Any trick for evaluating (\frac{\sqrt3}{2}\cos \theta+\fr

hadejada7x 2021-12-30 Answered
Any trick for evaluating (32cosθ+i2sin(θ))7?
Expressions of the form (acos(θ)+bisin(θ))n come up from time to time in applications of complex analysis, but to my knowledge the De Moivre's formula can only be applied with a=b. Is there some trick to deal with the case of ab, for example when the expression is (32cosθ+i2sinθ)7?
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Answers (3)

nick1337
Answered 2022-01-08 Author has 575 answers

In general, from de Moivre's theorem, we have:
cosx=eix+eix2
and
sinx=eixeix2
Then,
S=(acosx+ibsinx)n=((a+b)eix+(ab)eix)n2n
Hence, from the Binomial Theorem,
S=r=0n((n),(r))(a+b)nr(ab)rei(n2r)x2n
Since ei(n2r)x=cos((n2r)x)+isin((n2r)x) we have:
S=(r=0n((n),(r))(a+b)nr(ab)rcos((n2r)x)2n)+i(r=0n((n),(r))(a+b)nr(ab)rsin((n2r)x)2n)
Thus it is evident what this particular case shall simplify to.

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user_27qwe
Answered 2022-01-08 Author has 230 answers

Long Comment:
For your particular problem there is a small tiny tiny algebraic simplification as
T=sin(π3)cosθicosπ3sinθ=123cosθ12isinθ (1)
or
T=cos(π6)cosθisin(π6)sinθ=123cosθ12isinθ (2)
Rewriting (1) as
S=(sinωcosθicosωsinθ)n (3)
we can get to
S=(12+i2)n(sin(θ+ω)+isin(θω))n (4)
and likewise for (2)
S=(cosθcos(ω2)+isinθsin(ω2))n
we get
S=(12i2)n(cos(θ+ω2)+icos(θω2))n
Using @Buraian 's methods we have the interesting general result for (3)
S=(sinωcosθicosωsinθ)n (5)
=(12+i2)n(1cos2θcos2ω)n2e(intan1(sin(θ+ω),sin(θω)))
where tan1(x,y)=tan1(yx) gives the angle of the point [x,y] and where sin2(θω)+sin2(θ+ω) simplifies to 1cos(2θ)cos(2ω)

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karton
Answered 2022-01-08 Author has 439 answers

How about this:
32cosθ=cos30cosθ=12[cos(30θ)+cos(30+θ)]=12(b+a)
12sinθ=sin(30)sinθ=12[cos(30θ)cos(30+θ)]=12(ba)
We have:
127[(a+b)+i(ba)]7
Call a+b=u=3cosθ ,ba=v=sinθ, then for the complicated part:
[u+iv]7=(u2+v2)72[u+iv(u2+v2)12]7=(u2+v2)72ei7(tan1vu)
Now back substitute and simplify.

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