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.