Question

(cos@+isin@)^4 solve by demoviers therem​

Answer 1

Answer:

By De Moivre’s theorem,

(cosθ+isinθ)4=cos4θ+isin4θ.

On the other hand

(a+b)4=a4+4a3b+6a2b2+4ab3+b4

Expand the binomial and equate the real and imaginary parts. Where you find sin2θ, substitute 1−cos2θ.

You have, almost correctly,

cos4θ+isin4θ=cos4θ+4icos3θsinθ+6i2cos2θsin2θ+4i3cosθsin3θ+i4sin4θ

Now i2=−1, i3=−i and i4=1, so, by equating the real and imaginary parts, we get

cos4θ=cos4θ−6cos2θsin2θ+sin4θsin4θ=4cos3θsinθ−4cosθsin3θ

Now it's just a matter of substituting sin2θ=1−cos2θ, so, for example, we have

sin4θ=4sinθ(cos3θ−cosθsin2θ)=4sinθ(cos3θ−cosθ+cos3θ)=8cos3θsinθ−4cosθsinθ.

Do similarly for cos4θ.

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