De Moivre's Formula

This week, we learn about De Moivre’s Formula.

You may first choose to read Polar Form if you have not already done so.

How would you use De Moivre's Formula to solve the following?

>

For a positive integer nn, express tan(nθ) \tan (n \theta) in terms of t=tanθ t = \tan \theta.

For those who want a coding challenge, use this to determine tan1 tan \, 1^\circ to 10 decimal places. How does this compare to π180 \frac{\pi}{180} , which would be the naive approximation tanθθ \tan \theta \approx \theta ?

Note by Calvin Lin
6 years ago

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I always wonder if De Moivre's formula came before Euler's formula. I mean it is very easy to derive De Moivre's formula using Euler's formula.

(cosx+isinx)n=(eix)n=einx=cos(nx)+isin(nx){(\cos x + i\sin x)}^n = {(e^{ix})}^n = e^{in x} = \cos(n x) + i \sin(n x)

jatin yadav - 6 years ago

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Actually, Euler's Formula states that cosx+isinx=eix \cos x + i \sin x = e^{ix} for some e e . In order to prove Euler's Formula, we have to prove that cis \text{cis} is an exponential function, which we do through de Moivre's Theorem.

Josh Petrin - 6 years ago

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Euler's formula is actually usually proven using the Taylor expansion, not de Moivre's Theorem. In fact, Euler himself used the Taylor expansion proof in the paper in which he published the formula. Though what you're suggesting would work, proving eix=cosx+isinx e^{ix} = \cos x + i \sin x by Taylor expansion is much simpler and de Moivre's is a natural extension of Euler's formula, as Jatin showed.

Carl Denton - 6 years ago

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@Carl Denton The whole thing sort of depends on what functions you choose to define and what you then try to show. For example, the introduction of Walter Rudin's text, "Real and Complex Analysis" begins by defining ez=k=0zkk!, e^z = \sum_{k=0}^\infty \frac{z^k}{k!}, from which he efficiently progresses through a number of assertions, beginning with the convergence of the series for all zC z \in \mathbb{C} , defining the trigonometric functions sin \sin and cos \cos and the value of π \pi from the series expansion, and then showing that these definitions satisfy the necessary properties. It is one of the most elegant introductory passages I have ever read in any mathematics textbook, and it reveals why Rudin is so renowned for his expository elegance.

hero p. - 6 years ago

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We can write tan(nθ)tan(n \theta)=cos(nθ)sin(nθ)\frac{cos(n \theta)}{sin(n \theta)}. So, we have to calculate the real and imaginary parts of cos(nθ)+isin(nθ)cos(n \theta)+isin(n \theta) separately.

cos(nθ)+isin(nθ)cos(n \theta)+isin(n \theta)=(cosθ+isinθ)n(cos \theta + isin \theta)^n

=cosnθcos^n\theta+(n1)(cosn1θ)(isinθ){n \choose 1}(cos^{n-1} \theta)(isin \theta)+(n2)(cosn2θ)(isin2θ){n \choose 2}(cos^{n-2} \theta)(isin^2 \theta)+..........+(nn1)(cosθ)(isinn1θ){n \choose n-1}(cos \theta)(isin^{n-1} \theta)+isinnθisin^n \theta

We see that the terms which are in even position are real (and equal to cos(nθ)cos(n \theta)) and the terms which are in odd position are imaginary (and equal to sin(nθ)sin(n \theta)). So, when n is even,

cos(nθ)cos(n \theta)=k=1n2(n2k)cosn2kθsin2kθ(1)k\sum_{k=1}^{\frac{n}{2}} {n \choose 2k}cos^{n-2k} \theta sin^{2k} \theta (-1)^k

sin(nθ)sin(n \theta)=k=1n2(n2k+1)cosn2k1θsin2k+1θ(1)k\sum_{k=1}^{\frac{n}{2}} {n \choose 2k+1}cos^{n-2k-1} \theta sin^{2k+1} \theta (-1)^k

tan(nθ)tan(n \theta) can be obtained by cos(nθ)sin(nθ)\frac{cos(n \theta)}{sin(n \theta)}

But the problem I am facing here is how to convert the terms into tanθtan \theta?

Maharnab Mitra - 6 years ago

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Hint: tanθ=sinθcosθ \tan \theta = \frac{ \sin \theta } { \cos \theta } .

Note that you quote the tangent formula wrongly.

Calvin Lin Staff - 6 years ago

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Sorry, this really was a big mistake. Thank you for pointing this out.

So, my new answer is tan(nθ)tan(n \theta)=k=0n2(n2k+1)cosn2k1θsin2k+1θ(1)kk=0n2(n2k)cosn2kθsin2kθ(1)k\frac{\sum_{k=0}^{\frac{n}{2}} {n \choose 2k+1}cos^{n-2k-1} \theta sin^{2k+1} \theta (-1)^k}{\sum_{k=0}^{\frac{n}{2}} {n \choose 2k}cos^{n-2k} \theta sin^{2k} \theta (-1)^k}.

We can write tanθtan \theta=sinθcosθ\frac{sin \theta}{cos \theta}

Thus, my answer reduces to tan(nθ)tan(n \theta)=k=0n2(n2k+1)cosnθtan2k+1θ(1)kk=0n2(n2k)cosnθtan2kθ(1)k\frac{\sum_{k=0}^{\frac{n}{2}} {n \choose 2k+1}cos^{n} \theta tan^{2k+1} \theta (-1)^k}{\sum_{k=0}^{\frac{n}{2}} {n \choose 2k}cos^{n} \theta tan^{2k} \theta (-1)^k}.

So, cosnθcos^n \theta can be taken out and cancelled.

The answer becomes tan(nθ)tan(n \theta)=k=0n2(n2k+1)tan2k+1θ(1)kk=0n2(n2k)tan2kθ(1)k\frac{\sum_{k=0}^{\frac{n}{2}} {n \choose 2k+1} tan^{2k+1} \theta (-1)^k}{\sum_{k=0}^{\frac{n}{2}} {n \choose 2k} tan^{2k} \theta (-1)^k}.

Replace tanθtan \theta with tt.

tan(nθ)tan(n \theta)=k=0n2(n2k+1)t2k+1(1)kk=0n2(n2k)t2k(1)k\frac{\sum_{k=0}^{\frac{n}{2}} {n \choose 2k+1} t^{2k+1} (-1)^k}{\sum_{k=0}^{\frac{n}{2}} {n \choose 2k} t^{2k} (-1)^k}.

This was for n=even. For n=odd, replace then2\frac{n}{2} in the limits by n12\frac{n-1}{2}.

Is this answer right? Can it be simplified any further?

Maharnab Mitra - 6 years ago

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@Maharnab Mitra Yes this is correct as I too derived the same result but used another technique (Tangents of sum) (see it here.). Although I couldn't possibly simplify it further.

Rahul Nahata - 6 years ago

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Isn't cos(nθ)=cosnθ+k=1n2(n2k)cosn2kθsin2kθ(1)kcos(nθ)=\cos^nθ+\sum_{k=1}^{\frac{n}{2}} {n \choose 2k}\cos^{n-2k}θ\sin^{2k}θ(-1)^k ???

Kunal Rmth - 6 years ago

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Yes, you are right. The mistake I had done in the first comment was that k should have started from 0. My second comment rectifies that mistake.

Maharnab Mitra - 6 years ago

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