Using De-Moivre’s Theory
Find Cos5θ and Sin5θ
I am finding it difficult to solve.
By Moivre:
$\displaystyle (\cos\theta +i\sin\theta)^5=\cos 5\theta +i\sin 5\theta$
By Newton:
$\displaystyle (\cos\theta+i\sin\theta)^5=$
$\displaystyle =\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta +(5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta)i$
Then,
$\displaystyle \cos 5\theta=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta$
$\displaystyle \sin 5\theta=5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta$
Hints:
$\displaystyle 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)
$
$\displaystyle \cos (5\theta ) = {\mathop{\rm Re}\nolimits} \left( {e^{5i\theta } } \right)$
$\displaystyle {\mathop{\rm Re}\nolimits} \left( {\left[ {x + iy} \right]^5 } \right) = x^5 - 10x^3 y^2 + 5xy^4 $
Let $\displaystyle x = \cos (\theta )\,\& \,y = \sin (\theta )$