Question

Express sin^5 x as a linear combination of sin kx and coskx, k belongs to Z. using De Moivre's theorem.

Answer 1

Let's have a look on $\iota$ complex number (Cos(x) + $\iota$*Sin(x)). And calculate the fifth power of this number

(Cos(x)+$\iota$*Sin(x))⁵ = Cos⁵(x) + 5$\iota$*Cos⁴(x)Sin(x) + 10$\iota$²Cos³(x)Sin²(x) + 10$\iota$³*Cos²(x)Sin³(x) + 5$\iota$⁴*Cos(x)Sin⁴(x) + $\iota$⁵Sin⁵(x) [using binomial formula] = (Cos5(x) - 10*Cos³(x)Sin²(x) + 5*Cos(x)Sin⁴(x)) + $\iota$*(5*Cos⁴(x)Sin(x) - 10*Cos²(x)Sin³(x) + Sin⁵(x)).

From the other side, (Cos(x)+$\iota$*Sin(x))⁵ = Cos(5*x) + $\iota$*Sin(5*x) [using Moivre's theorem]

It means that both numbers are equal.

So Sin(5x) = 5*Cos⁴(x)Sin(x) - 10*Cos²(x)Sin³(x) + Sin⁵(x) $\implies$

Sin⁵(x) = Sin(5x) - 5*Cos⁴(x)Sin(x) + 10*Cos²(x)Sin³(x) = Sin(5x) + 10*(1 - Sin²(x))Sin³(x) - 5*(1 - Sin²(x))²Sin(x) = Sin(5x) + 10*Sin³(x) - 10*Sin⁵(x) - 5*Sin(x) + 10*Sin³(x) - 5*Sin⁵(x) $\implies$

16*Sin⁵(x) = Sin(5x) + 20*Sin³(x) - 5*Sin(x)

/// [using the fornula for 4*Sin³(x) = 3*Sin(x) - Sin(3x)]

We get 16*Sin⁵(x) = Sin(5x) + 15*Sin(x) - 5*Sin(3x) - 5*Sin(x) = Sin(5x) - 5*Sin(3x) + 10*Sin(x)

So, the result is Sin⁵(x) = (1/16)*Sin(5x) - (5/16)*Sin(3x) + (5/8)*Sin(x)