Question

some help om de-moivres theorem guys

Answer 1

what is the problem?

Answer 2

show that Sin3\[\theta\]=3sin\[\theta\]-4sin\[^3\]\[\theta\]

Answer 3

kant tell wats up with this editor just try to write down as it is...

Answer 4

From De Moivre’s theorem (n = 3 ) we have
(cos θ + i sin θ)^3 = cos 3θ + i sin 3θ
However, expanding the left-hand side
(using: (x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3) we have:\[ \cos^3 θ + 3i \cos^2 θ \sin θ − 3\cos θ \sin^2 θ − i \sin^3 θ = \cos 3θ + i \sin 3θ\]
now equate the complex terms on both sides
\[3\cos^2 θ \sin θ − \sin^3 θ = \sin 3θ\]
now use cos^2θ =1-sin^2 θ \[3(1-\sin ^2 \theta)\sin \theta -\sin^3 \theta=\sin 3\theta\]
\[3\sin \theta -3\sin^3 \theta - \sin^3 \theta = \sin3\theta \]
so \[\sin 3\theta=3 \sin \theta -4\sin^3 \theta\]

Answer 5

wow.....should be one of your fans