# Math Help - Proof by induction

1. ## Proof by induction

This problem recently came up in an FP3 (old P6) textbook, and none of my 7-strong further maths class have been able to solve it

Prove by induction that:

$
\sum_{r=1}^n\ \tan(r\theta)\tan((r+1)\theta)=\tan((n+1)\theta) \cot(\theta)-n-1
$

Thanks for any help you can give; this problem has really started to irritate us!

2. Originally Posted by rabywebb
This problem recently came up in an FP3 (old P6) textbook, and none of my 7-strong further maths class have been able to solve it

Prove by induction that Σ from r = 1 to n of tan rΘ tan (r + 1)Θ = tan (n + 1)Θ cotΘ - n - 1

Thanks for any help you can give; this problem has really started to irritate us!
This will only be a sketch of the proof, I will leave the demonstration of the
truth of the base case to the reader, as well as the weasle words needed
to make this a proof by induction.

$
\sum_{r=1}^n\ \tan(r\theta)\tan((r+1)\theta)=\tan((n+1)\theta) \cot(\theta)-n-1
$

Let (for convienience only):

$
L_n(\theta)=\sum_{r=1}^n\ \tan(r\theta)\tan((r+1)\theta)
$

and

$
R_n(\theta)=\tan((n+1)\theta) \cot(\theta)-n-1
$

So we now need to prove that:

$
L_n(\theta)=R_n(\theta)
$

Suppose this is true for $n=k$ then we need to prove that it is true
for $n=k+1$.

Now:

$
L_{k+1}(\theta)=L_k(\theta) +\tan((k+1)\theta)\tan((k+2)\theta)$
$=R_k(\theta) +\tan((k+1)\theta)\tan((k+2)\theta)
$

and:

$
R_{k+1}(\theta)=\tan((k+2)\theta) \cot(\theta)-k-2
$

Thus to prove what is to be proven for $k+1$ is equivalent to proving that:

$
\tan((k+2)\theta) \cot(\theta)-k-2=$
$R_k(\theta) +\tan((k+1)\theta)\tan((k+2)\theta)
$
,

or:

$
\tan((k+2)\theta) \cot(\theta)-k-2=$
$\tan((k+1)\theta) \cot(\theta)-k-1 +\tan((k+1)\theta)\tan((k+2)\theta)
$
.

Which may be simplified to:

$
\tan((k+2)\theta) \cot(\theta)-1=$
$\tan((k+1)\theta) \cot(\theta) +\tan((k+1)\theta)\tan((k+2)\theta)\ \ \ \ \dots\ (1)
$
.

To prove this (there must be many other ways) use the trig-identity for
the $\tan$ of a sum to obtain:

$
\tan((k+2) \theta)=\frac{\tan((k+1)\theta)+\tan(\theta)}{1-\tan((k+1)\theta)\tan(\theta)}
\ \ \ \ \ \dots\ (2)
$
.

Now substitute the RHS of $(2)$ for $\tan((k+2)\theta)$ on both the RHS and LHS of $(1)$,
on simplification these are both:

$
\frac{\tan((k+1)\theta)[\cot(\theta)+\tan(\theta)]}{1+\tan((k+1)\theta)\tan(\theta)}
$
.

Which proves:

$
L_{k+1}(\theta)=R_{k+1}(\theta)
$

from the assumption:

$
L_k(\theta)=R_k(\theta)
$
.

RonL

3. Thank you very much!

4. Originally Posted by rabywebb
Thank you very much!