
#1
May3108, 08:09 PM

P: 89

The equation cos(ax)=0 has a closed form solution in terms of x
so does cos(ax)+cos(bx)=0 but I am sure that cos(ax)+cos(bx)+cos(cx)=0 does not I first encountered this problem back in late 2003 and began working on it again 



#2
May3108, 10:04 PM

P: 352

[tex]ax = bx +(2n+1)\frac{\pi}{2}[/tex] 



#3
Jun208, 12:36 AM

P: 89

Wikipedia doesn't seem to have any identities for triplecosinesumtoproduct




#4
Jun308, 09:38 PM

P: 201

very difficult math problem 



#5
Jun308, 11:57 PM

P: 724

Your question seemed a bit ambiguous to me, so I hope this is what you are asking for:
We start off with: Cos(ax) + Cos(bx) + Cos(cx) = 0 Cos(bx) + Cos(cx) = Cos(ax) ArcCos[Cos(bx) + Cos(cx)] = ArcCos[Cos(ax)] Looking at the right side of the equation after taking the inverse Cos, it will be 180  ax. It will be 180  ax because the Cosine there is negative. Now, we want to write the sum of the Cosines on the left side as a single Cosine so we can take the inverse of it and get rid of Cosine all together. We can rewrite: Cos(bx) + Cos(cx) as Cos(ax) We can do this because the right side of the equation, before taking the inverse Cosine, is Cos(ax). The left side must be the additive inverse of this, and so must be Cos(ax). However, the left side is Cos(bx) + Cos(cx), which means that Cos(bx) + Cos(cx) must be the additive inverse of Cos(ax), which is Cos(ax). So now, we have ArcCos[Cos(ax)] = 180  ax ax = 180  ax 2ax = 180 a = 90/x So, we now know a can be written as 90/x. Let's substitute that into the original equation: Cos(ax) + Cos(bx) + Cos(cx) = 0 Cos( 90/x * x) + Cos(bx) Cos(cx) = 0 Cos(90) + Cos(bx) + Cos(cx) = 0 But Cos90 = 0, so: Cos(bx) + Cos(cx) = 0 Cos(bx) = Cos(cx) ArcCos[Cos(bx)] = ArcCos[Cos(cx)] bx = 180  cx bx + cx = 180 x(b+c) = 180 b+c = 180/x So, we now have two equations: a = 90/x and b + c = 180/x So, pick any value of x, substitute it into the first equation, you will get a value for a. Substitute a value for x into the second equation, then pick values of b and c to satisfy that equation and you will your values for a, b, c, and x that will satisfy your original equation (as long as x does not equal 0). I hope that is what you were asking for, and sorry if the format is hard to follow. I don't know how to use math symbols on the computer. 



#6
Jun408, 12:05 AM

P: 270

I think what elfboy is asking for is a closedform expression for the set of all x that satisfy the equation, given some a, b and c. For a single cosine, this is trivial: it's just the set of all integer multiples of [itex]\frac{\pi}{a}[/itex]. For a sum of two cosines, you can apply a trig identity to get a product of two cosines, and then apply the same reasoning as before to each one. This gives the solution as all integer multiples of either [itex]\frac{2\pi}{a+b}[/itex] or [itex]\frac{2\pi}{ab}[/itex]. For the case of three cosines, it's not so obvious, at least when [itex]c \neq 0 [/itex]...




#7
Jun408, 03:23 AM

P: 124

Cosbx+Coscx=cosax cosax=cosax 0=0 Im not sure though that understand what you mean, but anyway what you have done is not the answer the the thread starters question. I think hes question was: Given real numbers a,b and c, find the solution of Cosax+Cosbx+Coscx=0 in terms of a,b and c. 



#8
Jun408, 09:46 AM

P: 724

In that part I'm not substituting a value from one equation into itself. I am saying that the left side must be the additive inverse of the right side in this situation, so we can rewrite Cos(ax) + Cos(bx) as Cos(ax). The two equations I came up with work in any situation, except where x = 0. But I'm not 100% sure I properly answered the threadmakers question, I agree with you there. Any input from the thread maker? 



#9
Jun508, 01:46 AM

P: 89

[QUOTE=quadraphonics;1754556]I think what elfboy is asking for is a closedform expression for the set of all x that satisfy the equation, given some a, b and c. ..[QUOTE]
thats right The cos(ax)+cos(bx)=0 roots can be derived by adding cos(ax+bx) and cos(axbx) but the expanded form of cos(ax+bx+cx) doesn't yield a sumtoproduct identity as the case above does. the closest I have gotten is: cos(a+b+c)+cos(a)+cos(b)+cos(c)=4*cos((b+c)/2)*cos((a+b)/2)*cos((a+c)/2) 


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