# Mathematical Induction HELP!

• Mar 29th 2010, 10:52 PM
Zel
Mathematical Induction HELP!
Use the principle of mathematical induction to prove the following:
cosθ + cos3θ + cos5θ ... + cos(2n-1)θ = [sin 2nθ]/[2sinθ] for all positive n

sinθ + sin3θ + sin5θ ... + sin(2n-1)θ = [1-cos2nθ]/[2sinθ] for all positive n

1 + cisθ + cis2θ ... + cis nθ = [1-cis(n+1)θ] / [1-cisθ] where cisθ doesn't equal 1, n is greater than or equal to 1

I have solved these for n=1, so i said P(k)=1 is true (n=k), but now I am onto P(k)= k+1, and I am really stuck!
Any help would be greatly appreciated!
• Mar 30th 2010, 01:48 AM
Hello Zel
Quote:

Originally Posted by Zel
Use the principle of mathematical induction to prove the following:
cosθ + cos3θ + cos5θ ... + cos(2n-1)θ = [sin 2nθ]/[2sinθ] for all positive n

sinθ + sin3θ + sin5θ ... + sin(2n-1)θ = [1-cos2nθ]/[2sinθ] for all positive n

1 + cisθ + cis2θ ... + cis nθ = [1-cis(n+1)θ] / [1-cisθ] where cisθ doesn't equal 1, n is greater than or equal to 1

I have solved these for n=1, so i said P(k)=1 is true (n=k), but now I am onto P(k)= k+1, and I am really stuck!
Any help would be greatly appreciated!

You sound as though you have the correct structure for the induction proofs, so here are the trig bits that you need to show that $P(k) \rightarrow P(k+1)$.

1) You'll need to show that:
$\frac{\sin 2k\theta}{2\sin\theta} + \cos(2k+1)\theta = \frac{\sin2(k+1)\theta}{2\sin\theta}$
So:
$\frac{\sin 2k\theta}{2\sin\theta} + \cos(2k+1)\theta = \frac{\sin2k\theta + 2\sin\theta\cos(2k+1)\theta}{2\sin\theta}$
$= \frac{\sin2k\theta + 2\sin\theta(\cos2k\theta\cos\theta - \sin2k\theta\sin\theta)}{2\sin\theta}$

$= \frac{\sin2k\theta(1-2\sin^2\theta) + 2\sin\theta\cos2k\theta\cos\theta}{2\sin\theta}$

$= \frac{\sin2k\theta\cos2\theta + \cos2k\theta\sin2\theta}{2\sin\theta}$

$= \frac{\sin2(k+1)\theta}{2\sin\theta}$

2) is very similar. Start in the same way, combining to a single fraction. Then expand $\sin(2k+1)\theta$ and group the resulting terms in $\cos2k\theta$.

3) is dead easy if you remember that:
$\cos n\theta+i\sin n\theta =(\cos \theta + i \sin \theta)^n$
and hence:
$\cos (n+1)\theta+i\sin (n+1)\theta =(\cos \theta + i \sin \theta)^{n+1}$
$=(\cos \theta + i \sin \theta)^n(\cos\theta + i\sin\theta)$
And you will need to use it in the form:
$\text{cis} (n+2)\theta = \text{cis} (n+1) \theta \text{ cis}\theta$
Can you complete it now?