You are not logged in.
Pages: 1
Find real solutions of the system
sin x + 2 sin(x+y+z) = 0,
sin y + 3 sin(x+y+z) = 0,
sin z + 4 sin(x+y+z) = 0.
Offline
Offline
but how JaneFairfax
Offline
So you have
The solution to this is sinx = siny = sinz = 0.
Offline
PLESE step by step
Offline
Take each individual equation and solve for sin(x + y + z).
So, from your first equation we get 2sin(x + y + z) = -sin(x) ---> sin(x + y + z) = -(1/2)(sin(x)). Doing the same for the other 2 equations gives you sin(x + y + z) = -(1/3)(sin(y)) and sin(x + y + z) = -(1/4)(sin(z)).
Now, since sin(x + y + z) appears by itself in all 3 equations, we know the right side of each equation is equal to each other, so we now have -(1/2)(sin(x)) = -(1/3)(sin(y)) = -(1/4)(sin(z)). Multiply through by -12 to get 6sin(x) = 4sin(y) = 3sin(z).
Take the first equation, 6sin(x) = 4sin(y). Since both sides are of the form a*sin(w), where a is a constant and w is a variable, the graphs of both with have the same period and initial position. The only difference is the amplitude, which will cause the graphs not to touch anywhere except where sin(w) = 0. This same argument can be used for 3sin(z).
This gives us our final equation of sin(x) = sin(y) = sin(z) = 0. As you probably know, this is true when x, y, and z are any integers multiplied by pi.
Wrap it in bacon
Offline
Take the first equation, 6sin(x) = 4sin(y). Since both sides are of the form a*sin(w), where a is a constant and w is a variable, the graphs of both with have the same period and initial position. The only difference is the amplitude, which will cause the graphs not to touch anywhere except where sin(w) = 0.
No, this is not correct. They will all have the same period, but they may not have the same initial position in which case they may intersect at places other than where sin(w) = 0. For example, if we consider just the equation 6sin(x) = 4sin(y) alone, we see that it can be satisfied by x = π⁄6, y = sin[sup]−1[/sup](0.75).
In fact, Ive just realized that I was rather hasty too to reach my conclusion in my working.
Last edited by JaneFairfax (2007-11-01 03:26:09)
Offline
I've been giving out a lot of wrong answers.
Going back to our equation 6sin(x) = 4sin(y) = 3sin(z), solve for both sin(x) and sin(y) in terms of z:
sin(x) = (1/2) sin(z)
sin(y) = (3/4) sin(z)
Now take the arcsin of these equations:
x = arcsin( (1/2) sin(z) )
y = arcsin( (3/4) sin(z) )
We know this operation is legal since the range of the sin function is -1 to 1, which is the domain of the arcsin function. By multiplying sin(z) by a value less than 1 we stay within arcsin's domain.
From here choose any real value for z and plug it into these equations for a legal triple. Stated more mathematically:
Note that there's probably a way of simplifying from here, but I don't know how.
Edit: Well, on second thought it doesn't seem to work out very well. Taking Jane's example of x = pi/6, y = arcsin(3/4), and z = pi/2 we get sin(x) = 1/2, sin(y) = 3/4, and sin(z) = 1. This means that we need sin(x + y + z) = -1/4 to solve the original 3 equations. Unfortunately, sin(x + y + z) in this example equals 0.1978..... It appears that something is wrong here.
Last edited by TheDude (2007-11-02 02:40:56)
Wrap it in bacon
Offline
We have
.If
In the same way if one of the others two factors from (1) is 0
Offline
Pages: 1