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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.
You still need to prove that it's impossible for X = 1/2.
To do this, consider that
Now we need to prove that
It's important to note that
Knowing this, we also know that
is always even, since all prime numbers after 2 are odd. This means that the prime factorization of the numberator will always have at least one 2 in it. We also know that obviously the denominator will never have a 2 in it's prime factorization, since all the denominators are prime numbers greater than 2. Since we have a prime factor in the numerator that can never be cancelled by the denominator we know that X can never reach 1/2, so that should complete your proof.Heh, your method is a lot easier.
Set f(a) = b. Substitute to get g(b) = g(f(a)), which we know from the definition of g is equal to a. Thus, g(b) = a.
Do the same for h to get h(f(a)). We know from the definition of h that f(h(f(a))) = f(a), which tells us that h(f(a)) = a. Substitute back to get h(b) = a = g(b).
You'd be amazed at how quickly you can forget even the simple things.
Like I said, I'm not good with matrices, so I had to use this Wikipedia page: http://en.wikipedia.org/wiki/Cofactor_%28linear_algebra%29
Cofactor(A) =
For example:
I leave the rest to you.
Your cofactor(A) doesn't look correct to me. I'm not great with matrices, but I think this is the cofactor:
Cofactor(A) = 1 -10 -7
-7 10 -11
17 10 1
= M
I think this is how you get to M from A.
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.
As always, wikipedia can get you started: http://en.wikipedia.org/wiki/Euler%27s_totient_function
Better late than never:
This isn't really a coding problem, but here's the method anyway. From the original equation we know a couple facts about A and h. First, we know that A > 0 (strictly greater than) because ln is undefined for values of A <= 0. For the same reason we know that h > 0. However, we also know that 1 - h > 0, which implies h < 1 (again, strictly). This gives us A > 0 and 0 < h < 1.
Now, we first simplify the equation by combining the logs on the right side. ln(a) - ln(b) = ln(a / b) is a logarithmic identity, so we get ln(A) = ln(h / (1 - h) ).
Now, since the exponential function is continuous over the real numbers we can take the exponential of both sides to remove the logs and get A = h / (1 - h).
From here on out it's just algebra, but for the sake of completeness I'll go step by step. Since we know both A and h are strictly greater than 0 we can take the reciprocal of both sides to give us 1 / A = (1 - h) / h. We then split up the fraction on the right side to get 1 / A = (1 / h) - 1. Add 1 to give us 1 + (1 / A) = 1 / h --------> (A + 1) / A = 1 / h. Again, A + 1 > 0, so take the reciprocal again to give us A / (A + 1) = h QED.