Use DeMoivre's theorem to prove
$\sin 4 \theta=4 \cos ^2 \theta \sin \theta-4 \cos \theta \sin ^2 \theta$

To use DeMoivre's theorem to prove the identity sin 4 θ = 4 cos 2 θ sin θ − 4 cos θ sin 2 θ , we start by understanding and applying DeMoivre's theorem which states:
( cos θ + i sin θ ) n = cos ( n θ ) + i sin ( n θ ) .
For n = 4 , we have:
( cos θ + i sin θ ) 4 = cos ( 4 θ ) + i sin ( 4 θ ) .
Expanding ( cos θ + i sin θ ) 4 using the binomial theorem gives:
cos 4 θ + 4 i cos 3 θ sin θ − 6 cos 2 θ sin 2 θ − 4 i cos θ sin 3 θ + i 4 sin 4 θ .
Recall that i 2 = − 1 , thus i 4 = 1 . The expression simplifies to:
cos 4 θ − 6 cos 2 θ sin 2 θ + sin 4 θ + i ( 4 cos 3 θ sin θ − 4 cos θ sin 3 θ ) .
By equating the imaginary parts on both sides, we have:
sin ( 4 θ ) = 4 cos 3 θ sin θ − 4 cos θ sin 3 θ .
This can be rearranged by factoring out 4 cos θ sin θ :
sin ( 4 θ ) = 4 cos θ sin θ ( cos 2 θ − sin 2 θ ) .
Now replacing cos 2 θ − sin 2 θ with cos ( 2 θ ) , we rewrite it as:
sin ( 4 θ ) = sin ( 2 ⋅ 2 θ ) ,
which aligns with the double angle formula for sine:
sin ( 2 α ) = 2 sin α cos α .
Therefore, the identity is proven through DeMoivre's theorem.

Answered by LiamAlexanderSmith | 2025-07-08