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HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Extension 1 => Topic started by: etranger101 on September 21, 2019, 01:12:17 pm

Title: How would I do this 3U Induction trigonometry question
Post by: etranger101 on September 21, 2019, 01:12:17 pm: Please help I don't know how to do this! Thank you
Title: Re: How would I do this 3U Induction trigonometry question
Post by: fun_jirachi on September 21, 2019, 07:22:45 pm: $\text{Firstly, to simplify a few things down the track, we're going to prove that} \\ \frac{\pi}{4} - \tan^{-1} \left(\frac{1}{2n+1}\right) = \tan^{-1} \left(\frac{n}{n+1}\right), \ n \in \mathbb{Z}^+ \\ \text{Also, note that since} \ \left|\frac{\pi}{4} - \tan^{-1} \left(\frac{1}{2n+1}\right)\right| \ \text{and} \ \left|\tan^{-1} \left(\frac{n}{n+1}\right)\right| \leq \frac{\pi}{2}\\ \text{It is sufficient to prove that} \ \tan(\text{LHS}) = \tan(\text{RHS}) \\ \begin{align*} \text{ie.} \ \tan(\text{LHS})&= \tan\left(\tan^{-1} (1) - \tan^{-1} \left(\frac{1}{2n+1}\right)\right) \\&= \frac{1-\frac{1}{2n+1}}{1+\frac{1}{2n+1}} \\&= \frac{2n}{2n+2} \\&= \frac{n}{n+1} \\&= \tan \left(\tan^{-1} \left(\frac{n}{n+1}\right)\right) \\&=\tan(\text{RHS}) \end{align*}$
$\text{Then for n} = 1 \\ \begin{align*} \text{LHS}&= \tan^{-1} \left(\frac{1}{2 \times 1^2}\right) \\&=\frac{\pi}{4} - \tan^{-1} \left(\frac{1}{3}\right) \ \text{(From the above result)} \\&= \frac{\pi}{4}-\tan^{-1} \left(\frac{1}{2(1)+1}\right) \\&=\text{RHS} \end{align*} \\ \text{ie. True for n}=1$
$\text{Now assume true for n} =k, \ k \in \mathbb{Z}^+ \\ \text{ie.} \tan^{-1} \left(\frac{1}{2 \times 1^2}\right) + \tan^{-1} \left(\frac{1}{2 \times 2^2}\right) + ... + \tan^{-1} \left(\frac{1}{2 \times k^2}\right) = \frac{\pi}{4}-\tan^{-1} \left(\frac{1}{2k+1}\right) \\ \text{Proving true for n} =k+1 \\ \text{ie.} \tan^{-1} \left(\frac{1}{2 \times 1^2}\right) + \tan^{-1} \left(\frac{1}{2 \times 2^2}\right) + \\ ... + \tan^{-1} \left(\frac{1}{2 \times k^2}\right) + \tan^{-1} \left(\frac{1}{2 \times (k+1)^2}\right) = \frac{\pi}{4}-\tan^{-1} \left(\frac{1}{2k+3}\right) \\ \text{It is sufficient to prove that} \ \frac{\pi}{4} - \tan^{-1} \left(\frac{1}{2k+1}\right) + \tan^{-1} \left(\frac{1}{2 \times (k+1)^2}\right) = \tan^{-1} \left(\frac{k+1}{k+2}\right) \\ \text{(From the first result, and our assumption)} \\ \text{ie.} \ \tan^{-1}\left(\frac{k}{k+1}\right) + \tan^{-1} \left(\frac{1}{2 \times (k+1)^2}\right) = \tan^{-1} \left(\frac{k+1}{k+2}\right) \\ \text{(From our first result)} \\ \text{Given also from the above lines that each side of the equation clearly has absolute value} \leq \frac{\pi}{2} \\ \text{It is sufficient to prove} \ \tan(\text{LHS}) = \tan(\text{RHS})$
$\begin{align*} \ \text{Then,} \ \tan(\text{LHS}) &=\frac{\frac{k}{k+1}+\frac{1}{2(k+1)^2}}{1-\frac{k}{2(k+1)^3}} \\&= \frac{\frac{2k(k+1)+1}{2(k+1)^2}}{\frac{2(k+1)^3-k}{2(k+1)^3}} \\&=\frac{(k+1)(2k(k+1)+1)}{2(k+1)^3-k} \\&= \frac{(k+1)(2k^2+2k+1)}{(2k^3+6k^2+5k+2)} \\&=\frac{(k+1)(2k^2+2k+1)}{(k+2)(2k^2+2k+1)} \\&= \frac{k+1}{k+2} \\&= \tan(\text{RHS}) \end{align*} \\ \text{Thus, true by induction for all positive integers n}$

Hope this helps :)
Title: Re: How would I do this 3U Induction trigonometry question
Post by: BakerDad12 on November 17, 2019, 09:43:37 am: Hello, we have finished our topic of induction a few weeks ago. However, I have never come across a question of this complexity in 3U maths. Is this normal? Also, I have no idea what some of the notation you used means. Is this recommended to learn or does this answer fall outside of the scope of Extension 1 Maths?

Thanks!
Title: Re: How would I do this 3U Induction trigonometry question
Post by: fun_jirachi on November 17, 2019, 10:20:32 am: Hey there!

Not sure what you mean by 'normal' - if you could explain on this a bit more I might be able to give you more of an accurate answer, but I'll do the best I can :)

This is a complex question, and it does take a lot of foresight and thinking ahead to simplify; I doubt this question would appear as a standalone in an exam (it should appear with previous parts to simplify the question, ie. part a) prove the first result, b) hence prove by induction the second part) (and from what I've seen, it hasn't!). This is after all just a tough textbook question, so if you're looking to advance your skills and whatnot, this is definitely the sort of level of question you'd be looking to be able to solve at some point down the line.

I think I can assume that you're implying that the notation you don't understand is $\in \mathbb{Z}^+$; the first symbol just means 'is a member of', the second means the set of positive integers. Combined, it just means such that n is a member of the set of positive integers. You definitely don't have to know this (you can write out where n is a positive integer every single time), but I choose not to, because it's extra characters that I don't need :) If there's any other notation you don't understand, let us know and we'll gladly help you out :)

Hope this helps :)