/sci/ - Science & Math

If you fail at the latter part, then you have the determinant in expanded form right?
The [1*(stuff)+cos(a)*(more stuff) + cos(2a)*(even more stuff)]?

>>2172265
Thanks for the reply.

Yes, I tried calculating the determinant like that, but I don't quite see it working. Trigonometry is my problem here, I think.
There must be some way to simplify it, that's what the exercise says.

Well you can get cos 4a in terms of cos 2a with double/half angle formulas(wikipedia is nice), so the first term would be something like
<span class="math">1*(\cos^2{3a}-\cos{2a}f(cos{2a}))[/spoiler]
Basically I think you need to apply sum/difference and half/double angle formulas until everything is in terms of cos(a).
Sounds like a pain.

>>2172327

I see how this might work. Thank you. However, this method doesn't really make use of any determinant properties, does it? Like, multiplying rows/columns with a nonzero factor, adding a row/column to another one, etc.

>>2172349
No, it doesn't, but there might be some series way to represent it, since 1=cos(0*a).
the coefficients in front of a are 0 1 2, 1 2 3, 2 3 4, so there might be some trick, but it's probably too specific to be that important. Using determinant properties to do row operations wouldn't yield that much since cos(n*a) doesn't really play that well with different values of n. It's not a linear function or anything.

>>2172355

Alright. I will test this out to see how it works. Meanwhile, if anyone else has anything to contribute, please go right ahead.

>>2172360
>Meanwhile, if anyone else has anything to contribute, please go right ahead.
Thanks for your permission to post in this thread.

...

I realize it's just being nice or a formality or whatever, but it's totally unnecessary and you'd save a lot of time not typing niceties here. It's 4chan and you're anonymous, no one really cares.

>>2172252
if you dont want to write it out and sum, i get some sort of fourier vibe from this, but cant put my finger on it

>>2172369

Haha. Yeah. What I meant to say is that I'd stick around for a bit.

Problem I remember from the Putnam Competition that looked kind of similar.

The answer mentions that cos(2+a) is a multiple of cos(a)
Perhaps you can find something like that.

>>2172252
Probably not the fastest way, but you can always convert your sines to powers of e if you understand those better.

bumping for help

Looking at it from the linear Algebra view we have a symmetric 3x3 matrix but we dont know if it is regular, the set B={1,cosx,cos2x} is apparently independent (a base), if the second column was a linear combination of B then the determinant would be 0. Also, maybe we can diagonalize that shit since equivalent matrices have the same determinant.

I dunno man

>>2173386

I know the determinant is zero, calculators show as much. But I have to prove it with basic determinant properties (ie row or column transposition etc). I just can't see how it can be done

>>2173430
try using trig. to show that the columns or rows are not lineary independent. For instance, by multiplying second column with cos(a) and adding to first column, i see that you get the trig. identity for double angle, which is first element in last column. Maybe the other work out as well?

>>2173483

I don't see how that works.

>>2173527
It probably doesn't, i just wrote down my first thoughts looking at the problem. Too occupied with different shit to do the actual calculations.

>>2173539

Alright. Thanks anyway.

>>2173527

Double angle identities.

1*(cos2a * cos4a - cos3a * cos3a) -
cosa(cosa * cos4a - cos2a * cos3a) +
cos2a(cosa * cos3a - cos2a * cos2a)
expanding along first column

>>2173552

The point is to prove it elegantly in few lines knwoing that the determinant is, some columns are dependent on others, pure trigonometry

>>2173561
I'm stupid apparently and still fail to understand, could you please explain further?

OP here, best I've gotten so far is
1 2(cosa)^2-1 cos2a
cosa 2cosa*cos2a - cosa cos3a
cos2a 2cosa*cos3a - cos2a cos4a

And I can see that there, the second element of the first row equals the first element of the third column.

What now?

shameless bump

1. Switching two rows or columns changes the sign.

2. Scalars can be factored out from rows and columns.

3. Multiples of rows and columns can be added together without changing the determinant's value.

4. Scalar multiplication of a row by a constant multiplies the determinant by .

5. A determinant with a row or column of zeros has value 0.

6. Any determinant with two rows or columns equal has value 0.

Use prop 6 after rearranging a bit might work.

Anybody else?

bump

I can tell you a slick way but probably not the way your teacher inteded unless he required you to know about fourier analysis

>>2175157

No, that's not what I'm looking for, you're right.

So no one can help? I've tried everything in this thread without much success

It's easier if you convert the cosines to exponentials (cos x = e^{x i} ). From there you'll want to see if you can perform some elementary row operations that will make two rows equal to each other.

>>2176850

Do what he says then divide the second column by e^(ai), first and second column are now the same.