Finding a more efficient solution to a trigonometric identity problem.

Question

THIS IS NOT A HOMEWORK QUESTION

This question comes from a university entrance exam from around 1910. No solutions were ever published. My own solution is given here, but I get the feeling that a more efficient solution would be possible.

Question: Find the exact value of $\cos{\frac{2\pi}{15}}+\cos{\frac{4\pi}{15}}+\cos{\frac{8\pi}{15}}+\cos{\frac{14\pi}{15}}$

My solution: Re-write the functions as $\cos{\frac{5\pi-3\pi}{15}}+\cos{\frac{9\pi-5\pi}{15}}+\cos{\frac{5\pi+3\pi}{15}}+\cos{\frac{9\pi+5\pi}{15}}$

Which simplifies to $\frac{1}{2}\cos{\frac{\pi}{5}}+\frac{\sqrt{3}}{2}\sin{\frac{\pi}{5}}+\frac{1}{2}\cos{\frac{3\pi}{5}}+\frac{\sqrt{3}}{2}\sin{\frac{3\pi}{5}}+\frac{1}{2}\cos{\frac{\pi}{5}}-\frac{\sqrt{3}}{2}\sin{\frac{\pi}{5}}++\frac{1}{2}\cos{\frac{3\pi}{5}}-\frac{\sqrt{3}}{2}\sin{\frac{3\pi}{5}}$

$=\cos{\frac{\pi}{5}}+\cos{\frac{3\pi}{5}}$

$=2\cos{\frac{\pi}{5}}\cos{\frac{2\pi}{5}}$

Now, because I know that $\cos{\frac{\pi}{5}}=\frac{1+\sqrt{5}}{4}$ and $\cos{\frac{2\pi}{5}}=\frac{-1+\sqrt{5}}{4}$ I can simplify this to $\frac{1}{2}$.

However, I wonder:

1 - would students in 1910 have known such exact values?
2 - is there a more efficient solution?

Sum-to-product and product-to-sum identities seem to be commonplace in exams at the time, so I am wondering if there is a clever application of these that I have missed.

Thoughts appreciated as always.

Answer 1

I think that you have done it the way that a student in 1910 might have done it. I'm sure they'd know all about the geometry/trigonometry of the pentagon, so they'd know these values.

I think that nowadays it might be slicker to recall that $\cos\theta=\frac{1}{2}(e^{i\theta}+e^{-i\theta})$, and then observe that the quantity sought is half the sum of the primitive $15$-th roots of unity. Now that sum is, by inclusion/exclusion, the sum of all the $15$-th roots of unity less the sum of all the $5$-th roots of unity less the sum of all the $3$-rd roots of unity plus the sum of all the $1$-st roots of unity: that is $0-0-0+1=1$.

Answer 2

Once you have $2\cos(\pi/5)\cos(2\pi/5)$, continue:

$2\cos(\pi/5)\cos(2\pi/5)=\cos(3\pi/5)+\cos(\pi/5)$

$=-\cos(2\pi/5)+\cos(\pi/5)$

Now consider the sides of a regular pentagon $ABCDE$ as vectors arranged head to tail. Clearly these must have a zero resultant, and then taking the component parallel to side $AB$ gives

$1\underset{\text{sides BC, EA}}{\underbrace{+2\cos(2\pi/5)}}\underset{\text{sides CD, DE}}{\underbrace{-2\cos(\pi/5)}}=0.$

So

$2\cos(\pi/5)\cos(2\pi/5)=-\cos(2\pi/5)+\cos(\pi/5)=1/2,$

which is of course the answer. We did not need exact values of the pentagonal or decagonal angle cosines.

Students in 1910, and well before then, actually could have known the exact cosine values anyway. They may be derived from constructions involving the golden ratio, such as bisecting a base angle of an isosceles triangle when the base angles are each twice the apex angle, which have been known since the ancient Greeks. For that matter, the last equation above contains all the information needed to extract both cosine values; I will let the reader do so.

Answer 3

From your Question one can see that only coprime numbers to $15$ appear in the numerator $(2,4,8,14)$

Tempting to answer 2- is there a more efficient solution?

My thoughts suggest to look more towards number theory where for two primes $p,q$ we have for the sum:

\begin{align} \sum_{i \mid \gcd(p q , i)=1}^{\lfloor \frac{p q}{2} \rfloor}{\cos{\frac{2 i \pi }{p q}}} = \frac{1}{2} \;\; \mid p>q>2 \end{align}

For your example we set $p=5, q=3$

Though I don't know where you could look for a proof.

Answer 4

For the second question, apply the formula $$\cos\alpha+\cos\beta=2\cos \frac {\alpha+\beta}{2}\cos \frac {\alpha-\beta}{2}$$ to the first and third term and to the second and the forth term:

$$\cos{\frac{2\pi}{15}}+\cos{\frac{4\pi}{15}}+\cos{\frac{8\pi}{15}}+\cos{\frac{14\pi}{15}}=2\cos \frac {10\pi}{30}\cos \frac {6\pi}{30}+2\cos \frac {18\pi}{30}\cos \frac {10\pi}{30}$$ and then you get $$2\cos \frac {\pi}{3}\left(\cos \frac {\pi}{5}+\cos \frac {3\pi}{5}\right)=\cos \frac {\pi}{5}+\cos \frac {3\pi}{5}=\cos \frac {\pi}{5}-\cos \frac {2\pi}{5}.$$ Now, to evaluate $\cos \frac {\pi}{5}-\cos \frac {2\pi}{5}$, let $x$ be $\cos \frac {\pi}{5}$ and $y$ be $\cos \frac {2\pi}{5}$. From $\cos 2\alpha=2\cos^2 \alpha-1$ you get $y=2x^2-1$; observing that $\cos \frac{4\pi}{5}=-x$, from the same formula you get $x=1-2y^2$. Now $$x+y=2x^2-2y^2=2(x+y)(x-y).$$ As $x+y\neq 0$, then $1=2(x-y)$ and $$\cos \frac {\pi}{5}-\cos \frac {2\pi}{5}=\frac{1}{2}.$$