Question

What is the ordered pair of positive integers (a,b) for which a/b is a reduced fraction and x=[a(pi)]/b is the least positive solution of the equation: (2cos8x -1) (2cos4x -1) (2cos2x -1) (2cosx -1)=1?

Answer 1

is that \((2\cos^8x-1)(2\cos^4x-1)(2\cos^2x-1)(2\cos x-1)\)?

Answer 2

no \[(2\cos8x -1)(2\cos4x -1)(2\cos2x -1)(2cosx -1) =1\]

Answer 3

0 can definitely be a value for x, except that 0 cannot be expressed as a fraction...

Answer 4

this is damn tricky... I suppose it's \((1,1)\) by the fact that it yields \(x=\pi\) which is a trivial solution to the equation given \(\cos k\pi=0,k\in\mathbb{Z}\)

Answer 5

2 pi works too

Answer 6

cos 2pi = cos0

Answer 7

oops what the hell did I just write

Answer 8

I meant \(\cos k\pi=(-1)^k\) hehe

Answer 9

no worries, i just saw the cosine graph and realize that

Answer 10

since x=2pi, then (a,b) is (2,1)

Answer 11

but then 2/1 isn't a reduced fraction... ><

Answer 12

2/1 is a reduced fraction actually :-p

Answer 13

oh.... but 2/1 should be simplified to 2

Answer 14

Do you think that is correct? (a,b)--> (2,1)

Answer 15

yes, indeed, and \(2/1\) is the number \(2\) in lowest \(p/q\) form.

well let me think. I know that \(\cos(\pi n)=(-1)^n\) so let \(z=-1\) and we can write$$(2z^8-1)(2z^4-1)(2z^2-1)(2z-1)$$but hmm I wonder how to prove no \(\dfrac{a}b\pi\) works for \(a<b\)

Answer 16

Thanks for looking at this with me!! Appreciate it~ =D

Answer 17

I think I've got it

Answer 18

so obvious in retrospect ;-)$$(2\cos(8x)-1)(2\cos(4x)-1)(2\cos(2x)-1)(2\cos(x)-1)=1$$so$$(2\cos(8x)-1)(2\cos(4x)-1)(2\cos(2x)-1)(4\cos^2(x)-1)=2\cos(x)+1$$recall that \(\cos^2(x)=\dfrac12(1+\cos(2x))\) ergo \(4\cos^2(x)=2+2\cos(2x)\) hence \(4\cos^2(x)-1=2\cos(2x)+1\) and we have:$$(2\cos(8x)-1)(2\cos(4x)-1)(2\cos(2x)-1)(2\cos(2x)+1)=2\cos(x)+1\\(2\cos(8x)-1)(2\cos(4x)-1)(4\cos^2(2x)-1)=2\cos(x)+1$$observe this effect continues to cascade:$$(2\cos(8x)-1)(2\cos(4x)-1)(2\cos(4x)+1)=2\cos(x)+1\\(2\cos(8x)-1)(4\cos^2(4x)-1)=2\cos(x)+1\\(2\cos(8x)-1)(2\cos(8x)+1)=2\cos(x)+1\\4\cos(8x)^2-1=2\cos(x)+1\\2\cos(16x)+1=2\cos(x)+1\\\cos(16x)=\cos(x)$$

Answer 19

so we know that when \(\cos(16x)=\cos(x)\) it means that \(16x=2\pi k\pm x\) for integer \(k\). Observe:$$17x=2\pi k\\x=\frac{2\pi k}{17}$$or$$15x=2\pi k\\x=\frac{2\pi k}{15}$$in both cases it is obvious that we choose \(k=1\) and \(2/17<2/15\) ergo \((2,17)\) is our answer

Answer 20

I am very expressed by your solution! Well done!!

Answer 21

I was able to follow everything except I am a little confused on how did you get from step one to step two

Answer 22

Multiplying by \(2\cos(x)+1\):$$(2\cos(x)-1)(2\cos(x)+1)=4\cos^2(x)-1$$

Answer 23

that is I multiplied both sides by \(2\cos(x)+1\) and then simplified the LHS as above

Answer 24

okay, how did you figure out that you have to multiply (2cosx +1) ?

Answer 25

just wondering.....

Answer 26

It was an initial guess of mine but I didn't inspect too closely... came back later and realized it worked :-p

Answer 27

=D Wonderful~

Answer 28

Well, thank you so much for figuring this one out with me!!! Appreciate it!!! =D

Answer 29

NP -- glad I could help

Answer 30

@oldrin.bataku Sorry for the trouble
I wanted some clarification on the last 3 steps, as I was re-writing this on paper, I realize something different than your solution:
\[4\cos^28x -1 = 2\cos x +1\]
\[4(\cos^28x)-1=4(\frac{\cos 8x +1 }{ 2 }) =2\cos8x +1 \]
then should it be \[2\cos 8x +1 =2\cos x +1\]
then \[\cos 8x = \cos x\]
Thus the solution will be 2/9 or 2/7
and (a,b) ---> (2,9)

Answer 31

are you sure \(\cos^2(8x)=\dfrac12(\cos(8x)+1)\)? ;-)

Answer 32

...?

Answer 33

you will see that answer does not work while mine does:
http://www.wolframalpha.com/input/?i=%282cos%288*2%2F9+pi%29+-+1%29%282+cos%284*2%2F9+pi%29+-+1%29%282+cos%282*2%2F9+pi%29+-+1%29%282+cos%282%2F9+pi%29+-+1%29

Answer 34

I see my mistake now :P thanks again~

Answer 35

mmm.... just one last justification, please, @oldrin.bataku
I couldn't understand why 16x=2πk±x

Answer 36

whenever you have \(\cos(x)=\cos(y)\) recall that we know \(\cos(x)=\cos(-x)\) so \(y\) could be \(\pm x\) and further we know that \(\cos(x+2\pi k)=\cos(x),k\in\mathbb{Z}\) i.e. that y could also be \(2\pi k\pm x\)

Answer 37

essentially if we know two values \(x,y\) have the same cosine then we know they are (up to a difference that is an integer multiple of \(2\pi\) as \(\cos\cdot\) is periodic) equal up to sign change

Answer 38

draw it: