# Using De Moivre's Theorem

• Mar 24th 2011, 12:05 PM
alexgeek
Using De Moivre's Theorem
Hello,
Doing a practice paper and found this question, not sure what to do:
Quote:

Write down de Moivre's Theorem for n=5. hence show that for $\displaystyle \sin \theta \neq 0$

$\displaystyle \frac{\sin 5 \theta}{\sin \theta} = A cos^4 \theta + B \cos^2 \theta + C$

where A, B, C are constants to be determined.
Deduce the limiting value of $\displaystyle \frac{\sin 5 \theta}{\sin \theta}$ as $\displaystyle \theta$ tends to zero.
I've written de Moivre for n=5:

$\displaystyle (r(\cos \theta + i \sin \theta))^5 = r^5(\cos 5\theta + i \sin 5 \theta)$

But don't know where to go from there, at all.

Thanks
• Mar 24th 2011, 07:26 PM
mr fantastic
Quote:

Originally Posted by alexgeek
Hello,
Doing a practice paper and found this question, not sure what to do:

I've written de Moivre for n=5:

$\displaystyle (r(\cos \theta + i \sin \theta))^5 = r^5(\cos 5\theta + i \sin 5 \theta)$

But don't know where to go from there, at all.

Thanks

From De'moivre's theorem it follows that

$\displaystyle (\cos (\theta) + i \sin (\theta))^5 = \cos (5\theta) + i \sin (5 \theta)$.

Now I suggest you expand the left hand side of this identity and equate the imaginary part of it to $\displaystyle \sin (5 \theta)$.
• Mar 28th 2011, 01:32 AM
alexgeek
Finally got it, tried expanding it by hand then realised I could do it by binomial. Cheers