Author Topic: 4U Maths Question Thread (Read 665360 times) Tweet Share

RuiAce · « **Reply #1395 on:** August 26, 2017, 07:59:21 pm »

Quote from: bluecookie on August 26, 2017, 06:54:34 pm

Show that cos4θ =8(cosθ )^4-8(cosθ )^2+1.
a) Solve the equation 8x^4-8x^2+1=0 and deduce the exact values of cospi/8 and cos5pi/8.
b) Solve the equation 16x^4-16x^2+1=0 and deduce the exact v alues of cospi/12 and cos5pi/12.

Moderator action: Posts merged. At times like these, please resort to the modify at the top right corner of a post, to refrain from multi-posting.

$\text{This is fairly standard, albeit the tedious version}\\ \begin{align*}\cos 4\theta + i\sin 4\theta &= (\cos\theta + i\sin\theta)^4\\ &= \cos^4\theta + 4i\sin\theta\cos^3\theta - 6\sin^2\theta\cos^2\theta - 4i\sin^3\theta\cos\theta + \sin^4\theta \end{align*}$
$\text{Equating real parts,}\\ \begin{align*}\cos 4\theta &= \cos^4\theta -6\sin^2\theta\cos^2\theta + \sin^4\theta\\ &= \cos^4\theta - 6(1-\cos^2\theta)\cos^2\theta + (1-2\cos^2\theta+\cos^4\theta)\\ &= 8\cos^4\theta - 8\cos^2\theta + 1\end{align*}$
$\text{In b), let }x=\cos \theta. \text{ Then, }\\ \begin{align*}16x^4 - 16x^2 + 1 &= 0\\ 8x^4 - 8x^2 &= -\frac12\\ 8\cos^4\theta - 8\cos^2\theta + 1 &= \frac12\\ \cos 4\theta &=\frac12\end{align*}$
$\text{This has general solution}\\ \begin{align*}4\theta &= 2k\pi \pm \frac{\pi}{3}\\ \theta &= \frac{\pi}{2}k \pm \frac{\pi}{12}\end{align*}\\ \text{where }k\text{ is an integer.}$
$\text{Without loss of generality, take }k=-1,0,1,2\text{ to obtain}\\ \theta = -\frac{7\pi}{12}, -\frac{\pi}{12}, \frac{5\pi}{12}, \frac{11\pi}{12}\\ \text{and hence the roots are }x=\cos\left(-\frac{7\pi}{12}\right), \dots\\ \text{However, from using relevant trigonometric identities, we find that the roots are,}\\ \text{more simply, }x=\pm \cos \frac{\pi}{12}, \pm \cos \frac{5\pi}{12}$
______________________________________
$\text{Going back to the original equation, from the quadratic formula}\\ \begin{align*}16x^4 - 16x^2 + 1 &=0\\ x^2 &= \frac{16 \pm \sqrt{256-64}}{32}\\ &= \frac{2\pm \sqrt{3}}4\\ \therefore x &= \pm \sqrt{\frac{2\pm \sqrt{3}}4}\end{align*}$
$\text{Naturally, the negatives correspond to negatives and positives correspond to positives}\\ \text{So }\cos \frac{\pi}{12}\text{ and }\cos \frac{5\pi}{12}\text{ are either }\sqrt{\frac{2+ \sqrt{3}}4}\text{ or }\frac{2- \sqrt{3}}4$
$\text{Exploiting the fact that cos is monotonic decreasing over }0< x \le \frac{\pi}{2}\\ \frac{\pi}{12}< \frac{5\pi}{12} \implies \cos \frac{\pi}{12} > \cos \frac{5\pi}{12}\\ \text{so }\cos \frac{\pi}{12} = \sqrt{\frac{2+ \sqrt{3}}4}\text{ and }\cos \frac{5\pi}{12}=\sqrt{\frac{2- \sqrt{3}}4}$
$\text{This could potentially be simplified further by exploiting the fact}\\ \sqrt{2+\sqrt3} = \sqrt{\frac12 (4+2\sqrt3)}= \sqrt{\frac12 (1+2\sqrt3+3)} = \sqrt{\frac12 (1+\sqrt3)^2} = \frac{1}{\sqrt2}(1+\sqrt3)$

Kle123 · « **Reply #1396 on:** August 27, 2017, 12:29:20 pm »

Could i get help with part 3. Thanks!

RuiAce · « **Reply #1397 on:** August 27, 2017, 12:45:57 pm »

Quote from: Kle123 on August 27, 2017, 12:29:20 pm

Could i get help with part 3. Thanks!

In the future, please provide progress on the previous parts. Here, part i would've been useful.
$\text{We know that the roots of }z^7-1\text{ are }z_k=\cos \frac{2k\pi}{7} + i\sin \frac{2k\pi}{7}\\ \text{where }k=0,1,2,3,4,5,6\\ \text{Because }z^7-1 =z^3(z-1)\left(\left(z+\frac1z\right)^3+\left(z+\frac1z\right)^2-2\left(z+\frac1z\right)-1\right)\\ \text{we must compare which is the root of what}$
$\text{Naturally, }z_0 = 1\text{ and is the root of }(z-1)\\ \text{Hence, }z_1,z_2,\dots, z_6\text{ must be the roots of }\left(\left(z+\frac1z\right)^3+\left(z+\frac1z\right)^2-2\left(z+\frac1z\right)-1 \right)$
$\textit{Carefully note that the root of the }z^3\text{ is }z=0\\ \textbf{but }P(z)\neq 0\text{ as we have the dividing by zero problem from }\frac1z$
________________________________________
$\text{Let }x = z+\frac1z\\ \text{Then }\left(z+\frac1z\right)^3+\left(z+\frac1z\right)^2-2\left(z+\frac1z\right)-1=0\text{ becomes }x^3 + x^2 - 2x - 1=0\\ \text{which is precisely the cubic in question}$
$\textbf{What are the roots of this cubic equation?}\\ \text{We know that the roots of }\left(z+\frac1z\right)^3+\left(z+\frac1z\right)^2-2\left(z+\frac1z\right)-1\\\text{ were }z = \text{ cis }\frac{2\pi}{7}, = \text{ cis }\frac{4\pi}{7}, \dots, \text{ cis }\frac{12\pi}{7}\\ \text{As }x=z+\frac1z,\text{ our roots of the cubic will be }\\ x=\text{cis }\frac{2\pi}{7} + \frac{1}{\text{cis }\frac{2\pi}7}, \dots, \text{cis }\frac{12\pi}{7} + \frac{1}{\text{cis }\frac{12\pi}7}$
$\text{From explicit computation, we realise that the roots are }x=2\cos \frac{2\pi}{7}, 2\cos \frac{4\pi}{7}, 2\cos \frac{6\pi}{7}\\ \text{noting a few cases overlap.}\\ \text{The computation is done with the aid of the observation }\frac{1}{\text{cis }\theta} = \overline{\text{cis }\theta} = \text{cis }(-\theta)\\ \text{and also some relevant trigonometric identities involving cos}$

Kle123 · « **Reply #1398 on:** August 27, 2017, 12:54:35 pm »

THANKSSSS Rui. Ill make sure to include the working for previous parts in the post next time!

bluecookie · « **Reply #1399 on:** August 27, 2017, 03:52:50 pm »

Quote from: RuiAce on August 26, 2017, 07:01:07 pm

$\text{Note that }P(a\cos \theta,b\sin \theta)\text{ lies on }x=ae\\ \text{Hence equating coordinates, we must have }a\cos\theta = ae \implies \boxed{\cos \theta = e}$
$\text{Noting }\cos(-\theta)=\cos\theta, \sin(-\theta)=-\sin\theta\\ \text{we find}\\ \begin{align*}PQ &= \sqrt{(a\cos \theta - a \cos \theta)^2 + (b\sin \theta + b\sin \theta)^2}\\ &= \sqrt{4b^2\sin \theta}\\ &= 2 |b\sin\theta|\end{align*}$
$\text{Now, given }\cos \theta = e\\ \begin{align*}\sin^2\theta &= 1-\cos^2\theta\\ &= 1-e^2\\ \implies |\sin\theta |&= \sqrt{1-e^2}\end{align*}$
$\text{So subbing back in, }PQ = 2b \sqrt{1-e^2}\\ \text{but we know that }b^2 = a^2 (1-e^2) \iff \frac{b}{a} = \sqrt{1-e^2}\\ \text{Hence }\boxed{PQ = \frac{2b^2}{a}}$

Thanks

About the post mergings - I was actually considering doing that before I multi-posted (I think I've been told of for that before and know it's something generally not appreciated in this thread) but I ran into a problem where, if I spent time typing all those questions before, by the time I posted them all at once, there may have already been a solution offered to the first, you get what I mean? (Geezus I feel like I'm doing such a terrible job explaining sorry, but it's just that, if I post 1 problem, in the time it takes for me to type another, someone (like you) can read it and give me a solution - effectively saving time) so that's why I multi-posted. What should I do about this problem in the future?

RuiAce · « **Reply #1400 on:** August 27, 2017, 05:28:43 pm »

Quote from: bluecookie on August 27, 2017, 03:52:50 pm

Thanks About the post mergings - I was actually considering doing that before I multi-posted (I think I've been told of for that before and know it's something generally not appreciated in this thread) but I ran into a problem where, if I spent time typing all those questions before, by the time I posted them all at once, there may have already been a solution offered to the first, you get what I mean? (Geezus I feel like I'm doing such a terrible job explaining sorry, but it's just that, if I post 1 problem, in the time it takes for me to type another, someone (like you) can read it and give me a solution - effectively saving time) so that's why I multi-posted. What should I do about this problem in the future?

Questions like these are very unlikely to fall under the umbrella of "quick" questions; their solutions require over 10 minutes to be typed up properly, making this unlikely. If you have a series of small questions there's nothing wrong with simply asking them all at once, otherwise preferably wait for one (or if it isn't overly excessive, maybe two) to be answered already before posting more.

bluecookie · « **Reply #1401 on:** August 28, 2017, 12:59:32 pm »

Quote from: RuiAce on August 26, 2017, 07:59:21 pm

$\text{This is fairly standard, albeit the tedious version}\\ \begin{align*}\cos 4\theta + i\sin 4\theta &= (\cos\theta + i\sin\theta)^4\\ &= \cos^4\theta + 4i\sin\theta\cos^3\theta - 6\sin^2\theta\cos^2\theta - 4i\sin^3\theta\cos\theta + \sin^4\theta \end{align*}$
$\text{Equating real parts,}\\ \begin{align*}\cos 4\theta &= \cos^4\theta -6\sin^2\theta\cos^2\theta + \sin^4\theta\\ &= \cos^4\theta - 6(1-\cos^2\theta)\cos^2\theta + (1-2\cos^2\theta+\cos^4\theta)\\ &= 8\cos^4\theta - 8\cos^2\theta + 1\end{align*}$
$\text{In b), let }x=\cos \theta. \text{ Then, }\\ \begin{align*}16x^4 - 16x^2 + 1 &= 0\\ 8x^4 - 8x^2 &= -\frac12\\ 8\cos^4\theta - 8\cos^2\theta + 1 &= \frac12\\ \cos 4\theta &=\frac12\end{align*}$
$\text{This has general solution}\\ \begin{align*}4\theta &= 2k\pi \pm \frac{\pi}{3}\\ \theta &= \frac{\pi}{2}k \pm \frac{\pi}{12}\end{align*}\\ \text{where }k\text{ is an integer.}$
$\text{Without loss of generality, take }k=-1,0,1,2\text{ to obtain}\\ \theta = -\frac{7\pi}{12}, -\frac{\pi}{12}, \frac{5\pi}{12}, \frac{11\pi}{12}\\ \text{and hence the roots are }x=\cos\left(-\frac{7\pi}{12}\right), \dots\\ \text{However, from using relevant trigonometric identities, we find that the roots are,}\\ \text{more simply, }x=\pm \cos \frac{\pi}{12}, \pm \cos \frac{5\pi}{12}$
______________________________________
$\text{Going back to the original equation, from the quadratic formula}\\ \begin{align*}16x^4 - 16x^2 + 1 &=0\\ x^2 &= \frac{16 \pm \sqrt{256-64}}{32}\\ &= \frac{2\pm \sqrt{3}}4\\ \therefore x &= \pm \sqrt{\frac{2\pm \sqrt{3}}4}\end{align*}$
$\text{Naturally, the negatives correspond to negatives and positives correspond to positives}\\ \text{So }\cos \frac{\pi}{12}\text{ and }\cos \frac{5\pi}{12}\text{ are either }\sqrt{\frac{2+ \sqrt{3}}4}\text{ or }\frac{2- \sqrt{3}}4$
$\text{Exploiting the fact that cos is monotonic decreasing over }0< x \le \frac{\pi}{2}\\ \frac{\pi}{12}< \frac{5\pi}{12} \implies \cos \frac{\pi}{12} > \cos \frac{5\pi}{12}\\ \text{so }\cos \frac{\pi}{12} = \sqrt{\frac{2+ \sqrt{3}}4}\text{ and }\cos \frac{5\pi}{12}=\sqrt{\frac{2- \sqrt{3}}4}$
$\text{This could potentially be simplified further by exploiting the fact}\\ \sqrt{2+\sqrt3} = \sqrt{\frac12 (4+2\sqrt3)}= \sqrt{\frac12 (1+2\sqrt3+3)} = \sqrt{\frac12 (1+\sqrt3)^2} = \frac{1}{\sqrt2}(1+\sqrt3)$

Thank you

Quote from: RuiAce on August 27, 2017, 05:28:43 pm

Questions like these are very unlikely to fall under the umbrella of "quick" questions; their solutions require over 10 minutes to be typed up properly, making this unlikely. If you have a series of small questions there's nothing wrong with simply asking them all at once, otherwise preferably wait for one (or if it isn't overly excessive, maybe two) to be answered already before posting more.

Okay, will do ^^

bluecookie · « **Reply #1402 on:** August 28, 2017, 01:46:18 pm »

Show that cos5θ=15(cosθ)^5-20(cosθ)^3+5cosθ. Hence
a) Solve the equation 16x^5-20x^3+5x-1=0 and deduce the exact values of cos(2pi/5) and cos(4pi/5)
b) Solve the equaion 32x^5-40x^3+10x-1=0 and deduce that
i) cospi/15+cos7pi/15+cos13pi/15+cos19pi/15=-1/2
ii) cospi/15cos7pi/15cos13pi/15cos19pi/15=1/16.

RuiAce · « **Reply #1403 on:** August 28, 2017, 03:22:19 pm »

Quote from: bluecookie on August 28, 2017, 01:46:18 pm

Show that cos5θ=15(cosθ)^5-20(cosθ)^3+5cosθ. Hence
a) Solve the equation 16x^5-20x^3+5x-1=0 and deduce the exact values of cos(2pi/5) and cos(4pi/5)
b) Solve the equaion 32x^5-40x^3+10x-1=0 and deduce that
i) cospi/15+cos7pi/15+cos13pi/15+cos19pi/15=-1/2
ii) cospi/15cos7pi/15cos13pi/15cos19pi/15=1/16.

$\text{Much of the working at the start follows the }\textbf{exact}\text{ same method}\\ \text{as for an earlier question, therefore I will not those parts}\\ \text{and only assume what is necessary to continue.}$
If you have trouble replicating the same procedure you should post up your working instead for guidance/feedback.
_______________________________________
$\text{The roots to the equation in a) are (as always with trig identities)}\\ x=1, \cos \frac{2\pi}{5}, \cos \frac{2\pi}{5}, \cos \frac{4\pi}{5}, \cos \frac{4\pi}{5}\\ \text{So by the sum of roots, we have}\\ 2\cos \frac{2\pi}{5} +2\cos \frac{4\pi}{5} + 1 = 0$
$\text{Using the relevant double-angle identity}\\ \begin{align*}2\cos \frac{2\pi}{5} + 4\cos^2 \frac{2\pi}{5} - 2 + 1& = 0\\ \cos \frac{2\pi}{5}&= \frac{-2\pm \sqrt{4+16}}{8}\\ &= -\frac14 \pm \frac{\sqrt5}{4}\end{align*}$
$\text{Clearly }\cos \frac{2\pi}{5} > 0\text{ and hence }\cos \frac{2\pi}{5} =-\frac14 + \frac{\sqrt5}{4}$
$\text{Applying the double angle identity again}\\ \cos \frac{4\pi}{5} = 2\left(-\frac14+\frac{\sqrt5}4\right)^2-1 = -\frac14 - \frac{\sqrt5}{4}$
The next question is done via a similar method (with differences being that no exact values make it faster, and the equation to solve is weirder). To be honest, these methods are all in my 4U Notes book; you need to take a look at it.

bluecookie · « **Reply #1404 on:** September 04, 2017, 11:42:39 am »

Quote from: RuiAce on August 28, 2017, 03:22:19 pm

$\text{Much of the working at the start follows the }\textbf{exact}\text{ same method}\\ \text{as for an earlier question, therefore I will not those parts}\\ \text{and only assume what is necessary to continue.}$
If you have trouble replicating the same procedure you should post up your working instead for guidance/feedback.
_______________________________________
$\text{The roots to the equation in a) are (as always with trig identities)}\\ x=1, \cos \frac{2\pi}{5}, \cos \frac{2\pi}{5}, \cos \frac{4\pi}{5}, \cos \frac{4\pi}{5}\\ \text{So by the sum of roots, we have}\\ 2\cos \frac{2\pi}{5} +2\cos \frac{4\pi}{5} + 1 = 0$
$\text{Using the relevant double-angle identity}\\ \begin{align*}2\cos \frac{2\pi}{5} + 4\cos^2 \frac{2\pi}{5} - 2 + 1& = 0\\ \cos \frac{2\pi}{5}&= \frac{-2\pm \sqrt{4+16}}{8}\\ &= -\frac14 \pm \frac{\sqrt5}{4}\end{align*}$
$\text{Clearly }\cos \frac{2\pi}{5} > 0\text{ and hence }\cos \frac{2\pi}{5} =-\frac14 + \frac{\sqrt5}{4}$
$\text{Applying the double angle identity again}\\ \cos \frac{4\pi}{5} = 2\left(-\frac14+\frac{\sqrt5}4\right)^2-1 = -\frac14 - \frac{\sqrt5}{4}$
The next question is done via a similar method (with differences being that no exact values make it faster, and the equation to solve is weirder). To be honest, these methods are all in my 4U Notes book; you need to take a look at it.

Thanks! Yeah I dunno why, I'm just not very good at those. For part b i) I got one of the roots as cos(-11pi/15), and the only one that doesn't match up to any of the other roots I got was cos19pi/15. I was wondering how you got from cos(-11pi/15) to cos(19pi/15).

RuiAce · « **Reply #1405 on:** September 04, 2017, 12:33:35 pm »

Quote from: bluecookie on September 04, 2017, 11:42:39 am

Thanks! Yeah I dunno why, I'm just not very good at those. For part b i) I got one of the roots as cos(-11pi/15), and the only one that doesn't match up to any of the other roots I got was cos19pi/15. I was wondering how you got from cos(-11pi/15) to cos(19pi/15).

$\text{Note that }\cos (2\pi + x) = \cos x\\ \text{and in fact, }\sin (2\pi+x) = \sin x \text{ and }\tan (2\pi + x) = \tan x\text{ as well}\\ \text{because every "ordinary" trig function is at least, periodic every }2\pi$
$\therefore \cos \left( -\frac{11\pi}{15} \right) = \cos \left(-\frac{11\pi}{15} + 2\pi\right) = \cos \frac{19\pi}{15}$

beau77bro · « **Reply #1406 on:** September 13, 2017, 04:43:03 pm »

can someone explain this both in words and equations - struggle with dummy variables

RuiAce · « **Reply #1407 on:** September 13, 2017, 05:07:38 pm »

Quote from: beau77bro on September 13, 2017, 04:43:03 pm

can someone explain this both in words and equations - struggle with dummy variables

$\text{Splitting the integral up gives}\\ \int_{-a}^a f(x)\,dx = \int_{-a}^0 f(x)\,dx + \int_0^a f(x)\,dx\\\text{Because the boundaries are 0 and }a\\ \text{it makes sense to perform a substitution on the second integral.}$
$\text{The only substitutions that can change the boundaries}\\ \text{as required are }u=-x\text{ and }u=x+a\\ \text{But the latter seems to be useless, judging by the final options}\\ \text{so we prefer }u=-x \implies du = dx$
$\text{Using this substitution,}\\ \int_{-a}^0 f(x)\,dx = \int_{a}^0 f(u)\, (-du) = \int_0^a f(-u)\, du$
____________________________________________________
$\textbf{All that the statement of 'dummy variables' states is that}\\ \boxed{\int_a^b f(x)\,dx = \int_a^b f(u)\,du}\\ \text{which furthermore equals }\int_a^b f(s)\,ds =\int_a^b f(y)\,dy =\int_a^b f(\zeta)\, d\zeta = \int_a^b f(banana)\, d(banana) = \dots$
$\text{Whilst this probably isn't immediately obvious, a bit of thought should make it}\\ \text{intuitively clear, because when performing a definite integral,}\\ \text{the variable we integrate with respect to always vanishes}\\ \text{e.g. in }\int_0^1 x^2\,dx = \frac13\text{, you never see }x\text{ again in your final answer.}$
$\text{And in a similar way, you can check that }\int_0^1 u^2\,du = \frac13\\ \text{So }\int_0^1 x^2\,dx = \int_0^1 u^2\,dx \text{ as well, giving us an example of how this works.}$
____________________________________________________
$\text{Applying this to our scenario, }\int _0^a f(-u) \,du = \int_0^a f(-x)\,dx$
$\text{So our original integral becomes}\\ I = \int_0^a f(x)\,dx + \int_0^a f(-x)\,dx = \boxed{\int_0^a (f(x)+f(-x))\,dx}$
____________________________________________________
$\text{So at this point none of our answers match up}\\ \text{but from here, doing the same thing with the substitution}\\ s=a-x\text{, it is then clear that D is the answer}$

justwannawish · « **Reply #1408 on:** September 14, 2017, 01:18:49 pm »

Hi!

I was wondering what are the best textbooks for 4U? I've heard that while cambridge is good for the harder questions, it isn't really realistic for the types of questions in the HSC? (Any recs for 3U as well? Is Fitzpatrick 'better' than Cambridge?)

Thank you

RuiAce · « **Reply #1409 on:** September 14, 2017, 10:24:43 pm »

Quote from: justwannawish on September 14, 2017, 01:18:49 pm

Hi!

I was wondering what are the best textbooks for 4U? I've heard that while cambridge is good for the harder questions, it isn't really realistic for the types of questions in the HSC? (Any recs for 3U as well? Is Fitzpatrick 'better' than Cambridge?)

Thank you

There is no such thing as best textbooks to rely on HSC questions/preparation, especially given that the ultimate focus is always past papers.

For some decently worthwhile mentions, if you're using the small Cambridge book that is only ever so decent. As for question types, they are either too easy or too hard.

On the other hand, things like Terry Lee's textbook are generally recommended for selective schools, due to the difficulty of the questions.

4U Fitzpatrick is decent, provided the newer version is chosen. The old version is quite useless.

What's generally regarded as a standout is the Sydney Grammar textbook, as it is most similar to the 3U Cambridge textbook. This textbook, however, must be purchased from the school. It was never published (presumably due to the new syllabus things), and comes as only a PDF.

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