# Thread: A little help with de Moivre's

1. ## A little help with de Moivre's

Hi, could someone just give me a pointer where to start on this question.

Use de Moivre's Theorem to show that $\sin{5\theta} = 16\sin^5{\theta} - 20\sin^3{\theta} + 5\sin{\theta}$.

I know that de Moivre's Theorem states that $(\cos{\theta} + i\sin{\theta})^n = \cos{n\theta} + i\sin{n\theta}$, just not sure where to start seeing as the question contains no $\cos$ or no $i$?

2. Originally Posted by craig
Hi, could someone just give me a pointer where to start on this question.

Use de Moivre's Theorem to show that $\sin{5\theta} = 16\sin^5{\theta} - 20\sin^3{\theta} + 5\sin{\theta}$.

I know that de Moivre's Theorem states that $(\cos{\theta} + i\sin{\theta})^n = \cos{n\theta} + i\sin{n\theta}$, just not sure where to start seeing as the question contains no $\cos$ or no $i$?

Hi

$(\cos{\theta} + i\sin{\theta})^5 = \cos{5\theta} + i\sin{5\theta}$

Expand the left hand side and take the imaginary part.
Then substitute $\cos^2 \theta$ by $1-\sin^2 \theta$

3. Got it! Expand and equate the imaginary parts.