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 i= -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, sinx + cos2 x = 1 rearranged as cos2 x = 1 - sinx, we can say

cos4 x = (1 - sinx)2= 1 - 2 sinx + 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 

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