Trig equation

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• Jul 1st 2008, 01:08 PM
phillips101
Trig equation
Sorry, I don't know if this is in the right section or not.

Basically, the question is:
Solve sin(x) + sin(y) + sin(z) = 0 , for all x,y and z.

How I attempted it:

I assumed the identity sin(x) + sin(x+2pi/3) + sin(x-2pi/3) = 0 , which is easily proved with the compound angle formulae.

Then I just compared the equation and the identity, and said that if y were of the form x+2pi/3 and z were of the form x-2pi/3, then the equation would hold. So effectively there are an infinite number of solutions.

The next question was:
Solve cos(x) + cos(y) + sin(z) = 0

FOr this, I re-wrote the cos's as sin(Pi/2-x) (or y, as the case may be), and solved it exactly the same way as above.

Are these methods correct? I'm filled wiht doubt because on another forum they were talking about putting them in the complex plane as vectors and rotating them and all sorts of things...
• Jul 1st 2008, 03:16 PM
mr fantastic
Quote:

Originally Posted by phillips101
Sorry, I don't know if this is in the right section or not.

Basically, the question is:
Solve sin(x) + sin(y) + sin(z) = 0 , for all x,y and z.

How I attempted it:

I assumed the identity sin(x) + sin(x+2pi/3) + sin(x-2pi/3) = 0 , which is easily proved with the compound angle formulae.

Then I just compared the equation and the identity, and said that if y were of the form x+2pi/3 and z were of the form x-2pi/3, then the equation would hold. So effectively there are an infinite number of solutions.

The next question was:
Solve cos(x) + cos(y) + sin(z) = 0

FOr this, I re-wrote the cos's as sin(Pi/2-x) (or y, as the case may be), and solved it exactly the same way as above.

Are these methods correct? I'm filled wiht doubt because on another forum they were talking about putting them in the complex plane as vectors and rotating them and all sorts of things...

Looks fine ..... But would another parametric solution to the first be y = x - 2pi/3 and z = x + 2pi/3 etc.
• Jul 1st 2008, 04:53 PM
ticbol
Quote:

Originally Posted by phillips101
Sorry, I don't know if this is in the right section or not.

Basically, the question is:
Solve sin(x) + sin(y) + sin(z) = 0 , for all x,y and z.

How I attempted it:

I assumed the identity sin(x) + sin(x+2pi/3) + sin(x-2pi/3) = 0 , which is easily proved with the compound angle formulae.

Then I just compared the equation and the identity, and said that if y were of the form x+2pi/3 and z were of the form x-2pi/3, then the equation would hold. So effectively there are an infinite number of solutions.

The next question was:
Solve cos(x) + cos(y) + sin(z) = 0

FOr this, I re-wrote the cos's as sin(Pi/2-x) (or y, as the case may be), and solved it exactly the same way as above.

Are these methods correct? I'm filled wiht doubt because on another forum they were talking about putting them in the complex plane as vectors and rotating them and all sorts of things...

Hard, Man, hard.

No relationships among x, y and z? Like x+y+z = 180deg, for example?
The problem says for all x, y, z. Man!

Your assuming y = x +2pi/3 and z = x -2pi/3) is a punch in the wind.

Hard.
• Jul 2nd 2008, 01:25 AM
phillips101
Quote:

Originally Posted by ticbol
Hard, Man, hard.

No relationships among x, y and z? Like x+y+z = 180deg, for example?
The problem says for all x, y, z. Man!

Your assuming y = x +2pi/3 and z = x -2pi/3) is a punch in the wind.

Hard.

To the first reply, yes that would be more solutions, and I suppose any permutations thereof as well.

To the guy I quoted: 'A punch in the wind'? Is that good or bad :S Anyway, no other information was given, and I'm not too interested in getting every single solution out - as long as my method -would- get every solution out I'll be happy.
• Jul 2nd 2008, 04:36 AM
ebaines
Your answer misses solutions such as:

x = y = z = 0,
x = -y, and z = 0,
x = -z and y =0,
y = -z and x = 0,

All of these are solutions to sin(x) + sin(y) + sin(z) = 0.
• Jul 2nd 2008, 04:39 AM
phillips101
Ah, yes, of course. There are lots more of them about as well... I think the person who posted the question missed something. (posted it originally, I'm not talking about me here)
• Jul 2nd 2008, 04:42 AM
ticbol
Quote:

Originally Posted by phillips101
To the first reply, yes that would be more solutions, and I suppose any permutations thereof as well.

To the guy I quoted: 'A punch in the wind'? Is that good or bad :S Anyway, no other information was given, and I'm not too interested in getting every single solution out - as long as my method -would- get every solution out I'll be happy.

A punch in the wind.
There is an almost impossible target. It is out there but you have no idea where it is. It might be beyond your reach or within your reach. You have to do something. Out of desperation, and with some prayer, you throw a punch in the wind, hoping you'd hit the target.
A billion punches, in all conceivable directions, might not even scratch your target.