Show, using de Moivre's theorem, that sin 5x = 16 sin^(5) x - 20 sin^(3) x + 5 sin x 

Using the question, first use de Moivre's theorem to say (cos x + i sin x)^5 = cos 5x + isin 5x. Now we can use a binomial expansion on the RHS to say 

cos 5x + isin 5x = cos⁵ x + 5i cos⁴ x sin x +10i² cos³ x sin² x + 10i³ cos² x sin³ x + 5i⁴ cos x sin⁴ x + i⁵sin⁵ x 

                          = cos⁵ x + 5i cos⁴ x sin x -10 cos³ x sin² x - 10i cos² x sin³ x + 5 cos x sin⁴ x + i sin⁵ x                    

since i² = -1 , i³ = -i , i⁴ = (i²)² = 1, 

Next, equating imaginary parts of the above expression we find:

sin 5x = 5 cos⁴ x sin x - 10 cos² x sin³ x + sin⁵ x                               (1)

Using the identity, sin² x + cos² x = 1 rearranged as cos² x = 1 - sin² x, we can say

cos⁴ x = (1 - sin² x)²= 1 - 2 sin² x + sin⁴ x , substituting this into (1) expanding and collecting terms we get the final answer:

sin 5x = 16 sin⁵ x - 20 sin³ x + 5 sin x