Solving for \(\displaystyle\alpha\) \(\displaystyle{\tan{\alpha}}+{2}{\tan{{2}}}\alpha+{4}{\tan{{4}}}\alpha+{8}{\tan{{8}}}\alpha+{16}{\tan{\alpha}}={\cot{\alpha}}\)

metalskaashw

metalskaashw

Answered question

2022-03-31

Solving for α
tanα+2tan2α+4tan4α+8tan8α+16tanα=cotα

Answer & Explanation

anghoelv1lw

anghoelv1lw

Beginner2022-04-01Added 19 answers

As cos2A=cos2Asin2A, sin2A+2sinAcosA
cotAtanA=cos2Asin2AcosAsinA=2cot2A
2tan2α+4tan4α+8tan8α+16tanα=cotαtanα=2cot2alph
4tan4α+8tan8α+16tanα=2(cot2αtan2α)=22cot4α
8tan8α+16tanα=4(2cot4αtan4α)=4(2cot8α)
16tanα=8(cot8αtan8α)=82cot16α
cot16α=tanα=cot(π2α)
16α=nπ+π2α
where n is any integer
α=(2n+1)π217
Cason Singleton

Cason Singleton

Beginner2022-04-02Added 13 answers

Iterating the identity cot(α)tan(α)=2cot(2α), we get
cot(α)tan(α)2tan(2α)4tan(4α)8tan(8α)=16cot(16α) (1)
Therefore, using (1), the equation in question becomes
cot(16α)=tan(α)
which, due to the periodicity of tan, is equivalent to
nπ+π216α=α
That is,
α=π34(2n+1)

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