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c12 2014 jan question

i did cos-1 (1/√ 2) and i got pi/4 ... in the ms there is also -pi/4 whereee did it come from????!!!!
(edited 7 years ago)
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Reply 1
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ARP1234
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think about the Cos(X) curve. It is symmetric about the x=0 line, hence both x=pi/4 and x=-pi/4 have the same value of Cos(x) = 1/root2. When you add pi/3 to -pi/4 youll get pi/12 which is in the range of 0<x<2pi. Sorry for being a bit confusing, but I hope you got it
Original post by ARP1234
think about the Cos(X) curve. It is symmetric about the x=0 line, hence both x=pi/4 and x=-pi/4 have the same value of Cos(x) = 1/root2. When you add pi/3 to -pi/4 youll get pi/12 which is in the range of 0<x<2pi. Sorry for being a bit confusing, but I hope you got it

but it said x is greater than zero
Reply 3
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ARP1234
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Original post by pondsteps
but it said x is greater than zero


I understand where your confusion is. But see, x-pi/3 is less than 0 when you think of -pi/4.
However, x-pi/3= -pi/4
so, x=pi/12. So X is greater than zero now right? So you have to look out for all possible values of x
Original post by ARP1234
I understand where your confusion is. But see, x-pi/3 is less than 0 when you think of -pi/4.
However, x-pi/3= -pi/4
so, x=pi/12. So X is greater than zero now right? So you have to look out for all possible values of x


thanks a lot! x
Original post by ARP1234
I understand where your confusion is. But see, x-pi/3 is less than 0 when you think of -pi/4.
However, x-pi/3= -pi/4
so, x=pi/12. So X is greater than zero now right? So you have to look out for all possible values of x

i just resolved it , arent we supposed to do pi/4 - 2pi to get the value?? what i got was -7pi/4 which is outside the range -pi/x < x <5pi/3 :/
Reply 6
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ARP1234
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Original post by pondsteps
i just resolved it , arent we supposed to do pi/4 - 2pi to get the value?? what i got was -7pi/4 which is outside the range -pi/x < x <5pi/3 :/


just remember that cos(-x) = cos(x). So, cos(pi/4)=cos(-pi/4). The value of -7/4pi which you received is one whole cycle behind pi/4, you dont have to go that far because then x=-17/12pi which is much outside of the range we're looking at. I think it'd be clearer if you searched up a y=cos(x) curve. Youll see between -2pi and 0, there are 2 angles for which the cos value is 1/root2, one of them being -7pi/4 while the other is -pi/4 (we are interested in this one only for this particular question). Sorry for the verbose explanation :s-smilie::s-smilie:
Original post by ARP1234
just remember that cos(-x) = cos(x). So, cos(pi/4)=cos(-pi/4). The value of -7/4pi which you received is one whole cycle behind pi/4, you dont have to go that far because then x=-17/12pi which is much outside of the range we're looking at. I think it'd be clearer if you searched up a y=cos(x) curve. Youll see between -2pi and 0, there are 2 angles for which the cos value is 1/root2, one of them being -7pi/4 while the other is -pi/4 (we are interested in this one only for this particular question). Sorry for the verbose explanation :s-smilie::s-smilie:


this is only for cos graphs right?
Reply 8
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ARP1234
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Original post by pondsteps
this is only for cos graphs right?


yeah this whole -pi/4 thing is contingent on the fact that a y=cos(x) curve is symmetric about the y-axis, which is not true for sine or tan curves.
sin(-x)=-sin(x), so these negative sine angles are slightly more difficult to work with.

If you dont mind my asking, are you sitting for C12 on 25th May?
Original post by ARP1234
yeah this whole -pi/4 thing is contingent on the fact that a y=cos(x) curve is symmetric about the y-axis, which is not true for sine or tan curves.
sin(-x)=-sin(x), so these negative sine angles are slightly more difficult to work with.

If you dont mind my asking, are you sitting for C12 on 25th May?

Ohh ok but we dont really need to apply on it do we?
And yesss i am, are u?
Reply 10
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ARP1234
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Original post by pondsteps
Ohh ok but we dont really need to apply on it do we?
And yesss i am, are u?


I don't remember there being a question like that, but I do recall a question involving the use of negative tan angles in one of the very recent papers.

Yeah I'm giving C12 as well. except for some tricky stuff like this and GP questions, I think all in all C12 isn't THAT hard

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