de Moivre's (1 Viewer)

[OLDMAN]

By now all ext 2 students should be familiar with de Moivre's^*, here's an interesting exercise on it.

Prove that cos(1^o), cos(2^o),cos(3^o),cos(4^o),cos(5^o) are all irrationals.



de Moivre's^* : is your teacher a real mathematician or not?- real ones pronounce this d'mwah's, otherwise they say "de moyberr's"

 

[abdooooo!!!]

Originally posted by OLDMAN
de Moivre's^* : is your teacher a real mathematician or not?- real ones pronounce this d'mwah's, otherwise they say "de moyberr's"

ya my teacher says d'mwah's... so i guess he is real.

why is that though???

 

[KeypadSDM]

I think it's some pompous mathematics arrogance thing.

 

[OLDMAN]

Sadly I am no real mathematician, and I use the form the student is used to. However, real mathematicians attend them fancy international conferences, and high level maths is populated with many French names : to pronounce them in less than the right way would nearly be offensive -trust me.

 

[Giant Lobster]

lol
Many many skillful math teachers and tutors are fobby middle aged asians with Masters and Doctorates (My tutor as an example) and they say De Moivre like how it's spelt.

 

[Grey Council]

i guess my teacher is a half half then. lol, he says it neither d'mwahs nor d'moi'vrais or de moyberr's



lol, what a fancy concept. telling the "pro'ness" of a teacher by his/her pronunciation of a some wierd french name. heheh

 

[freaking_out]

Originally posted by KeypadSDM
I think it's some pompous mathematics arrogance thing.

yes- like trying to use hard terms (as we discussed in the other thread. )

 

[KeypadSDM]

Originally posted by freaking_out
yes- like trying to use hard terms (as we discussed in the other thread. )

Oh god, not again.

 

[freaking_out]

Originally posted by KeypadSDM
Oh god, not again.

yep- thats why latin still exists these days.

 

[OLDMAN]

[ QUOTE Originally posted by KeypadSDM
Oh god, not again. ]

Might be time to get back on the learning track.

Using de Moivre's, and collecting real terms,

cos[n@]= cos^n[@] - nC2 cos^(n-2)[@]sin^2[@] + nC4cos^(n-4)[@]sin^4[@] - ....

Even powers of sin can be expressed in terms of polynomials of cos with integer coefficients.

Thus cos[n@]=polynomial in cos of degree n, with integer coefficients, nCr= integer.

Thus, if cos(@) is rational, then cos[n@]= rational.

An equivalent logical statement to this is

if cos[n@]= irrational, then cos[@] must be irrational.

Now cos(30)=sqrt(3)/2 irrational (easy to prove that the sqrt(prime) is irrational.

The rest is straightforward.

 

[KeypadSDM]

I don't know if you can assume that logical statement.

If Cos[A] is rational, Cos[nA] must be rational. I can accept that.

But does your proof also mean that

Iff Cos[A] is rational, Cos[nA] must be rational?

 

[OLDMAN]

Keypad:

If cos(n@) is irrational, then cos @ must be irrational, because if otherwise ie. cos @ is rational then cos(n@) is rational. Contradiction.

 

[KeypadSDM]

Originally posted by OLDMAN
Keypad:

If cos(n@) is irrational, then cos @ must be irrational, because if otherwise ie. cos @ is rational then cos(n@) is rational. Contradiction.

Oh yeah, whoops. Lol.

 

[freaking_out]

Originally posted by KeypadSDM
Oh yeah, whoops. Lol.

yep- he might b an oldman, but not an easy victim to pick on i tell u!

 

[OLDMAN]

Might just mention the equivalence of :

If A then B = If not B then not A.

 

[Grey Council]

Does this type of thing ever come up in the HSC? like that "iff" thing, or these type of statements.

 

[OLDMAN]

Originally posted by Grey Council
Does this type of thing ever come up in the HSC? like that "iff" thing, or these type of statements.

The "iff" is equivalent to two statements :
if A then B, and, if B then A.

Yes logical statements are done all the time in maths. For example to prove that SQRT(P) is irrational is to prove just one statement : If P is prime then SQRT(P) is irrational. Proving the equivalent ,if SQRT(P) is not irrational then P is not Prime, gets you the desired result. You might know this as proving by contradiction. While we're at it, here's a well known proof:
if SQRT(P) is not irrational ie. SQRT(P)=q/r where q,r are relatively prime (no common factors)
then P =(q/r)^2=(q/r)(q/r)
Thus q/r must be an integer, because no fraction multiplied by itself gives a whole number if that fraction is made up of two numbers with no common factor. Thus P must not be Prime.

 

[Grey Council]

hang on:

SQRT(9/4) = 3/2 (relatively prime, no common factors)

9/4 = (3/2)² = (3/2)(3/2)
however, 9/4 isn't an integer. Aren't integers counting numbers?

hrm, I have this gut feeling that 9/4 is irrational. hang on, no its not. 2.25
what am I doing wrong?

 

[KeypadSDM]

You forgot, p is necessarily an integer.

 

[Grey Council]

oh, right.