# Thread: Pls solve this for me now.

1. ## Pls solve this for me now.

Using De-Moivre’s Theory

Find Cos5θ and Sin5θ

I am finding it difficult to solve.

2. By Moivre:

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

By Newton:

$(\cos\theta+i\sin\theta)^5=$
$=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta +(5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta)i$

Then,

$\cos 5\theta=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta$

$\sin 5\theta=5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta$

3. Hints:
$e^{5i\theta } = \left( {e^{i\theta } } \right)^5 = \left( {\cos (\theta ) + i\sin (\theta )} \right)^5 = \left( {\cos (5\theta ) + i\sin (5\theta )} \right)
$

$\cos (5\theta ) = {\mathop{\rm Re}\nolimits} \left( {e^{5i\theta } } \right)$
${\mathop{\rm Re}\nolimits} \left( {\left[ {x + iy} \right]^5 } \right) = x^5 - 10x^3 y^2 + 5xy^4$

Let $x = \cos (\theta )\,\& \,y = \sin (\theta )$

4. Originally Posted by golo102
Using De-Moivre’s Theory

Find Cos5θ and Sin5θ

I am finding it difficult to solve.
Funny. Jhevon gave you a hint in your other thread a while ago.

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-Dan