A little help with de Moivre's

**craig** · June 15th 2009, 10:54 AM

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 $i$ ?

Thanks in advance

**running-gag** · June 15th 2009, 11:08 AM

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 $i$ ?

Thanks in advance

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 $1-\sin^2 \theta$

**craig** · June 15th 2009, 11:24 AM

Got it! Expand and equate the imaginary parts.

Thanks for the reply

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