\tan( \alpha ) + 2\tan(2 \alpha ) + 4 \tan(4 \alpha ) + 8 \cot(8 \alpha ) = \cot( \alpha ) Help plz

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asked Feb 22, 2020 65.6k views

1 Answer

Ahmet Recep Navruz · Answer 1 · 2020-02-25T23:00:04+0000

I think proving a more general form will actually be easier than this specific case - it appears to be true that

$2^(n+1)\cot(2^(n+1)\alpha)+\displaystyle\sum_(i=0)^n2^i\tan(2^i\alpha)=\cot\alpha$

for
$n=0,1,2,3,\ldots$ .

Let's consider a proof by induction. The base case
n=0 gives

$2\cot2\alpha+\tan\alpha=2(\cos2\alpha)/(\sin2\alpha)+(\sin\alpha)/(\cos\alpha)$

$=(\cos^2\alpha-\sin^2\alpha)/(\sin\alpha\cos\alpha)+(\sin\alpha)/(\cos\alpha)$

$=(\cos^2\alpha-\sin^2\alpha+\sin^2\alpha)/(\sin\alpha\cos\alpha)$

$=(\cos\alpha)/(\sin\alpha)=\cot\alpha$

as desired.

Suppose the identity holds for
n=k , so that

$2^(k+1)\cot(2^(k+1)\alpha)+\displaystyle\sum_(i=0)^k2^i\tan(2^i\alpha)=\cot\alpha$

For
n=k+1 , we have

$2^(k+2)\cot(2^(k+2)\alpha)+\displaystyle\sum_(i=0)^(k+1)2^i\tan(2^i\alpha)$

$=2^(k+2)\cot(2^(k+2)\alpha)+2^(k+1)\tan(2^(k+1)\alpha)+\displaystyle\sum_(i=0)^k2^i\tan(2^i\alpha)$

$=2^(k+2)\cot(2^(k+2)\alpha)+2^(k+1)\tan(2^(k+1)\alpha)+(\cot\alpha-2^(k+1)\cot(2^(k+1)\alpha))$

So we ultimately need to show that

$2^(k+2)\cot(2^(k+2)\alpha)+2^(k+1)\tan(2^(k+1)\alpha)-2^(k+1)\cot(2^(k+1)\alpha)=0$

or

$2\cot(2^(k+2)\alpha)+\tan(2^(k+1)\alpha)=\cot(2^(k+1)\alpha)$

If we replace
$\beta=2^(k+1)\alpha$ , we get(!) the base case, which we've shown to be true,

$2\cot2\beta+\tan\beta=\cot\beta$

and thus the identity is proved.