Proving \(\displaystyle{\sin{{4}}}\theta={4}{{\cos}^{{3}}\theta}{\sin{\theta}}-{4}{\cos{\theta}}{{\sin}^{{3}}\theta}\) Using De Moivre's formula: \(\displaystyle{\left({\cos{\theta}}+{i}{\sin{\theta}}\right)}^{{4}}={\cos{{4}}}\theta+{i}{\sin{{4}}}\theta\)

aidinacol7

aidinacol7

Answered question

2022-03-29

Proving sin4θ=4cos3θsinθ4cosθsin3θ
Using De Moivre's formula:
(cosθ+isinθ)4=cos4θ+isin4θ

Answer & Explanation

Regan Gallegos

Regan Gallegos

Beginner2022-03-30Added 9 answers

We can also do this without the de-Moivres formula, only using the double angle identities for sin and cos. Using cos(2θ)=cos2(θ)sin2(θ) and sin(2θ)=2sin(θ)cos(θ), we may calculate
4cos3(θ)sin(θ)4cos(θ)sin3(θ)=4cos(θ)sin(θ)(cos2(θ)sin2(θ))
=2(2sin(θ)cos(θ))cos(2θ)=2sin(2θ)cos(2θ)=sin(4θ)
as desired.

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