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.
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