# a little help here ...plzWatch

#1
The question is on the microsoft words file .. i dunno how to type it out ..
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#2
emm ... help me plz .........
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#3
anyone ? it's a product of circular functions question .. juz one question only ..
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12 years ago
#4
assuming that is really a question , couldn't you just type them up on your calc ?.
p.s : i find those powers really strange.
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#5
emm .. it's a math olympiad question .. no calculators allowed ..
anyway, i tried it with my graphical calculator .. it says overflow ..
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#6
can someone type the question out ?
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#7
how come no body answer ?? is the question too hard ?
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12 years ago
#8
i'm not promising anything, but i think what i've done so far yields promise:

look at terms from opposite ends of the multiplication. note that sin(pi/2 - x) = cos(x), hence if we multiply these two terms together we get (16cos^2 x sin^2 x - 12sin^2 x - 12cos^2 x + 9) = (4sin^2 2x - 3).

hurrah, the thing has been halved in length. but now notice the things being sined span from 0 to pi. but also note the sin^2 x is symmetrical between 0 and pi, with pi/2 being the line of symmetry. so you can cut the multiplication in the middle, and apply the same method as before to each half of it, then times them together. then keep doing this again and again, hopefully spotting a pattern along the way.

there's probably a simpler way, and i'm not sure how the answer is going to come about, but oh well. reply with your thoughts/results.
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12 years ago
#9
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12 years ago
#10
If you look at terms like pi/2^2005, well that's effectively zero, and the majority of the terms will be like that. 2^2000.pi/2^2005 = 0.03125 sin of that is a number in terms of 10^-4. So effectively zero.
So, um, cheating a bit, couldn't you just say it's the products of all the -3s ? Something like (-3)^2006 ?
I'm assuming you'll never want/get an exact answer without a computer/calculator, so try an approximation, as above.
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12 years ago
#11
(Original post by NeilosRobbos)
If you look at terms like pi/2^2005, well that's effectively zero, and the majority of the terms will be like that. 2^2000.pi/2^2005 = 0.03125 sin of that is a number in terms of 10^-4. So effectively zero.
So, um, cheating a bit, couldn't you just say it's the products of all the -3s ? Something like (-3)^2006 ?
I'm assuming you'll never want/get an exact answer without a computer/calculator, so try an approximation, as above.
the question will require an exact answer.
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12 years ago
#12
(continuing from my post above...)

= (4sin^2 0 - 3)...(4sin^2 pi - 3)

( sin^2 x = sin^2 (pi-x) )

= ((4sin^2 0 - 3)...(4sin^2 pi/2))^2

by similar argument, we eventually arrive at:

= ((4sin^2 0 - 3)(4sin^2 pi/2))^(2^2004) = (-3)^(2^2004)

not sure though... maybe i mean (-3)^(2^2003)? maybe i'm completely wrong? who knows. normally you get an answer that's more manageable than this...
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#13
I think you might be right ... there are ppl telling me that i need to multiply the first term and the last term ..

i dont think you can use approximation ... it's maths olympiad question ..
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12 years ago
#14
hmmm, might have missed something, brb

edit: i still think + or - 1
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12 years ago
#15
i have little doubt you're right. care to expand?
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12 years ago
#16
I've changed my mind to +/- 3 ....moley >.<

Anywhoo;

All parts = all even . all odd

All parts = All even (as you showed) . middle one / last one

so

All even . all odd = -All even

=> all odd = -1

Since the thingies are divided into 2n segments, there will always be very every +8 a -8, for every + 16 a -16 (not explained very well, sorry, have a look at n=5, see what I mean) - so they all pair away, like in solitaire, until we are left with the first, middle and last values, whose product is 3....the + and - bit comes with the fact, I don't know how many times I would be doing the iteration.
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#17
why is it like so complicated ?? which chapter in maths is this ?
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12 years ago
#18
what module is this??
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12 years ago
#19
psycho said it was a (B)MO question...
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12 years ago
#20
(Original post by chewwy)
psycho said it was a (B)MO question...
what is BMO?
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