Question

4%20%5Calpha%20%29%20%20%20%2B%208%20%5Ccot%288%20%5Calpha%20%29%20%20%3D%20%20%5Ccot%28%20%5Calpha%20%29%20" id="TexFormula1" title=" \tan( \alpha ) + 2\tan(2 \alpha ) + 4 \tan(4 \alpha ) + 8 \cot(8 \alpha ) = \cot( \alpha ) " alt=" \tan( \alpha ) + 2\tan(2 \alpha ) + 4 \tan(4 \alpha ) + 8 \cot(8 \alpha ) = \cot( \alpha ) " align="absmiddle" class="latex-formula">
Help plz​

Answer 1

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

$2\cot2\alpha+\tan\alpha=2\dfrac{\cos2\alpha}{\sin2\alpha}+\dfrac{\sin\alpha}{\cos\alpha}$

$=\dfrac{\cos^2\alpha-\sin^2\alpha}{\sin\alpha\cos\alpha}+\dfrac{\sin\alpha}{\cos\alpha}$

$=\dfrac{\cos^2\alpha-\sin^2\alpha+\sin^2\alpha}{\sin\alpha\cos\alpha}$

$=\dfrac{\cos\alpha}{\sin\alpha}=\cot\alpha$

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.