Use de Moivre's theorem to calculate an expression for sin(5x) in terms of sin(x) only.

de Moivre's theorem gives us that (cos(x) + i sin(x))ⁿ = cos(nx) + i sin(nx), for integers n and real values x.Therefore cos(5x) + i sin(5x) = (cos(x) + i sin(x))⁵ = cos⁵(x) + 5i cos⁴(x)sin(x) - 10 cos³(x) sin²(x) - 10i cos²(x)sin³(x) + 5cos(x)sin⁴(x) + i sin⁵(x).Taking the imaginary part of both sides, we find sin(5x) = 5cos⁴(x)sin(x) - 10cos²(x)sin(x) + sin⁵(x).To remove the cosines, we use the substitution cos²(x) = 1 - sin²(x).This gives us sin(5x) = 5(1 - sin²(x))²sin(x) - 10(1 - sin²(x))sin(x) + sin⁵(x)Expanding the brackets carefully, we obtain sin(5x) = 5sin(x) - 10sin³(x) + 5sin⁵(x) - 10sin³(x) + 10sin⁵(x) + sin⁵(x).Adding all of the same powers together, we have sin(5x) = 5sin(x) - 20sin³(x) + 16sin⁵(x), which is our final answer.