this makes more sense, your question is actually asking the product of real parts of all nth roots of 1
Check your solution by substituting some outlandish values of n (such as n=4, 7, 11 etc.)
An issue that sean raised up from the question is that k isn't a random value, but instead k=n-1. And n must always remain even.
yes only good for even number n and k is half of n.
Good question and nice solution!Solution:
^{\frac{n}{2}}cos\frac{\pi}{n}.... )
When making it, it didnt seem that hard ahaha sorry bout thatGood question and nice solution!
I don't think you have a full solution though, as your "k" is only "n/2" (what if n is odd?), instead of it being (n-1)
So I would square that result you have, because
And then you would have to account for double counting and make separate cases for when n is even or odd
Also, these questions are better suited for the Advanced Level thread, if your question is too difficult for Q16 of the HSC, then it is better to put it in the Advanced Thread
The issue was not so much difficulty as you not saying that:
a)Franky edited that from the beginning. It was the first question we made so go easy on us
Neither of you confirmed his guess or corrected my post "well, the first (k+1) of them anyway".
.Here is my solution:There's still a serious problem in the question: say, ifthen the LHS will be zero. So I modify this question into
(1-z_2)...(1-z_{n-1})=n. )
bumpAnyways new question:
I added the area bit in lets see how this goes
