Problem 34: Trig sum ($\u2713$ $\u2713$ $\u2713$) 1999 Paper II

Prove that

$$\sum _{k=0}^{n}sink\mathit{\theta}=\frac{cos\frac{1}{2}\mathit{\theta}-cos\left(n+\frac{1}{2}\right)\mathit{\theta}}{2sin\frac{1}{2}\mathit{\theta}}\phantom{\rule{2.77695pt}{0ex}}.$$ | ($\ast $) |

- (i)
- Deduce that, for $n\ge 1\phantom{\rule{0.3em}{0ex}}$,
$$\sum _{k=0}^{n}sin\left(\frac{k\pi}{n}\right)=cot\left(\frac{\pi}{2n}\right)\phantom{\rule{2.77695pt}{0ex}}.$$
- (ii)
- By differentiating $\left(\ast \right)$
with respect to $\mathit{\theta}$,
or otherwise, show that, for $n\ge 1\phantom{\rule{0.3em}{0ex}}$,
$$\sum _{k=0}^{n}k{sin}^{2}\left(\frac{k\pi}{2n}\right)=\frac{{\left(n+1\right)}^{2}}{4}+\frac{1}{4}{cot}^{2}\left(\frac{\pi}{2n}\right)\phantom{\rule{2.77695pt}{0ex}}.$$

Comments

The very ﬁrst part can be done by multiplying both sides of ($\ast $) by $sin\frac{1}{2}\mathit{\theta}$ and using the identity

It can also be done by induction (worth a try even if you do it by the above method) or by considering the imaginary part of $\sum exp\left(ik\mathit{\theta}\right)$ (summing this as a geometric progression), if you know about de Moivre’s theorem.

Before setting pen to paper for part (ii), it pays to think very hard about simpliﬁcations of the right hand side of $\left(\ast \right)$ that might make the differentiation more tractable – perhaps bearing in mind the calculations required for part (i).

In the original question, you were asked to show that, for large $n$,

and

using the approximations, valid for small $\mathit{\theta}$, $sin\mathit{\theta}\approx \mathit{\theta}$ and $cos\mathit{\theta}\approx 1-\frac{1}{2}\phantom{\rule{0.3em}{0ex}}{\mathit{\theta}}^{2}\phantom{\rule{0.3em}{0ex}}$. Of course, these results follow quickly from the exact results; but if you only want approximate results you can save yourself a bit of work by approximating early to avoid doing some of the trigonometric calculations. I thought that the exact results were nicer than the approximate results so I changed the question a bit for this book (though I felt a bit guilty, because approximations are an important part of mathematics).

Solution to problem 34

For the ﬁrst part, we will show that

$$\sum _{k=0}^{n}2sink\mathit{\theta}sin\frac{1}{2}\mathit{\theta}=cos\frac{1}{2}\mathit{\theta}-cos\left(n+\frac{1}{2}\right)\mathit{\theta}\phantom{\rule{0.3em}{0ex}}.$$Starting with the left hand side, we have

which gives the result immediately, since almost all the terms in the sum cancel. (Write our a few terms if you are not certain of this.)

(i) Let $\mathit{\theta}=\pi \u2215n$. Then

as required.

(ii) Differentiating the left hand side of $\left(\ast \right)$ and using a double angle trig. formula gives

Before attempting to differentiate the right hand side of $\left(\ast \right)$ it is a good idea to write it in a form that gets rid of some of the fractions. Omitting for the moment the factor $1\u22152$, we have

and differentiating gives $-\frac{1}{2}\left(1-cosn\mathit{\theta}\right){cosec}^{2}\frac{1}{2}\mathit{\theta}+nsinn\mathit{\theta}cot\frac{1}{2}\mathit{\theta}+ncosn\mathit{\theta}\phantom{\rule{2.77695pt}{0ex}}.$ Now setting $\mathit{\theta}=\pi \u2215n$ gives

$$-{cosec}^{2}\frac{1}{2}\left(\pi \u2215n\right)-n\phantom{\rule{2.77695pt}{0ex}}.$$ | (‡) |

Setting $\mathit{\theta}=\pi \u2215n$ in equation $\left(\u2020\right)$ and comparing with $\left(\u2021\right)$ (remembering that there is a factor of $1\u22152$ missing) gives the required result.

Post-mortem

Now that I have had another go at this question it does not seem terribly interesting. At ﬁrst, I thought I might ditch
it. Then I thought that it perhaps was worthwhile: keeping cool under the pressure of the differentiation for part (ii)
– it should just be a few careful lines – and getting it out is a good conﬁdence booster. Anyway, now that I have
slogged^{25}
through it, I am going to leave it in.

(Next day) I recall now that the reason that I included this question in the ﬁrst place was because in its original form (with approximate answers as given in the comments section above), the result for part (ii) can be obtained very quickly from part (i) by differentiating both sides with respect to $\pi $ and using a double angle formula. Try it!

Unfortunately, you’d have a hard time justifying this approach, and anyway it is not going to work for the exact
result.^{26}

^{25 }

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