65.6k views
1 vote

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

Help plz​

User Daniel W
by
7.9k points

1 Answer

4 votes

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.

User Ahmet Recep Navruz
by
7.9k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories