BOS Trials 2013 Extension 2 Discussion Thread (1 Viewer)
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Makematics
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Q16???
Makematics
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carrot said none of the answers to Q2 were right? so is everyone getting a mark or what? i wouldnt mind a free mark.
integral95
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Yes carrot sticks will be holding another BOS trials next year
With different sets of question obviously
Makematics
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dont do it if you dont want to get hospitalised.
but jks, actually do it next yr man, great experience, and it's free due to carrot's generosity
plus i see you have high aims which is great! carrot's test will put you to the test.Makematics
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Just realised there was a typo in Q14. there are two different Q14(b)s. not that it really matters
iBibah
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how are B, C and D incorrect?
RealiseNothing
what is that?It is Cowpea
I can write up my solution to question 15 if anyone wants?
Makematics
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thatd be cool
first remember that the domain of y=arctanx is -pi/2 < x < pi/2
That rules out B and C, as anything in the second quadrant has an argument between pi/2 and pi, while anything in the third quadrant has an argument between -pi/2 and -pi.
In the actual paper we got, the option for D was different to what carrot put in the pdf which is linked in this thread. D is correct for the pdf version. In the hardcopy version, option D was arctan(-b/a), while in the pdf version it is arctan(b/a).
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RealiseNothing
what is that?It is Cowpea
15)a)i)  - \sin(A-B))
 \cos(B) + \sin(B) \cos(A) - (\sin(A) \cos(B) - \sin(B) \cos(A)))
 \sin(B))
ii) If
 = \frac{1}{2} + \sum_{k=1}^{n} \cos(0) = \frac{1}{2} + \sum_{k=1}^{n} 1 = n + \frac{1}{2})
If
we re-arrange the equation in part (i) to make )
the subject:
 = \frac{\sin(A+B) - \sin(A-B)}{2\sin(B)})
Let
and 
and substitute this into our summation:
 = \frac{1}{2} + \sum_{k=1}^{n} \frac{\sin(k\theta + \frac{\theta}{2}) - \sin(k\theta - \frac{\theta}{2})}{2\sin(\frac{\theta}{2})})
 = \frac{1}{2} + \sum_{k=1}^{n} \frac{\sin(\frac{(2k+1)\theta}{2}) - \sin(\frac{(2k-1)\theta}{2})}{2\sin(\frac{\theta}{2})})
This makes a telescoping sum and so we are left with:
 = \frac{1}{2} + \frac{\sin(\frac{(2n+1)\theta}{2})}{2\sin(\frac{ \theta}{2})} - \frac{\sin(\frac{\theta}{2})}{2\sin(\frac{ \theta}{2})})
 = \frac{\sin(\frac{(2n+1)\theta}{2})}{2\sin(\frac{ \theta}{2})})
ii) If
If
Let
This makes a telescoping sum and so we are left with:
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iBibah
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thatd be cool
first remember that the domain of y=arctanx is -pi/2 <x < pi/2
That rules out B and C, as anything in the second quadrant has an argument between pi/2 and pi, while anything in the third quadrant has an argument between -pi/2 and -pi.
In the actual paper we got, the option for D was different to what carrot put in the pdf which is linked in this thread. D is correct for the pdf version. In the hardcopy version, option D was arctan(-b/a), while in the pdf version it is arctan(b/a).
yep got it, thanks man
Makematics
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finally know what a telescoping sum is lolz. thought it was some complicated stuff that you guise learnt yourselves.15)a)i)
ii) If
If
Let
This makes a telescoping sum and so we are left with:
iBibah
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Well it's not explicitly taught, though it's still used without naming it.
RealiseNothing
what is that?It is Cowpea
b)  dx)
Let
so 
and )
and }{k})
}{k}] - \int_0^\pi \frac{\sin(kx)}{k} dx)
Now the first part is 0 and so we get:
}{k^2}])
from 
^k - 1}{k^2})
c)i) Letx}{2})}{2\sin(\frac{x}{2})} dx)
From part a)ii) we get:
 dx)
^k - 1}{k^2}])
Substituting in the limits for the first part gives the required result of:
^k - 1}{k^2}])
c)ii) Simply let
^{k} - 1}{(k)^2}])
If you expand this, you will notice that all the even reciprocal squares cancel out, as^{2k} - 1 = 1-1 = 0)
which leaves you with:
^2}])
Divide everything by
and take out the minus sign from the summation yields the result:
^2}])
Let
Now the first part is 0 and so we get:
c)i) Let
From part a)ii) we get:
Substituting in the limits for the first part gives the required result of:
c)ii) Simply let
If you expand this, you will notice that all the even reciprocal squares cancel out, as
Divide everything by
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Makematics
Well-Known Member
exactly! ive come across it heaps of times without knowing it's name. probably same thing with this squeeze theorem or whatever
Makematics
Well-Known Member
ahhh so it was a typo at the bottom of page 15? the top of the sigma sign should be 2n-1 not n, right?
