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:

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

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.

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

Let (for convienience only):

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

and

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

So we now need to prove that:

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

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

Now:

$\displaystyle L_{k+1}(\theta)=L_k(\theta) +\tan((k+1)\theta)\tan((k+2)\theta)$$\displaystyle =R_k(\theta) +\tan((k+1)\theta)\tan((k+2)\theta) and: \displaystyle R_{k+1}(\theta)=\tan((k+2)\theta) \cot(\theta)-k-2 Thus to prove what is to be proven for \displaystyle k+1 is equivalent to proving that: \displaystyle \tan((k+2)\theta) \cot(\theta)-k-2=$$\displaystyle R_k(\theta) +\tan((k+1)\theta)\tan((k+2)\theta)$,

or:

$\displaystyle \tan((k+2)\theta) \cot(\theta)-k-2=$$\displaystyle \tan((k+1)\theta) \cot(\theta)-k-1 +\tan((k+1)\theta)\tan((k+2)\theta) . Which may be simplified to: \displaystyle \tan((k+2)\theta) \cot(\theta)-1=$$\displaystyle \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 $\displaystyle \tan$ of a sum to obtain:

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

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

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

Which proves:

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

from the assumption:

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

RonL

3. Thank you very much!

4. Originally Posted by rabywebb
Thank you very much!