metalskaashw

2022-03-31

Solving for $\alpha$
$\mathrm{tan}\alpha +2\mathrm{tan}2\alpha +4\mathrm{tan}4\alpha +8\mathrm{tan}8\alpha +16\mathrm{tan}\alpha =\mathrm{cot}\alpha$

anghoelv1lw

As
$\mathrm{cot}A-\mathrm{tan}A=\frac{{\mathrm{cos}}^{2}A-{\mathrm{sin}}^{2}A}{\mathrm{cos}A\mathrm{sin}A}=2\mathrm{cot}2A$
$2\mathrm{tan}2\alpha +4\mathrm{tan}4\alpha +8\mathrm{tan}8\alpha +16\mathrm{tan}\alpha =\mathrm{cot}\alpha -\mathrm{tan}\alpha =2\mathrm{cot}2alph$
$⇒4\mathrm{tan}4\alpha +8\mathrm{tan}8\alpha +16\mathrm{tan}\alpha =2\left(\mathrm{cot}2\alpha -\mathrm{tan}2\alpha \right)=2\cdot 2\mathrm{cot}4\alpha$
$⇒8\mathrm{tan}8\alpha +16\mathrm{tan}\alpha =4\left(2\mathrm{cot}4\alpha -\mathrm{tan}4\alpha \right)=4\cdot \left(2\mathrm{cot}8\alpha \right)$
$⇒16\mathrm{tan}\alpha =8\cdot \left(\mathrm{cot}8\alpha -\mathrm{tan}8\alpha \right)=8\cdot 2\mathrm{cot}16\alpha$
$⇒\mathrm{cot}16\alpha =\mathrm{tan}\alpha =\mathrm{cot}\left(\frac{\pi }{2}-\alpha \right)$
$⇒16\alpha =n\pi +\frac{\pi }{2}-\alpha$
where n is any integer
$⇒\alpha =\frac{\left(2n+1\right)\pi }{2\cdot 17}$

Cason Singleton

Iterating the identity $\mathrm{cot}\left(\alpha \right)-\mathrm{tan}\left(\alpha \right)=2\mathrm{cot}\left(2\alpha \right)$, we get
$\mathrm{cot}\left(\alpha \right)-\mathrm{tan}\left(\alpha \right)-2\mathrm{tan}\left(2\alpha \right)-4\mathrm{tan}\left(4\alpha \right)-8\mathrm{tan}\left(8\alpha \right)=16\mathrm{cot}\left(16\alpha \right)$ (1)
Therefore, using (1), the equation in question becomes
$\mathrm{cot}\left(16\alpha \right)=\mathrm{tan}\left(\alpha \right)$
which, due to the periodicity of $\mathrm{tan}$, is equivalent to
$n\pi +\frac{\pi }{2}-16\alpha =\alpha$
That is,
$\alpha =\frac{\pi }{34}\left(2n+1\right)$

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