 b^2} = t = \frac{- 1 + 3 b c - b^2 c - b c^2}{b + c - 3 b c + b^2 c^2}$$ Ignoring $t$ (and assuming $c\neq a$), this gives $$1 = a b c = e^{2i(\alpha+\beta+\gamma)}$$ so that $$\alpha + \beta + \gamma = n \pi$$