Question #282428

Apply De Moivre’s formula to express cos 4θ and sin 4θ in terms of cos θ and sin θ.


1
Expert's answer
2021-12-27T03:23:46-0500

Let us apply De Moivre’s formula to express cos4θ\cos 4θ and sin4θ\sin 4θ in terms of cosθ\cos θ and sinθ.\sin θ. It follows from De Moivre’s formula that (cosθ+isinθ)4=cos4θ+isin4θ,(\cos\theta+i\sin\theta)^4=\cos4\theta+i\sin4\theta, where i2=1.i^2=-1.

On the other hand, the Binomial Theorem implies that (cosθ+isinθ)4=cos4θ+4cos3θisinθ+6cos2θi2sin2θ+4cosθi3sin3θ+i4sin4θ=cos4θ+4icos3θsinθ6cos2θsin2θ4icosθsin3θ+sin4θ=cos4θ6cos2θsin2θ+sin4θ+i(4cos3θsinθ4cosθsin3θ).(\cos\theta+i\sin\theta)^4 \\=\cos^4\theta+4\cos^3\theta\cdot i\sin\theta+6\cos^2\theta\cdot i^2\sin^2\theta +4\cos\theta\cdot i^3\sin^3\theta+i^4\sin^4\theta \\=\cos^4\theta+4i\cos^3\theta\cdot \sin\theta-6\cos^2\theta\cdot \sin^2\theta -4i\cos\theta\cdot \sin^3\theta+\sin^4\theta \\=\cos^4\theta-6\cos^2\theta\cdot \sin^2\theta +\sin^4\theta+i(4\cos^3\theta\cdot \sin\theta-4\cos\theta\cdot \sin^3\theta).


Taking into account that two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal, we conclude that


cos4θ=cos4θ6cos2θsin2θ+sin4θ,sin4θ=4cos3θsinθ4cosθsin3θ.\cos4\theta=\cos^4\theta-6\cos^2\theta\cdot \sin^2\theta +\sin^4\theta, \\ \sin4\theta=4\cos^3\theta\cdot \sin\theta-4\cos\theta\cdot \sin^3\theta.


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