"You must spread some Reputation around before giving it to InteGrand again."The fastest method of all: "By inspection, Q.E.D."
Thought there was something like that right under my nose.
you didn't think of chasing it? You dont have a HSC to worry about, you can prove it however you like so long as it's within the syllabus.
Oftentimes once one has found a solution, they type it up without first searching (too hard) for other ones, because there's no guarantee that another one will be easily found and they want to get their solution up.
Once it's been proven, it's been proven
true... I'm no different from that sometimes...
Tfw InteGrand is bored af~The fastest method of all: "By inspection, Q.E.D."
Hence why I prefer to take the imaginary part of the complex formulation of problems instead of dealing with conjugation and i's.
If I recall correctly, this result was proved geometrically in one of the Q's in the 2015 BOS HSC 4U trial (the proof was based on the angle bisector theorem (https://en.wikipedia.org/wiki/Angle_bisector_theorem) which the paper got you to prove as an earlier part).
thanks, but what about algebraically? can you please give me a hint or two? I started with by squaring both sides, then writing them as the products complex numbers and conjugates but im not sure if thats on the right track
you need to work with the real and imaginary parts of the complex numbers so make substitutions. Then complete squares. The algebra will work out at the end.
It's essentially just tedious algebra I think. You need to show two things: 1) any point satisfying that relationship between the moduli lies on the given circle, and 2) any point on that circle satisfies the relationship between the moduli. (Luckily the question tells you what the locus is, so you know what to work towards.)
alright, I know I was wrong last time I asked, but just to be sure: cot^-1 k^2/8 or cot^-1 k/8 ? Simplify into one term: $\cot^{-1}{\frac{2}{x+2}} - \cot^{-1}{\frac{2}{x-2}} \\\\ $b) Evaluate the following partial sum: $\sum_{k=1}^n \cot^{-1}{\frac{k^2}{8}} \\\\ $c) Hence, evaluate $\sum_{k=1}^\infty \cot^{-1}{\frac{k^2}{8}})