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# 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?

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

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

Note by Calvin Lin
3 years, 11 months 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.

$${(\cos x + i\sin x)}^n = {(e^{ix})}^n = e^{in x} = \cos(n x) + i \sin(n x)$$ · 3 years, 11 months ago

Actually, Euler's Formula states that $$\cos x + i \sin x = e^{ix}$$ for some $$e$$. In order to prove Euler's Formula, we have to prove that $$\text{cis}$$ is an exponential function, which we do through de Moivre's Theorem. · 3 years, 11 months ago

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 $$e^{ix} = \cos x + i \sin x$$ by Taylor expansion is much simpler and de Moivre's is a natural extension of Euler's formula, as Jatin showed. · 3 years, 11 months ago

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 $e^z = \sum_{k=0}^\infty \frac{z^k}{k!},$ from which he efficiently progresses through a number of assertions, beginning with the convergence of the series for all $$z \in \mathbb{C}$$, defining the trigonometric functions $$\sin$$ and $$\cos$$ and the value of $$\pi$$ 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. · 3 years, 10 months ago

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

$$cos(n \theta)+isin(n \theta)$$=$$(cos \theta + isin \theta)^n$$

=$$cos^n\theta$$+$${n \choose 1}(cos^{n-1} \theta)(isin \theta)$$+$${n \choose 2}(cos^{n-2} \theta)(isin^2 \theta)$$+..........+$${n \choose n-1}(cos \theta)(isin^{n-1} \theta)$$+$$isin^n \theta$$

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

$$cos(n \theta)$$=$$\sum_{k=1}^{\frac{n}{2}} {n \choose 2k}cos^{n-2k} \theta sin^{2k} \theta (-1)^k$$

$$sin(n \theta)$$=$$\sum_{k=1}^{\frac{n}{2}} {n \choose 2k+1}cos^{n-2k-1} \theta sin^{2k+1} \theta (-1)^k$$

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

But the problem I am facing here is how to convert the terms into $$tan \theta$$? · 3 years, 11 months ago

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

Note that you quote the tangent formula wrongly. Staff · 3 years, 11 months ago

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

So, my new answer is $$tan(n \theta)$$=$$\frac{\sum_{k=0}^{\frac{n}{2}} {n \choose 2k+1}cos^{n-2k-1} \theta sin^{2k+1} \theta (-1)^k}{\sum_{k=0}^{\frac{n}{2}} {n \choose 2k}cos^{n-2k} \theta sin^{2k} \theta (-1)^k}$$.

We can write $$tan \theta$$=$$\frac{sin \theta}{cos \theta}$$

Thus, my answer reduces to $$tan(n \theta)$$=$$\frac{\sum_{k=0}^{\frac{n}{2}} {n \choose 2k+1}cos^{n} \theta tan^{2k+1} \theta (-1)^k}{\sum_{k=0}^{\frac{n}{2}} {n \choose 2k}cos^{n} \theta tan^{2k} \theta (-1)^k}$$.

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

The answer becomes $$tan(n \theta)$$=$$\frac{\sum_{k=0}^{\frac{n}{2}} {n \choose 2k+1} tan^{2k+1} \theta (-1)^k}{\sum_{k=0}^{\frac{n}{2}} {n \choose 2k} tan^{2k} \theta (-1)^k}$$.

Replace $$tan \theta$$ with $$t$$.

$$tan(n \theta)$$=$$\frac{\sum_{k=0}^{\frac{n}{2}} {n \choose 2k+1} t^{2k+1} (-1)^k}{\sum_{k=0}^{\frac{n}{2}} {n \choose 2k} t^{2k} (-1)^k}$$.

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

Is this answer right? Can it be simplified any further? · 3 years, 11 months ago

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. · 3 years, 11 months ago

Isn't $$cos(nθ)=\cos^nθ+\sum_{k=1}^{\frac{n}{2}} {n \choose 2k}\cos^{n−2k}θ\sin^{2k}θ(−1)^k$$ ??? · 3 years, 10 months ago