2005 Independant, Q7 c) iii) (1 Viewer)

[king_of_boredom]

2005 Independent trial, Q7 c) iii)

what's the go here? need help!!

this year's 2u independent trial paper, question 7, c) iii)


how does one show LHS = RHS?

 

[who_loves_maths]

hi king_of_boredom,

i'm not sure if this is a 3u question or just 2u... but you do need to know the trig. limit thingy, you know, the one that involves sin(x):

Theory:

lim(x -> 0) [Sin(m.x)/(m.x)] = 1 ; where 'm' is any real number.

Your Question:

let 2/n = x ; ie. as n -> infinity, x = 2/(infinity) -> 0

hence, lim(n -> infinity) [(n/2).sin(2pi/n)] = lim(x -> 0) [sin(pi.x)/x] = pi.lim(x -> 0 [sin(pi.x)/(pi.x)]

---> = pi.1 = pi {ie. it's as if you let pi = m ...}

Therefore: lim(n -> infinity) [(n/2).sin(2pi/n)] = Pi


hope that helps

P.S. if this is 3u stuff, and it appeared in a 2u test paper, then it is either misplaced, or there is a lead-in part to the question that you have not shown... but the theory behind it is same.

Edit: maybe i should've shown an intermediary step between the transition from the 'n's to the 'x's above ^ to make the algebra clearer: it's simply to note that since x = 2/n , then n/2 = 1/x ... the '2/n' part was inside the Sine function, and the 'n/2' part was outside the Sine function.

 

[Emma-Jayde]

Geez.
That flew so far over the top of my head that it isn't funny!
Maybe I should go revise that topic...

And yes, it was in a 2u paper, and it was part iii) of a question, but the 2 preceeding parts, as far as I can see, have nothing to do with this question.
(It has to do with a polygon constructed inside a unit circle, with n sides.)

Btw, I tried to do it tonight, several times, different ways, and I kept getting zero too. :s

 

[acmilan]

Geez.
That flew so far over the top of my head that it isn't funny!
Maybe I should go revise that topic...

And yes, it was in a 2u paper, and it was part iii) of a question, but the 2 preceeding parts, as far as I can see, have nothing to do with this question.
(It has to do with a polygon constructed inside a unit circle, with n sides.)

Btw, I tried to do it tonight, several times, different ways, and I kept getting zero too. :s

Its possible to prove that limit geometrically, so who knows, it may have been relevant.

 

[who_loves_maths]

Originally Posted by Emma-Jayde
And yes, it was in a 2u paper, and it was part iii) of a question, but the 2 preceeding parts, as far as I can see, have nothing to do with this question.
(It has to do with a polygon constructed inside a unit circle, with n sides.)

AHHH... riteo... thanks for mentioning that Emma-Jayde

although i haven't seen the question, i suspect this is the gist of it:

let's construct a regular polygon of n-sides inside a circle of fixed radius r and centre O.

then join every vertices of this polygon to the centre O of the circle to make 'n' congruent (and isosceles) triangles within the polygon.

since there are 'n' such triangles, then each of the angle subtended by the n-sides at centre O is = 2pi/n {there are 2pi radians in one revolution}

hence, the area of each triangle = (1/2)r^2.Sin(2pi/n) .................. this is the Sine area formula applied.

and so the total area of the polygon = (n/2)r^2.Sin(2pi/n)

as 'n' grows larger and larger, this area will continue to increase - but since the polygon is constructed inside the circle, its area will never exceed that of the circle - ie. the area of the circle is a limiting value.

area of circle = pi.r^2

Therefore: Lim(n -> infinity) [(n/2)r^2.Sin(2pi/n)] = pi.r^2

the 'r^2' on both sides cancel, leaving:

Lim(n -> infinity) [(n/2).Sin(2pi/n)] = Pi


hope you can see it now, and thanks for mentioning the rest of the question again Emma

 

[Emma-Jayde]

hence, the area of each triangle = (1/2)r^2.Sin(2pi/n) .................. this is the Sine area formula applied.

*Smacks self in head and looks at question again*
I soooo should have seen that!!
I think I need a brain transplant!
Ahh well, I guess it's lucky I didn't have to do the 2u paper...

 

[who_loves_maths]

and just as a passing note:

if, instead of considering converging area, you take into account of the perimeter of a (or any) regular polygon of n-sides and as 'n' grows larger and larger, you'll arrive at the more basic limiting evaluation:

Lim(n -> infinity) [n.Sin(pi/n)] = Pi

 

[king_of_boredom]

hi king_of_boredom,

i'm not sure if this is a 3u question or just 2u... but you do need to know the trig. limit thingy, you know, the one that involves sin(x):

Theory:

lim(x -> 0) [Sin(m.x)/(m.x)] = 1 ; where 'm' is any real number.

Your Question:

let 2/n = x ; ie. as n -> infinity, x = 2/(infinity) -> 0

hence, lim(n -> infinity) [(n/2).sin(2pi/n)] = lim(x -> 0) [sin(pi.x)/x] = pi.lim(x -> 0 [sin(pi.x)/(pi.x)]

---> = pi.1 = pi {ie. it's as if you let pi = m ...}

Therefore: lim(n -> infinity) [(n/2).sin(2pi/n)] = Pi


hope that helps

P.S. if this is 3u stuff, and it appeared in a 2u test paper, then it is either misplaced, or there is a lead-in part to the question that you have not shown... but the theory behind it is same.

Edit: maybe i should've shown an intermediary step between the transition from the 'n's to the 'x's above ^ to make the algebra clearer: it's simply to note that since x = 2/n , then n/2 = 1/x ... the '2/n' part was inside the Sine function, and the 'n/2' part was outside the Sine function.

wow that is so awesome. thanks for that who_loves_maths!

see my problem was that i'd never seen a question of that nature before. it would have never crossed my mind to let 2/n = x, that just makes things twice as easy.

so....this answer would get me the full ONE mark that it's worth?!?!? i believe that a question of this calibre should be worth at least 2 marks...

[excuse the writing, i'm using MSPaint here.]

 

[who_loves_maths]

^ hi king_of_boredom,

yes it would certainly get you that one mark - however, to be on the safe side, i'd show the transition between the 'n's into the 'x's rather than just putting the first line of LHS in terms of 'x' straight away, because some examiners mightn't immediately see what you are doing (esp. since this question flows on from previous parts that involves a geometric argument.)

also, the second line of your working out {counting the first line as LHS = ...} is redundant... and so is your third line.
in fact, out of the first three lines, just one of them can replace all three - they all say exactly the same thing.

and of course, one last thing, this method is ONLY applicable IF the question says "hence or otherwise, prove..." - unless this is specified, you must use the geometric approach implied in previous parts of the question {which i have also outlined and shown in my second post. take a look.}


Edit: btw, that's some accurate writing using MSPaint there ... lol

 

[Shael]

That solution is kind of over kill. Take a look at the solution I have. It uses the topic "Small angles".

The basic idea is that in radians - "the sine/tan of the small angle = small angle"

 

[acmilan]

That's not entirely correct though, you are saying that for small angles sin(2pi/n) = 2pi/n, which is not the case, it is just an approximation. Its only when you have that limit that you can use the equality sign, because in that case its true.

 

[rama_v]

The actual solution they give in the independent paper solutions is much, much simpler although perhaps mathematically invalid. The previous question shows a diagram of a unit circle with a hexagon or some other polygon transcribed within it. They simply ask what each of the angles in the centre are equal to (in fact they are equal to 2pi/n).

Then the next question is the one above, the limit one. They simply say that as the number of sides of the polygon in the circle increases to infinity it becomes a circle, with area pir². = pi(1)² = pi. They do the solution in words