Hello Zel 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!

Thanks in advance =]

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 $\displaystyle P(k) \rightarrow P(k+1)$.

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

So:$\displaystyle \frac{\sin 2k\theta}{2\sin\theta} + \cos(2k+1)\theta = \frac{\sin2k\theta + 2\sin\theta\cos(2k+1)\theta}{2\sin\theta}$$\displaystyle = \frac{\sin2k\theta + 2\sin\theta(\cos2k\theta\cos\theta - \sin2k\theta\sin\theta)}{2\sin\theta}$

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

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

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

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

3) is dead easy if you remember that:$\displaystyle \cos n\theta+i\sin n\theta =(\cos \theta + i \sin \theta)^n$

and hence:$\displaystyle \cos (n+1)\theta+i\sin (n+1)\theta =(\cos \theta + i \sin \theta)^{n+1}$$\displaystyle =(\cos \theta + i \sin \theta)^n(\cos\theta + i\sin\theta)$

And you will need to use it in the form:$\displaystyle \text{cis} (n+2)\theta = \text{cis} (n+1) \theta \text{ cis}\theta$

Can you complete it now?

Grandad