# Thread: Proof by induction stuck on trigonometric sum :help:

1. ## Proof by induction stuck on trigonometric sum :help:

How do I convert:

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

$\displaystyle =tan(k+1)\theta[cot\theta+tan(k+2)\theta]-k-1$(As far as I could get)

Into:

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

Code:
Original sum:
Use the method of induction to prove that:

$\displaystyle \displaystyle\sum_{r=1}^n tanr\theta tan(r+1)\theta=tan(n+1)\theta cot\theta-n-1$
Help will be appreciated big time

How do I convert:

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

$\displaystyle =tan(k+1)\theta[cot\theta+tan(k+2)\theta]-k-1$(As far as I could get)

Into:

$\displaystyle tan(k+2)\theta cot\theta-{\color{red}(k+1)}-1$

Code:
Original sum:
Use the method of induction to prove that:

$\displaystyle \displaystyle\sum_{r=1}^n tanr\theta tan(r+1)\theta=tan(n+1)\theta cot\theta-n-1$
Notice that for the inductive step you need that (k+1) in place of k.

Use the formula $\displaystyle \tan(\phi-\psi) = \frac{\tan\phi-\tan\psi}{1+\tan\phi\tan\psi}$, taking $\displaystyle \phi=(k+1)\theta$, $\displaystyle \psi=\theta$. That tells you that $\displaystyle \tan(k+1)\theta = \frac{\tan(k+2)\theta-\tan\theta}{1+\tan\theta\tan(k+2)\theta}$.

Next, $\displaystyle \tan(k+1)\theta\bigl(\cot\theta+\tan(k+2)\theta\bi gr) = \cot\theta\tan(k+1)\theta\bigl(1+\tan\theta\tan(k+ 2)\theta\bigr)$. Substitute the above fractional expression for $\displaystyle \tan(k+1)\theta$ into that, and it will become $\displaystyle \tan(k+2)\theta\cot\theta-1$, which is what you want.

3. Originally Posted by Opalg
Notice that for the inductive step you need that (k+1) in place of k.

Use the formula $\displaystyle \tan(\phi-\psi) = \frac{\tan\phi-\tan\psi}{1+\tan\phi\tan\psi}$, taking $\displaystyle \phi=(k+1)\theta$, $\displaystyle \psi=\theta$. That tells you that $\displaystyle \tan(k+1)\theta = \frac{\tan(k+2)\theta-\tan\theta}{1+\tan\theta\tan(k+2)\theta}$.

Next, $\displaystyle \tan(k+1)\theta\bigl(\cot\theta+\tan(k+2)\theta\bi gr) = \cot\theta\tan(k+1)\theta\bigl(1+\tan\theta\tan(k+ 2)\theta\bigr)$. Substitute the above fractional expression for $\displaystyle \tan(k+1)\theta$ into that, and it will become $\displaystyle \tan(k+2)\theta\cot\theta-1$, which is what you want.
This trigonmetric stuff is tough, i may need some time to get the hang of it properly.
Many thanks for your effort, even if it is childplay for you