Author Topic: 4U Maths Question Thread (Read 665228 times) Tweet Share

arii · « **Reply #1785 on:** March 23, 2018, 07:50:30 pm »

Quote from: RuiAce on March 23, 2018, 06:59:07 pm

$\text{You technically }\textbf{risked}\text{ losing solutions when you jumped to}\\ \boxed{\frac{\cos 2x}{2\sin x} - \sin x = 1}.\\ \text{The reason for this is because you ended up dividing top and bottom by }\cos x.$
$\text{So what you should have checked was that }\cos x = 0\text{ yields no solutions.}\\ \text{Indeed, }\cos x = 0\text{ yields }x = \frac\pi2, \frac{3\pi}2, \dots\\ \text{but they are }\textbf{not}\text{ solutions, since your original equation was }\frac{1}{\cos 3x} = \frac{1}{\sin 2x}\\ \text{and }\frac{1}{\sin 2x}\text{ is undefined at }x = \frac\pi2, \frac{3\pi}2, \dots$
At no point anywhere else in the proof did you attempt to 'cancel things out', so that was the only instance that was risky.

I wasn't entirely sure how to continue from that last line in order to get to the answer they were expecting.

The answers had it as:
$x=\frac{\pi}{10} - \frac{2n\pi}{5}$

but technically, I only have the first two solutions...

RuiAce · « **Reply #1786 on:** March 23, 2018, 08:23:50 pm »

Quote from: arii on March 23, 2018, 07:50:30 pm

I wasn't entirely sure how to continue from that last line in order to get to the answer they were expecting.

The answers had it as:
$x=\frac{\pi}{10} - \frac{2n\pi}{5}$

but technically, I only have the first two solutions...

Hint: Double/Triple angles were overkill. Just use the identity $ \cos \alpha = \sin \left( \frac\pi2 - \alpha \right) $

kaustubh.patel · « **Reply #1787 on:** March 25, 2018, 01:02:15 pm »

Hey guys need some help with confusing trig questions.

RuiAce · « **Reply #1788 on:** March 25, 2018, 01:22:12 pm »

Quote from: kaustubh.patel on March 25, 2018, 01:02:15 pm

Hey guys need some help with confusing trig questions.

The question is to solve for x where 0< x < 2pi

Hints: the second one is a quadratic in tan(x) after expanding and dividing everything by cos^2(x). The last one is just 4sin^2(2x)=1.

I’ll get back to the first one when I’m at home, but if you’re stuck on any of the equations you should post up your progress thus far.

kaustubh.patel · « **Reply #1789 on:** March 25, 2018, 01:46:52 pm »

oh yeah i see how u used double angel formula for sin x.
And also thank you i got the quadratic after making the 5 cos^2x + sin^2x =3sin2x into 4 cos^2x +1= 6 sinx cosx and like u said divide by cos^2x

RuiAce · « **Reply #1790 on:** March 25, 2018, 04:47:31 pm »

(Of course, when I say $n$-gon, I really mean a regular $n$-gon as stated in the question.)

clovvy · « **Reply #1791 on:** March 28, 2018, 03:55:15 pm »

Question 13

jazzycab · « **Reply #1792 on:** March 28, 2018, 05:04:57 pm »

Quote from: clovvy on March 28, 2018, 03:55:15 pm

Question 13

Where is Question 13?

clovvy · « **Reply #1793 on:** March 28, 2018, 05:43:56 pm »

Lol just realised the number was covered, well the extension question where I have to proof by de moivre that sin(2n+1)= some nasty expression

RuiAce · « **Reply #1794 on:** March 28, 2018, 05:51:49 pm »

arii · « **Reply #1795 on:** March 28, 2018, 10:14:02 pm »

I need help with (c). I've put what I've found in (a) and (b) since the question says "Hence".

RuiAce · « **Reply #1796 on:** March 28, 2018, 10:41:28 pm »

Quote from: arii on March 28, 2018, 10:14:02 pm

I need help with (c). I've put what I've found in (a) and (b) since the question says "Hence".

$\text{This is a telescoping sum.}$
$\text{Let }S_n=\sum_{r=1}^n \tan^{-1} \frac{2}{r^2}.\\ \text{Using part a), }S_n = \sum_{r=1}^n \left( \tan^{-1} (r+1) - \tan^{-1}(r-1) \right)$
$\text{When we expand the sum out, we see that several terms cancel.}\\ \begin{align*}S_n &= ( \tan^{-1}2 - \tan^{-1}0 ) + (\tan^{-1}3 - \tan^{-1} 1) + (\tan^{-1}4 - \tan^{-1}2) + (\tan^{-1}5 - \tan^{-1} 3)\\ &\qquad + \dots + (\tan^{-1}(n-1) - \tan^{-1}(n-3)) + (\tan^{-1} n - \tan^{-1}(n-2)) + (\tan^{-1}(n+1) - \tan^{-1}(n-1))\\ S_n &= \tan^{-1}(n+1) + \tan^{-1} n - \tan^{-1} 1 - \tan^{-1}0\end{align*}$
$\text{We know for a fact that }\tan^{-1} 1 = \frac\pi4 \text{ and }\tan^{-1}0 = 0.\\ \text{Also, using the formula in part b),}\\ \begin{align*}\tan^{-1}(n+1) + \tan^{-1} n &= \pi + \tan^{-1} \left( \frac{(n+1) + n}{1 - (n+1)n} \right)\\ &= \pi + \tan^{-1} \left( \frac{2n+1}{1-n-n^2} \right).\end{align*}$
$\text{So putting everything back together, we have}\\ S = \pi + \tan^{-1} \left( \frac{2n+1}{1-n-n^2} \right) - \frac\pi4 - 0\\ \text{which gives the required result.}$

clovvy · « **Reply #1797 on:** March 29, 2018, 11:41:32 am »

Quote from: clovvy on March 28, 2018, 03:55:15 pm

Question 13

The extension question have multiple parts, the sin(2n+1) is part a, part b with the cot^2 is easy, however I don't know how to do part c and d which is the last two questions

RuiAce · « **Reply #1798 on:** March 29, 2018, 03:30:53 pm »

$\text{Part b) tells us that the roots are}\\ \cot^2 \left(\frac\pi{2n+1} \right), \cot^2 \left(\frac{k\pi}{2n+1} \right), \dots, \cot^2 \left(\frac{k \pi}{2n+1} \right)$
$\text{Therefore }\cot^2 \left(\frac\pi{2n+1} \right), \cot^2 \left(\frac{2\pi}{2n+1} \right), \dots, \cot^2 \left(\frac{n\pi}{2n+1} \right)\\ \text{is actually nothing more but the }\textbf{sum of roots}\\ \text{of the polynomial given in part b).}$
$\text{So by quoting the sum of roots formula, the expression equals}\\ \begin{align*}\text{Sum of roots }&=- \frac{-\binom{2n+1}{3}}{\binom{2n+1}{1}}\\ &= \frac{\frac{(2n+1)!}{3!(2n-2)!}}{\frac{(2n+1)!}{1!(2n)!}}\\ &= \frac{\frac{1}{3!(2n-2)!}}{\frac{1}{1!(2n)!}} \\ &= \frac{(2n)!} {6 (2n-2)!}\\ &= \frac{2n(2n-1)(2n-2)!}{6(2n-2)!}\\ &= \frac{n(2n-1)}{3}\end{align*}$
______________________________________________________
$\text{Observe that }\frac{k\pi}{2n+1} < \frac\pi2\text{ for all }k=1,2,\dots, n.\\ \text{Hence, using the given inequality,}\\ \begin{align*}\cot \left(\frac{k\pi}{2n+1}\right) &< \frac{2n+1}{k\pi}\\ \therefore \cot^2\left( \frac{k\pi}{2n+1}\right) &< \frac{(2n+1)^2}{k^2\pi^2}\end{align*}$
$\text{So we can do this to the expression in part b).}\\ \begin{align*}&\qquad\cot^2 \left(\frac{\pi}{2n+1}\right) + \cot^2 \left(\frac{2\pi}{2n+1}\right) + \dots + \cot^2 \left(\frac{k\pi}{2n+1}\right)\\ & < \frac{(2n+1)^2}{1^2\pi^2} + \frac{(2n+1)^2}{2^2\pi^2} + \dots + \frac{(2n+1)^2}{n^2\pi^2}\\ &= \left( \frac{1}{1^2} + \frac{1}{2^2} + \dots + \frac{1}{n^2}\right) \frac{(2n+1)^2}{\pi^2} \end{align*}$
$\text{So combining this with what we know in part c) gives}\\ \begin{align*}\left( \frac{1}{1^2} + \frac{1}{2^2} + \dots + \frac{1}{n^2}\right) \frac{(2n+1)^2}{\pi^2} &> \frac{n(2n-1)}{3}\\ \left( \frac{1}{1^2} + \frac{1}{2^2} + \dots + \frac{1}{n^2}\right) \frac{(2n+1)^2}{2n(2n-1)}&> \frac{\pi^2}{6}\end{align*}$

clovvy · « **Reply #1799 on:** March 30, 2018, 04:55:00 pm »

Question 24 is insane

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