- #1

- 89

- 1

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

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- Thread starter elfboy
- Start date

- #1

- 89

- 1

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

- 350

- 1

so does cos(ax)+cos(bx)=0

Those two cos's are exactly out of phase (to cancel like that) so

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

- #3

- 89

- 1

Wikipedia doesn't seem to have any identities for triple-cosine-sum-to-product

- #4

- 200

- 0

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

To clarify, when you say it has a closed form solution in terms of x, it seems like you're asking if given any x we can find a,b,c such that the equation is satisfied. Surely, you do not mean that since that problem is simple. I'm guessing you're asking it it is possible to find a,b, and c such that the equation holds for all x?

- #5

- 726

- 1

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.

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.

Last edited:

- #6

quadraphonics

- #7

- 143

- 0

No, you cant substitute a value from one equation into itself. You will get: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)

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

- 726

- 1

No, you cant substitute a value from one equation into itself. You will get:

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.

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?

Last edited:

- #9

- 89

- 1

thats right

The cos(ax)+cos(bx)=0 roots can be derived by adding cos(ax+bx) and cos(ax-bx)

but the expanded form of cos(ax+bx+cx) doesn't yield a sum-to-product 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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