# Proof by induction stuck on trigonometric sum :help:

• December 15th 2008, 11:31 AM
Proof by induction stuck on trigonometric sum :help:
How do I convert:

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

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

Into:

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

Code:

Original sum:   Use the method of induction to prove that:     <br /> img.top {vertical-align:15%;}<br /> $\displaystyle\sum_{r=1}^n tanr\theta tan(r+1)\theta=tan(n+1)\theta cot\theta-n-1$
Help will be appreciated big time(Thinking)
• December 15th 2008, 12:09 PM
Opalg
Quote:

Originally Posted by ssadi
How do I convert:

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

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

Into:

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

Code:

Original sum:   Use the method of induction to prove that:     <br /> img.top {vertical-align:15%;}<br /> $\displaystyle\sum_{r=1}^n tanr\theta tan(r+1)\theta=tan(n+1)\theta cot\theta-n-1$

Notice that for the inductive step you need that (k+1) in place of k.

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

Next, $\tan(k+1)\theta\bigl(\cot\theta+\tan(k+2)\theta\bi gr) = \cot\theta\tan(k+1)\theta\bigl(1+\tan\theta\tan(k+ 2)\theta\bigr)$. Substitute the above fractional expression for $\tan(k+1)\theta$ into that, and it will become $\tan(k+2)\theta\cot\theta-1$, which is what you want.
• December 15th 2008, 12:45 PM
Use the formula $\tan(\phi-\psi) = \frac{\tan\phi-\tan\psi}{1+\tan\phi\tan\psi}$, taking $\phi=(k+1)\theta$, $\psi=\theta$. That tells you that $\tan(k+1)\theta = \frac{\tan(k+2)\theta-\tan\theta}{1+\tan\theta\tan(k+2)\theta}$.
Next, $\tan(k+1)\theta\bigl(\cot\theta+\tan(k+2)\theta\bi gr) = \cot\theta\tan(k+1)\theta\bigl(1+\tan\theta\tan(k+ 2)\theta\bigr)$. Substitute the above fractional expression for $\tan(k+1)\theta$ into that, and it will become $\tan(k+2)\theta\cot\theta-1$, which is what you want.