Submitted by jtedds on Fri, 04/21/2017 - 16:30
I don't know which year the question is from so I can't check my solution, but this is what I've ended up with:
The 6 solutions to the equation are $t = tan(\frac{n\pi}{7})$ where $n \in [2,4,6,8,10,12]$ this is obtained from the binomial expansion of $(cos(\theta) + isin(\theta))^7$ then factorising out $sin(\theta)cos^6(\theta)$
The solution of the equation where $t = tan(\theta)$ are the same as $(cos(\theta) + isin(\theta))^7 = 1$ excluding the solution $\theta = 0$ since we factorised out the $sin(\theta)$ Thus we have the aforementioned solutions.
From $sin(\theta) = -sin(-\theta)$ and $cos(\theta) = cos(-\theta)$ we obtain $tan(\frac{2\pi}{7}) = -tan(\frac{12\pi}{7}$ etc
Therefore $tan(\frac{2\pi}{7})tan(\frac{4\pi}{7})tan(\frac{6\pi}{7}) = \pm\sqrt{7}$
Which from looking at the graphs we can take the postive squareroot.
Testing and forming a similar equation for n = 9 we find our answer to be $\sqrt{9}$ and so we assume for n = 11 it's $\sqrt{11}$
(I tested the entire case for n = 9 as I knew the equation formed would have a positive constant in so could be slightly different to the n = 7 case - for the n = 11 case I just tested the solutions of the equation to see if they were positive or negative)
It's actually STEP II, 1988,
It's actually STEP II, 1988, Q5.
Surely you get $\tan 7\theta =0 \Rightarrow 7\theta = n\pi$ so the roots are $t = \tan (frac{n\pi}{7})$ for $n \in \{1,2,3,4,5,6\}$ and not the evens as you have. That said, I think by periodicity it ends up being the same thing.
But otherwise, you seem to have the hang of things. You haven't mentioned 'product of the roots' which I believe is what you used, but I'm guessing that's just because you haven't typed it all up as you would have written it on paper.
One quick note, you want to use $n\in \{1,2,\ldots\}$ not $n \in [1,2,\ldots]$ - of course, nothing you'd be penalised for but best to get out of bad habits whilst you can (if you did that simply for typographical reasons then \{ \} is how you typeset curly bracekts in LaTeX).
Also: for $n=11$ you know that it must be negative because there are an odd number of 'negative' tans.
I took it to be the solutions
I took it to be the solutions of $(cos(\theta) + isin(\theta))^7 = 1$ so taking $|z| = 1$ all 7 solutions are on the circle and differ by an angle of $\frac{2\pi}{7}$ so that's where I got the solutions from.
What's the difference between the two types of brackets? I wish I could say it was typographical but for some reason, I thought it was the right one to use.
Missing the negative for $n =11$ was just down to thinking $\frac{6}{11} < \frac{1}{2}$
Fair enough, it all works out
Fair enough, it all works out anyway.
Well $n \in [2,5]$ would mean that $n$ could take on any real value in between $2$ and $5$, like $n=2.718\ldots$, whereas $n \in [2,5,7]$ means nothing in particular. $n \in \{1,2,3\}$ means that $n$ can only take on the values $1$, $2$, or $3$.
Ah, I see - yeah that's an easy mistake to make. Anyway, good work all around.
Thanks for the pointers!
Thanks for the pointers!
No worries!
No worries!