# Proof by induction stuck on a sum:help:

• December 15th 2008, 04:04 AM
Proof by induction stuck on a sum:help:
Original sum:
Use the mathematical induction to prove:
$\displaystyle\sum_{r=1}^n sin (2r-1)\theta=\frac{sin^2 n\theta}{sin\theta}$

My problem:
How do I convert

$\frac{sin^2 k\theta}{sin\theta}+sin (2(k+1)-1)\theta$

$=\frac{sin^2 k\theta}{sin\theta}+sin (2k+1)\theta$
to

$\frac{sin^2 (k+1)\theta}{sin\theta}$ as required by the question. Help me, I am stuck here since yesterday.(Doh)
• December 15th 2008, 06:34 AM
Opalg
Quote:

My problem:
How do I convert

$\frac{sin^2 k\theta}{sin\theta}+sin (2(k+1)-1)\theta$

$=\frac{sin^2 k\theta}{sin\theta}+sin (2k+1)\theta$
to

$\frac{sin^2 (k+1)\theta}{sin\theta}$

It pays to know your trig formulas! Use these:

$\sin\theta\sin\phi = \tfrac12\bigl(\cos(\theta-\phi) - \cos(\theta+\phi)\bigr),\qquad[1]$

$\cos2\theta = 1-2\sin^2\!\theta.\qquad[2]$

Then $=\frac{\sin^2 \!k\theta}{\sin\theta}+\sin (2k+1)\theta = \frac{\sin^2 \!k\theta+\sin (2k+1)\theta\sin\theta}{\sin\theta}$. The second term in the numerator is

$\tfrac12\bigl(\cos(2k\theta) - \cos(2k+2)\theta\bigr)$ by [1]

$= \sin^2(k+1)\theta - \sin^2\!k\theta$ by [2].
• December 15th 2008, 11:27 AM
Quote:

Originally Posted by Opalg
It pays to know your trig formulas! Use these:

$\sin\theta\sin\phi = \tfrac12\bigl(\cos(\theta-\phi) - \cos(\theta+\phi)\bigr),\qquad[1]$

$\cos2\theta = 1-2\sin^2\!\theta.\qquad[2]$

Then $=\frac{\sin^2 \!k\theta}{\sin\theta}+\sin (2k+1)\theta = \frac{\sin^2 \!k\theta+\sin (2k+1)\theta\sin\theta}{\sin\theta}$. The second term in the numerator is

$\tfrac12\bigl(\cos(2k\theta) - \cos(2k+2)\theta\bigr)$ by [1]

$= \sin^2(k+1)\theta - \sin^2\!k\theta$ by [2].

Where are they when I need them?
Thanks a lot.