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March 28, 2024, 10:47:08 pm

Author Topic: Specialist 1/2 Question Thread!  (Read 119947 times)  Share 

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sogreatsosad

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Re: Specialist 1/2 Question Thread!
« Reply #360 on: August 13, 2020, 05:34:33 pm »
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What are you struggling with? What have you tried yourself?

I tried multiplying it repeatedly, utilising the fact that i^2 is a real number. Can't get more minute than that, need help

keltingmeith

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Re: Specialist 1/2 Question Thread!
« Reply #361 on: August 13, 2020, 05:37:44 pm »
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I tried multiplying it repeatedly, utilising the fact that i^2 is a real number. Can't get more minute than that, need help

Yeah, so as you've seen, that'll quickly become an issue because it only works for when n is an integer - which it might not be. Have you learned about using De Moivre's theorem? Because that'll be integral to solving this question

sogreatsosad

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Re: Specialist 1/2 Question Thread!
« Reply #362 on: August 13, 2020, 05:39:58 pm »
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Yeah, so as you've seen, that'll quickly become an issue because it only works for when n is an integer - which it might not be. Have you learned about using De Moivre's theorem? Because that'll be integral to solving this question

I haven't learn that, no. I'll learn out the theorem then try the problem again, if that doesn't work then I'll come back here. Thanks for telling me about De Moivre's theorem!

keltingmeith

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Re: Specialist 1/2 Question Thread!
« Reply #363 on: August 13, 2020, 05:43:11 pm »
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I haven't learn that, no. I'll learn out the theorem then try the problem again, if that doesn't work then I'll come back here. Thanks for telling me about De Moivre's theorem!

All good - information on the internet can be confusing, but there should be information in your textbook - and here's a quick summary:

Once you've converted your complex number, z, to polar form, then it has a modulus (r) and an argument (theta), and has the form:



Then, the form of z^n is:


sogreatsosad

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Re: Specialist 1/2 Question Thread!
« Reply #364 on: August 13, 2020, 07:31:56 pm »
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All good - information on the internet can be confusing, but there should be information in your textbook - and here's a quick summary:

Once you've converted your complex number, z, to polar form, then it has a modulus (r) and an argument (theta), and has the form:



Then, the form of z^n is:



Hi, I tried using DM's theorem. I converted the given problem to polar form, and expanded n. My approach has been that basically sin((-pi*n)/3 cannot equal i, as if it did then i times i means a real number. However, I tried to use the CAS to solve for n, but it's simply returning "false". Where to now?

S_R_K

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Re: Specialist 1/2 Question Thread!
« Reply #365 on: August 13, 2020, 08:38:29 pm »
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Hi, I tried using DM's theorem. I converted the given problem to polar form, and expanded n. My approach has been that basically sin((-pi*n)/3 cannot equal i, as if it did then i times i means a real number. However, I tried to use the CAS to solve for n, but it's simply returning "false". Where to now?

The argument of an imaginary number is \(\frac{\pi}{2} + \pi k\) where k is an integer. Equate that to the argument of \((2-2\sqrt{3}i)^n\) and solve for the smallest positive integer n.

M-D

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Re: Specialist 1/2 Question Thread!
« Reply #366 on: August 30, 2020, 08:15:30 pm »
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Hi,
I need help with the following question:

Find the values of p and q such that is parallel to the x-axis.

I have worked out that:



I don't know how to get specific values for p and q so is parallel to the x-axis.

Any assistance will be much appreciated.

keltingmeith

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Re: Specialist 1/2 Question Thread!
« Reply #367 on: August 30, 2020, 08:19:21 pm »
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Hi,
I need help with the following question:

Find the values of p and q such that is parallel to the x-axis.

I have worked out that:



I don't know how to get specific values for p and q so is parallel to the x-axis.

Any assistance will be much appreciated.

Well, if the vector is parallel to the x-axis, then it should be equal to some multiple of the i unit vector - does that help?

redset8

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Re: Specialist 1/2 Question Thread!
« Reply #368 on: November 04, 2020, 08:08:31 pm »
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Hey,
A little guidance on the attached question would be much appreciated! I'm definitely missing something here...
« Last Edit: November 04, 2020, 08:11:29 pm by redset8 »

keltingmeith

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Re: Specialist 1/2 Question Thread!
« Reply #369 on: November 04, 2020, 08:13:00 pm »
+1
Hey,
A little guidance on the attached question would be much appreciated! I'm definitely missing something here.

Absolutely - before I help, I'm curious. What do you notice in this picture? Is there anything that jumps out at you? The 'trick' with circle theorem questions is to not focus on what you need to find, and to instead think what you notice about everything else in the question, then slowly build through smaller links until you eventually find one which tells you an unknown. So, what smaller things (maybe related to the pronumerals - maybe not) have you noticed about this particular circle?

redset8

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Re: Specialist 1/2 Question Thread!
« Reply #370 on: November 04, 2020, 08:22:16 pm »
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Absolutely - before I help, I'm curious. What do you notice in this picture? Is there anything that jumps out at you? The 'trick' with circle theorem questions is to not focus on what you need to find, and to instead think what you notice about everything else in the question, then slowly build through smaller links until you eventually find one which tells you an unknown. So, what smaller things (maybe related to the pronumerals - maybe not) have you noticed about this particular circle?

Uh all I notice is that it's not stated that BC is a straight line. Oh and triangle DOC is isosceles. So is triangle BOD. (due to radius) If, however, BC is diameter, then <BDC is 90, but I don't know if that helps.

keltingmeith

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Re: Specialist 1/2 Question Thread!
« Reply #371 on: November 04, 2020, 08:29:04 pm »
+1
Uh all I notice is that it's not stated that BC is a straight line. Oh and triangle DOC is isosceles. So is triangle BOD. (due to radius) If, however, BC is diameter, then <BDC is 90, but I don't know if that helps.

That's awesome! So, I think you can assume BC is a straight line - and if it is, that also means that ABCD is a cyclic quadrilateral. So, this should tell you what the value of a is - and hopefully using the facts you've already noticed, you should be able to get the rest.

EDIT: Also, that picture is so not to scale it hurts lmao. I don't think that shape can actually exist if the angle BAD is 65 degrees

redset8

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Re: Specialist 1/2 Question Thread!
« Reply #372 on: November 04, 2020, 08:36:28 pm »
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That's awesome! So, I think you can assume BC is a straight line - and if it is, that also means that ABCD is a cyclic quadrilateral. So, this should tell you what the value of a is - and hopefully using the facts you've already noticed, you should be able to get the rest.

So if ABCD is a cyclic quadrilateral (opposites sum to 180), then a=180-65=115. But that would mean b = 180 -2a = -50...? If triangle DOC is isosceles.
Or have I managed to stuff up the angles of a triangle?

keltingmeith

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Re: Specialist 1/2 Question Thread!
« Reply #373 on: November 04, 2020, 08:42:32 pm »
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So if ABCD is a cyclic quadrilateral (opposites sum to 180), then a=180-65=115. But that would mean b = 180 -2a = -50...? If triangle DOC is isosceles.
Or have I managed to stuff up the angles of a triangle?

Yeah - the fact that this shape CANNOT exist makes me think it's just a dud question. Notice how a looks acute, but the angle BAD looks obtuse? That's not just because it looks that way, it's because the angles HAVE to be that way for this specific shape to exist. I'm curious if you have answers for this question? If not, why don't you assume that a=65 degrees. In the spoiler, I'll tell you what b and c are, so that you can work it out yourself and have something to check against:

Spoiler
b=180-2a=180-130=50
2c+(180-b)=180 <==> c=(1/2)b=25
Alternatively,
a+c+90=180 <==> c=90-a=25

redset8

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Re: Specialist 1/2 Question Thread!
« Reply #374 on: November 04, 2020, 08:47:48 pm »
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Yeah - the fact that this shape CANNOT exist makes me think it's just a dud question. Notice how a looks acute, but the angle BAD looks obtuse? That's not just because it looks that way, it's because the angles HAVE to be that way for this specific shape to exist. I'm curious if you have answers for this question? If not, why don't you assume that a=65 degrees. In the spoiler, I'll tell you what b and c are, so that you can work it out yourself and have something to check against:
Unfortunately I don't have answers; it's an old school 1/2 Specialist exam, but the answers are just missing for this section A.
If we let a=65, I got the same answers as you.
Side note, how does your 'alternatively' section work? As in how can we say a+c+90=180? Is it still to do with the opposite angles of cyclic quadrilateral sum to 180?
« Last Edit: November 04, 2020, 08:57:52 pm by redset8 »