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September 21, 2019, 02:43:50 pm

AuthorTopic: 3U Maths Question Thread  (Read 549864 times) Tweet Share

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annabeljxde

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Re: 3U Maths Question Thread
« Reply #4155 on: July 27, 2019, 12:01:47 am »
0
Hey!

Need a bit of help with this question. The solutions went straight from (1-sin2θ) to the solution for tan2θ, which has left me confused. Do you use the condition 0 < θ < π/4 anywhere in the working out?

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Re: 3U Maths Question Thread
« Reply #4156 on: July 27, 2019, 01:28:58 am »
+2
Hey!

Need a bit of help with this question. The solutions went straight from (1-sin2θ) to the solution for tan2θ, which has left me confused. Do you use the condition 0 < θ < π/4 anywhere in the working out?

Start off with
$LHS = \sqrt{\frac{1 - sin2\theta}{1 + sin2\theta}}$
$= \sqrt{\frac{sin^2\theta + cos^2\theta - 2sin\theta cos\theta}{sin^2\theta + cos^2\theta + 2sin\theta cos\theta}}$
$\text{Using}\hspace{2mm}(sin\theta + cos\theta)^2 = sin^2\theta + cos^2\theta + 2sin\theta cos\theta$
$LHS = \frac{\pm(sin\theta - cos\theta)}{\pm(sin\theta + cos\theta)}$
$\text{Multiplying top and bottom by}\hspace{2mm}\frac{1}{cos\theta}\hspace{2mm}$$\text{and obseving that taking the negative one for the top one will give the desired answer we get}$
$LHS = \frac{-(tan\theta - 1)}{tan\theta + 1} = \frac{1 - tan\theta}{1 + tan\theta}$
$= RHS$
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RuiAce

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Re: 3U Maths Question Thread
« Reply #4157 on: July 27, 2019, 08:55:00 am »
+3
$LHS = \frac{\pm(sin\theta - cos\theta)}{\pm(sin\theta + cos\theta)}$
$\text{Multiplying top and bottom by}\hspace{2mm}\frac{1}{cos\theta}\hspace{2mm}$$\text{and obseving that taking the negative one for the top one will give the desired answer we get}$
Note: Not really the recommended justification here. Firstly (and somewhat less importantly) note that saying $\sqrt{a^2} = \pm a$ is technically incorrect, because that means you're claiming that $\sqrt{a^2}$ could potentially equal two different values. This would be problematic when saying something like $\sqrt{5^2} = \pm 5$. In general, the safest bet is to always say that $\sqrt{a^2} = |a|$ when doing these problems correctly.

But it's less of an issue here because you attempted to justify your way around it.

The justification behind the "$\pm$'s cancelling out into a negatives" isn't sufficient. The subtlety is that you haven't convinced me between which "$\sqrt{\frac{a^2}{b^2}}=\frac{\sqrt{a^2}}{\sqrt{b^2}}=\frac{\pm a}{\pm b}$" or "$\sqrt{\frac{a^2}{b^2}}=\frac{\sqrt{a^2}}{\sqrt{b^2}}=\frac{\pm a}{\mp b}$" is the favourable one. Saying that "observing that taking the negative one for the top one will give the desired answer" is not convincing because it looks like you're assuming what you're trying to prove,

A better justification here would be that $\sqrt{\frac{(\sin\theta-\cos\theta)^2}{(\sin\theta+\cos\theta)^2}} = \left| \frac{\sin\theta-\cos\theta}{\sin\theta+\cos\theta} \right| = \left| \frac{\tan\theta-1}{\tan\theta+1} \right| = -\frac{\tan\theta-1}{\tan\theta+1}$, justifying the last equality here by the fact that $\boxed{0\leq\theta\leq\frac\pi4}$.

We know that over this domain, $0 \leq \tan\theta \leq 1$. Hence on one hand, it is clear that $\tan\theta+1 > 0$. On the other hand, rearranging $\tan\theta \leq 1$ gives $\tan\theta - 1\leq 0$. Therefore $\frac{\tan\theta-1}{\tan\theta+1} \leq0$ (non-positive number divided by positive number), so $\left|\frac{\tan\theta-1}{\tan\theta+1}\right| = -\frac{\tan\theta-1}{\tan\theta+1}$. This now relies on the actual definition of the absolute value
$|x| = \begin{cases}x&x\geq 0\\ -x&x<0\end{cases}$
instead of intuition that cannot be justified.
« Last Edit: July 27, 2019, 06:08:57 pm by RuiAce »

mani.s_

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Re: 3U Maths Question Thread
« Reply #4158 on: July 27, 2019, 07:08:34 pm »
0
Hello, I just wanted some help on my assignment. We have to make a graph of desmos using parametric equations: x = cos 2t and y = sin t, (0 ≤ t ≤ 2pi).
We then have to choose 6 points on the graph including critical points. Now, this is the part which I don't understand. They ask us for a reason for each coordinate and how we obtained that coordinate from the parametric equations. I don't understand what that means, like what are we supposed to do???

Thanks

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Re: 3U Maths Question Thread
« Reply #4159 on: July 27, 2019, 07:56:51 pm »
+2
Hello, I just wanted some help on my assignment. We have to make a graph of desmos using parametric equations: x = cos 2t and y = sin t, (0 ≤ t ≤ 2pi).
We then have to choose 6 points on the graph including critical points. Now, this is the part which I don't understand. They ask us for a reason for each coordinate and how we obtained that coordinate from the parametric equations. I don't understand what that means, like what are we supposed to do???

Thanks
Without risking giving away a lot, this is my interpretation. (However note that usually it's worth double checking with your teachers if you don't get the wording of an assignment.)

Since this is a parametric plot, every point on the plot corresponds to at least one value for $t$. For example, the point represented by $(\cos (2\times 0), \sin 0)$ corresponds to at least the value $t=0$. It may be possible that it corresponds to another value of $t$, but that can be something for you to investigate.

So my guess would be that relating certain values of $t$ to the points on the curve may be something important here.

Judging by your assignment's context, however, $t$ may as well be a specific parameter that represents some sense of "time". So potentially you could talk about how at a certain time $t$, we are at this point. (Note that since the question also asks you to examine critical points, it is likely that they also want to see some, let's say, specific values of $t$ in your analysis.)

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Re: 3U Maths Question Thread
« Reply #4160 on: July 27, 2019, 11:31:18 pm »
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Hey,
Can I please have some help for this binomial question?

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Re: 3U Maths Question Thread
« Reply #4161 on: July 28, 2019, 08:21:23 pm »
+1
Hey,
Can I please have some help for this binomial question?
Since nobody else has posted a solution, I'll post mine.

Out of curiosity, is this a HSC question? Because I've never seen a binomial question involving square roots. It can be done, just that I've never seen it asked in the HSC.

First, multiply the top and bottom of the LHS by $1+x$, so that
$\sqrt{\frac{1+x}{1-x}}=\frac{1+x}{\sqrt{1-x^2}}$
We then use the following binomial expansion:
$
(1+x)^n = 1 + nx + O(x^2)
$

where $O(x^2)$ means terms of order $x^2$ and higher. The assumption of $|x|\ll1$ then implies that:
$(1+x)^n\approx1+nx$
Hence,
$(1-x^2)^{-1/2}\approx1+\frac{1}{2}x^2$
The LHS is now
$\sqrt{\frac{1+x}{1-x}}=\frac{1+x}{\sqrt{1-x^2}}\approx(1+x)\left(1+\frac{1}{2}x^2\right)$
which, when expanded, gives
$\sqrt{\frac{1+x}{1-x}}\approx1+x+\frac{1}{2}x^2$
which is the required second-order approximation.
« Last Edit: July 29, 2019, 01:04:42 pm by blyatman »
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mani.s_

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Re: 3U Maths Question Thread
« Reply #4162 on: July 28, 2019, 08:26:22 pm »
0
I just wanted to double-check a solution. I have to convert the following parametric equation to a cartesian equation: x = cos 2t y = sin t, 0<_ t <_ 2pi.

x = cos (2t)
cos (2t) = x
arccos (x) = 2t
t = arccos (x)/2

so the cartesian equation would be:
y = sin ( arccos (x) ) / 2

but apparently its:
y = plus-minus sine( arccos (x) ) / 2

Why is it plus-minus???

RuiAce

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Re: 3U Maths Question Thread
« Reply #4163 on: July 28, 2019, 08:30:30 pm »
+2
Is this a HSC question? Because I've never seen a binomial question involving square roots. It can be done, just that I've never seen it asked in the HSC.
I consulted with the poster later on and we found that the source of the question was well beyond the HSC. The question can be buzz-killed using their method, which is actually the generalised binomial theorem. (The generalised binomial theorem can be used, and then the usual method of equating coefficients is sufficient.)

They were just trying to look for hard binomial theorem questions for practice, and unintentionally stumbled somewhere too far. We do not need to worry about it.

blyatman

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Re: 3U Maths Question Thread
« Reply #4164 on: July 28, 2019, 08:42:59 pm »
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I consulted with the poster later on and we found that the source of the question was well beyond the HSC. The question can be buzz-killed using their method, which is actually the generalised binomial theorem. (The generalised binomial theorem can be used, and then the usual method of equating coefficients is sufficient.)

They were just trying to look for hard binomial theorem questions for practice, and unintentionally stumbled somewhere too far. We do not need to worry about it.

Yeh I thought it didn't seem like HSC material. As you said, I used the generalised binomial theorem in the above proof in the expansion of (1+x)^n, which is more or less the Taylor series expansion.

An interesting problem nonetheless.
« Last Edit: July 28, 2019, 08:48:55 pm by blyatman »
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fun_jirachi

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Re: 3U Maths Question Thread
« Reply #4165 on: July 28, 2019, 08:59:18 pm »
0
I just wanted to double-check a solution. I have to convert the following parametric equation to a cartesian equation: x = cos 2t y = sin t, 0<_ t <_ 2pi.

x = cos (2t)
cos (2t) = x
arccos (x) = 2t
t = arccos (x)/2

so the cartesian equation would be:
y = sin ( arccos (x) ) / 2

but apparently its:
y = plus-minus sine( arccos (x) ) / 2

Why is it plus-minus???

Without the plus-minus, sin(arccos(x))/2 is only positive. The range of arccos x is 0<=y<=pi, and hence the range of sin(arccos(x))/2 is 0<=y<=1/2. The plus-minus effectively remedies this situation by 'making' the range -1/2<=y<=1/2, since for values of t between +-2pi, sine and cosine can be both positive and negative, effectively mapping out the cartesian equation across the four quadrants instead of the two the inverse trig functions are usually limited to.

Hope this helps
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mani.s_

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Re: 3U Maths Question Thread
« Reply #4166 on: July 28, 2019, 09:01:37 pm »
0
Without the plus-minus, sin(arccos(x))/2 is only positive. The range of arccos x is 0<=y<=pi, and hence the range of sin(arccos(x))/2 is 0<=y<=1/2. The plus-minus effectively remedies this situation by 'making' the range -1/2<=y<=1/2, since for values of t between +-2pi, sine and cosine can be both positive and negative, effectively mapping out the cartesian equation across the four quadrants instead of the two the inverse trig functions are usually limited to.

Hope this helps
So is my working out and answer correct???

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Re: 3U Maths Question Thread
« Reply #4167 on: July 28, 2019, 09:13:23 pm »
+3
So is my working out and answer correct???
cos (2t) = x
arccos (x) = 2t
It is incorrect (only partially correct) as this line is unjustified. You haven't done a quadrants analysis on your condition $0\leq t \leq 2\pi$.
$\text{Note that consequently }0\leq 2t \leq 4\pi.$
That is, we have solutions for $2t$ in the 1st to 8th quadrants.
$\text{First suppose }x \geq 0.\\ \text{Then we need to consider the 1st, 4th, 5th and 8th quadrants.}$
$\text{Solving }\cos 2t = x\text{ then gives}\\ \boxed{2t = \arccos x, \, 2\pi - \arccos x, \, 2\pi + \arccos x, \, 4\pi - \arccos x}.$
You can then plug each of these back into $y=\sin t$. You will find that two of the solutions give $y = -\sin \left( \frac{\arccos x}{2} \right)$.

A similar thing happens for the case $x < 0$.

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Re: 3U Maths Question Thread
« Reply #4168 on: July 30, 2019, 08:15:56 pm »
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Hi, I'm having trouble converting this parametric equation: x= sin t and y= cos 2t, where 0<=t<=2pi. I'm supposed to use the double angle formula for cos 2t: cos 2t = 1 - 2sin^2 t.

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Re: 3U Maths Question Thread
« Reply #4169 on: July 30, 2019, 08:51:59 pm »
+1
Hey there!

It's a lot simpler than you think If x=sin t, and you have y=cos 2t = 1-2sin^2 t, then logically y=1-2x^2, substituting in the first thing.

Hope this helps
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