November 21, 2019, 12:28:21 pm

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#### spnmox

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##### Re: 3U Maths Question Thread
« Reply #4170 on: July 31, 2019, 11:01:46 am »
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1. The function y=secx for 0 <= y < pi/2, and is not defined for other values of x. State the domain and range of the inverse function.

I need help finding the domain.

2. Let p(a) = a^2(b+c)+b^2(c+a)+c^2(a+b)+2abc. Use the factor theorem to show that a+b is a factor and hence factorise P(a)

I cannot factorise P(a).
« Last Edit: July 31, 2019, 03:28:11 pm by spnmox »

#### mani.s_

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##### Re: 3U Maths Question Thread
« Reply #4171 on: July 31, 2019, 05:12:21 pm »
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Hi, I had another question on parametric equations. I have to convert x= cos t and y = sin 2t where 0<=t<=2pi.

This was the way I approached the question:
x= cos t
y= sin 2t

x^2= cos^2 t

y=sin 2t = 2sin t cos t
y^2 = 4sin^2 t cos^2 t

Now using cos^2 t + sin^2 t = 1

This is where I get stuck. I don't know what to do after.

#### DrDusk

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##### Re: 3U Maths Question Thread
« Reply #4172 on: July 31, 2019, 05:52:41 pm »
+2
Hi, I had another question on parametric equations. I have to convert x= cos t and y = sin 2t where 0<=t<=2pi.

This was the way I approached the question:
x= cos t
y= sin 2t

x^2= cos^2 t

y=sin 2t = 2sin t cos t
y^2 = 4sin^2 t cos^2 t

Now using cos^2 t + sin^2 t = 1

This is where I get stuck. I don't know what to do after.
You have the right idea. If we want to find sint in terms of x, you can just construct a triangle with side length x, hypotenuse 1 and angle t, because we have x = cos(t) = adjacent/hypotenuse.
Using this you get
$sin(t) = \sqrt{1-x^2}$
$\therefore y = 2x\sqrt{1-x^2}$
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#### fun_jirachi

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##### Re: 3U Maths Question Thread
« Reply #4173 on: July 31, 2019, 07:10:48 pm »
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You have the right idea. If we want to find sint in terms of x, you can just construct a triangle with side length x, hypotenuse 1 and angle t, because we have x = cos(t) = adjacent/hypotenuse.
Using this you get
$sin(t) = \sqrt{1-x^2}$
$\therefore y = 2x\sqrt{1-x^2}$

$\text{Note that if} \ x=\cos^2 t \ \text{then} \ \sin t = \pm \sqrt{1-x^2} \\ \text{And hence} \ y=\pm 2x\sqrt{1-x^2}$
We must have the plus-minus unless we are clearly allowed to disregard it (time, distance, etc.)

1. The function y=secx for 0 <= y < pi/2, and is not defined for other values of x. State the domain and range of the inverse function.

I need help finding the domain.

2. Let p(a) = a^2(b+c)+b^2(c+a)+c^2(a+b)+2abc. Use the factor theorem to show that a+b is a factor and hence factorise P(a)

I cannot factorise P(a).

1. For any inverse function, the range of the normal function is the domain of the new function, and vice versa. Hence the domain of the inverse function is 0<= x <= pi/2

2. On inspection it looks like the expansion of ((a+b+c)^3-a^3-b^3-c^3)/3, with which you can factor out (a+b). The factor theorem essentially states that (x-k) is a factor of P(x) if P(k) = 0, but I really don't see any logical way through yet. (might somewhere down the track though)

Hope this helps
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#### DrDusk

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##### Re: 3U Maths Question Thread
« Reply #4174 on: July 31, 2019, 09:02:19 pm »
+2

2. Let p(a) = a^2(b+c)+b^2(c+a)+c^2(a+b)+2abc. Use the factor theorem to show that a+b is a factor and hence factorise P(a)

I cannot factorise P(a).
In this part because it's a polynomial in terms of a, I'm just going to let a = x so it looks more like a polynomial. So we've already proved (a+b) is a factor, i.e.(x+b) is a factor. This allows us to establish this relationship:
$(x+b)Q(x) = (b+c)x^2+b^2(x+c)+c^2(x+b)+2bcx$
$\text{However by inspection we can tell Q(x) must have the form}\hspace{2mm} Q(x) = dx+e, (d,e)\in R \\ \text{as we need the highest power to be a 2. Now lets manipulate the RHS}$
$(x+b)(dx+e) = (b+c)x^2+b^2(x+c)+c^2(x+b)+2bcx$
$\therefore (x+b)(dx+e) = (b+c)x^2 + \underbrace{(b^2+2bc+c^2)}_{ = (b+c)^2}x+bc(b+c) \Rightarrow (1)$
$\text{Now by equating the coefficients of }\hspace{2mm}x^2\hspace{2mm} \text{etc, we can find out the value of d and e}$
$x^2: d = (b+c) \\ x: e + bd = (b+c)^2 \\ Constant: be = bc(b+c) \Rightarrow e = c(b+c)$
$\therefore (x+b)\left((b+c)x+c(b+c)\right) =(b+c)x^2 + (b^2+2bc+c^2)x+bc(b+c)$
$\therefore P(a) = (b+c)(a+b)(a+c) \hspace{2mm} \text{as we factorize out (b+c) from the LHS and replace 'x' with 'a' again}$

Sorry I only just noticed this question. Also you could've just let Equation (1) equal zero, but since the question said "hence", I was more inclined to use this method.
« Last Edit: July 31, 2019, 09:22:15 pm by DrDusk »
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#### spnmox

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##### Re: 3U Maths Question Thread
« Reply #4175 on: July 31, 2019, 11:26:03 pm »
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$\text{Note that if} \ x=\cos^2 t \ \text{then} \ \sin t = \pm \sqrt{1-x^2} \\ \text{And hence} \ y=\pm 2x\sqrt{1-x^2}$
We must have the plus-minus unless we are clearly allowed to disregard it (time, distance, etc.)

1. For any inverse function, the range of the normal function is the domain of the new function, and vice versa. Hence the domain of the inverse function is 0<= x <= pi/2

2. On inspection it looks like the expansion of ((a+b+c)^3-a^3-b^3-c^3)/3, with which you can factor out (a+b). The factor theorem essentially states that (x-k) is a factor of P(x) if P(k) = 0, but I really don't see any logical way through yet. (might somewhere down the track though)

Hope this helps

Sorry, I meant range haha!  I am not sure how to find the range of the inverse function. I subbed y=0 in the original function, but I get 0=1/cos0, which isn't right since cos0 is 1? i then subbed in y=pi/2 and got cosx=2/pi, but that doesn't give an exact value.
« Last Edit: July 31, 2019, 11:30:46 pm by spnmox »

#### fun_jirachi

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##### Re: 3U Maths Question Thread
« Reply #4176 on: July 31, 2019, 11:45:13 pm »
+1
Sorry, I meant range haha!  I am not sure how to find the range of the inverse function. I subbed y=0 in the original function, but I get 0=1/cos0, which isn't right since cos0 is 1? i then subbed in y=pi/2 and got cosx=2/pi, but that doesn't give an exact value.

Remember, y=sec x doesn't exist for -1<=y<=1 precisely because that's the range of cos x. Consider what would happen to the value of sec x as you took values of cos x for x=(n*pi)/2, where n is an integer. This would pose some serious problems, and you'd get total nonsense like you've got above (like 0=1) As a result, I've actually made a mistake in my previous solution though; the domain should be 1<=x<=pi/2.

Now for the range, we need to find the largest section of sec x {0<=y<=pi/2} that can be inverted ie. passes the horizontal line test. This 'section' of the curve exists from 0<=x<=arccos (2/pi), and hence the range will be 0<=y<=arccos (2/pi). I don't think the exact value should matter too much; most of the time in exams you're asked for simplest exact form anyway.

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#### spnmox

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##### Re: 3U Maths Question Thread
« Reply #4177 on: July 31, 2019, 11:50:16 pm »
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In this part because it's a polynomial in terms of a, I'm just going to let a = x so it looks more like a polynomial. So we've already proved (a+b) is a factor, i.e.(x+b) is a factor. This allows us to establish this relationship:
$(x+b)Q(x) = (b+c)x^2+b^2(x+c)+c^2(x+b)+2bcx$
$\text{However by inspection we can tell Q(x) must have the form}\hspace{2mm} Q(x) = dx+e, (d,e)\in R \\ \text{as we need the highest power to be a 2. Now lets manipulate the RHS}$
$(x+b)(dx+e) = (b+c)x^2+b^2(x+c)+c^2(x+b)+2bcx$
$\therefore (x+b)(dx+e) = (b+c)x^2 + \underbrace{(b^2+2bc+c^2)}_{ = (b+c)^2}x+bc(b+c) \Rightarrow (1)$
$\text{Now by equating the coefficients of }\hspace{2mm}x^2\hspace{2mm} \text{etc, we can find out the value of d and e}$
$x^2: d = (b+c) \\ x: e + bd = (b+c)^2 \\ Constant: be = bc(b+c) \Rightarrow e = c(b+c)$
$\therefore (x+b)\left((b+c)x+c(b+c)\right) =(b+c)x^2 + (b^2+2bc+c^2)x+bc(b+c)$
$\therefore P(a) = (b+c)(a+b)(a+c) \hspace{2mm} \text{as we factorize out (b+c) from the LHS and replace 'x' with 'a' again}$

Sorry I only just noticed this question. Also you could've just let Equation (1) equal zero, but since the question said "hence", I was more inclined to use this method.

Thank you!! That was a very clear explanation and totally cleared up any confusion I had

#### mani.s_

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##### Re: 3U Maths Question Thread
« Reply #4178 on: August 01, 2019, 05:59:34 pm »
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Hi, thanks for all the help recently you guys are awesome!!!

I have another question. We have to investigate the general form of the Lissajous curves given by: 𝑥 = 𝐴cos(𝑎𝑡 + 𝑐), 𝑦 = 𝐵sin( 𝑏𝑡 + 𝑑), 0 ≤ 𝑡 ≤ 2𝜋.
The question that we have to answer is, what effect do each element on the Lissajous curve have on the curve. What I mean is how changing values of A, B, a, b, c, and d have on the curve. These are all the elements in the general form. We have to use graphing technology (Desmos) to see what effect it has and give justification and evidence for each of the elements. For what I know is that changing A, changes the ellipticity of the curve along the x-axis and changing B, changes the ellipticity of the curve along the y-axis.

#### mani.s_

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##### Re: 3U Maths Question Thread
« Reply #4179 on: August 05, 2019, 09:56:46 pm »
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How do make the ABC logo using lissajous curve on desmos???

Thanks

#### ^^^111^^^

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##### Re: 3U Maths Question Thread
« Reply #4180 on: August 05, 2019, 10:36:40 pm »
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How do make the ABC logo using lissajous curve on desmos???

Thanks
Use these equations for lissajous figures:

x= A sin(at+ s)
y = B sin(bt + y)
Note that y is gamma symbol
and s is delta symbol
Also, it can be either sine or cosine.
e.g.
x = 3 cos t
y = cos(t+4)

The ABC logo's equation would then be:
x = cos(t/3)
y = sin (t)

#### spnmox

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##### Re: 3U Maths Question Thread
« Reply #4181 on: August 09, 2019, 05:43:36 pm »
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hey atarnotes people!!

i have a question about general solutions. when they say things like cosx=1, find the general solution, i never know if i should use x=0 or x=2pi?

also, if it says cos(2x-pi/4)=0, since there is a +- in the general solution, how are you meant to just get x out? do you find the positive and negative case and simplify two different equations?

#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #4182 on: August 09, 2019, 06:33:18 pm »
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hey atarnotes people!!

i have a question about general solutions. when they say things like cosx=1, find the general solution, i never know if i should use x=0 or x=2pi?

also, if it says cos(2x-pi/4)=0, since there is a +- in the general solution, how are you meant to just get x out? do you find the positive and negative case and simplify two different equations?

The general solution to $\cos x = a$ is $x = 2n\pi \pm \cos^{-1}a$, so when you sub $x=1$ you obtain $x=2n\pi \pm \cos^{-1}1$. But $\cos^{-1}1 = 0$, so it is recommended that you'd use 0.
\text{The idea for the other question is to simply leave everything inside the cos alone}\\ \text{until the next line, where you can start rearranging.}\\ \begin{align*} \cos \left(2x+\frac\pi4\right) &= 0\\ 2x+\frac\pi4 &= 2n\pi \pm \cos^{-1}0\\ 2x+\frac\pi4 &= 2n\pi \pm \frac\pi2\end{align*}
« Last Edit: August 09, 2019, 07:43:13 pm by RuiAce »

#### spnmox

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##### Re: 3U Maths Question Thread
« Reply #4183 on: August 09, 2019, 07:25:08 pm »
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The general solution to $\cos x = a$ is $x = 2n\pi \pm \cos^{-1}a$, so when you sub $x=1$ you obtain $x=2n\pi \pm \cos^{-1}1$. But $\cos^{-1}1 = 0$, so it is recommended that you'd use 0.
\text{The idea for the other question is to simply leave everything inside the cos alone}\\ \text{until the next line, where you can start rearranging.}\\ \begin{align*} \cos \left(2x+\frac\pi4\right) &= 0\\ 2x+\frac\pi4 &= 2n\pi \pm \cos^{-1}0\\ 2x+\frac\pi4 &= 2n\pi \pm \frac\pi4\end{align*}

yes, but cos inverse 1 also equals 2pi, so how do you know which one to use?

for the second one, my question was how do you then isolate x from your last line?

#### fun_jirachi

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##### Re: 3U Maths Question Thread
« Reply #4184 on: August 09, 2019, 07:41:59 pm »
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$\text{The idea here is that for} \ \cos x = a \\ x= 2n\pi \pm \cos^{-1}a \ (\text{for} \ n \in \mathbb{Z})\\ \text{Now consider} \ \cos x = 1 \\ x=2n\pi \pm \cos^{-1} (1) = 2n\pi, n \in \mathbb{Z} \\ \text{or for 'using'} \ 2\pi \ x=2n\pi \pm 2\pi = (2n+1)\pi, n \in \mathbb{Z}$
So as you can see it doesn't actually matter if you we use 0 or 2pi since 2n and 2n+1 are both integers. Just looks nicer and cleaner to use 0 (EDIT: also it's correct unlike using 2pi [refer to next post ]).

To isolate x:
$\cos \left(2x+\frac{\pi}{4}\right)=0 \\ 2x+\frac{\pi}{4} = 2n\pi \pm \frac{\pi}{2} \\ 2x=2n\pi + \frac{\pi}{4}, 2n\pi - \frac{3\pi}{4} \\ x= n\pi + \frac{\pi}{8}, n\pi -\frac{3\pi}{8} \\ \text{Which combines to form} \ x=\frac{(4n+1)\pi}{8}, n \in \mathbb{Z}$
« Last Edit: August 09, 2019, 07:48:22 pm by fun_jirachi »
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