- #1
Gwilim
- 126
- 0
Doing a past paper where long story short I have a set of equations and I have to prove the variables are equal to some trig function. Namely:
a^2 + c^2 = b^2 + d^2 = 1
ad - bc = ab + cd = 0
Now I know that sin^2 + cos^2 = 1 and that sin(A+B)=sinAcosB + cosAsinB, etc
and the conclusion is that these equations have a general solution expressing a, b, c and d as functions of some parameter *, i.e. a=cos* b=sin* c=-sin* d=cos*.
Now I can easily see that this is true when I put the results back into the equation, but I am wondering how I know that only these functions will do.
For example if we are told that a^2+b^2=1, it is certainly true for all a:=sin*, b:=cos*, but might there not be other solutions? Is it simply common knowledge that there aren't?
Well, I have a quadratic I can solve to check?
a^2-1+b^2=0
Oh it's a function of two variables
a^2=-(b+1)(b-1)
a=-((b+1)(b-1))^(1/2)
Okay so now I suppose we have a in terms of b, we have a similar argument for c in terms of d, and we can plug those into ab+cd=0 and ad-bc=1, and we will have two equations relating a to c.
I suspect this will still lead to an equation which looks like a trig identity. But even so, it would not serve as a proof that all solutions to a^2+b^2=1 must take the form a=sin* b=cos*.
I could jump to that conclusion to get the marks in an exam but I would like to see a proof. I'm probably being blind to something very obvious.
a^2 + c^2 = b^2 + d^2 = 1
ad - bc = ab + cd = 0
Now I know that sin^2 + cos^2 = 1 and that sin(A+B)=sinAcosB + cosAsinB, etc
and the conclusion is that these equations have a general solution expressing a, b, c and d as functions of some parameter *, i.e. a=cos* b=sin* c=-sin* d=cos*.
Now I can easily see that this is true when I put the results back into the equation, but I am wondering how I know that only these functions will do.
For example if we are told that a^2+b^2=1, it is certainly true for all a:=sin*, b:=cos*, but might there not be other solutions? Is it simply common knowledge that there aren't?
Well, I have a quadratic I can solve to check?
a^2-1+b^2=0
Oh it's a function of two variables
a^2=-(b+1)(b-1)
a=-((b+1)(b-1))^(1/2)
Okay so now I suppose we have a in terms of b, we have a similar argument for c in terms of d, and we can plug those into ab+cd=0 and ad-bc=1, and we will have two equations relating a to c.
I suspect this will still lead to an equation which looks like a trig identity. But even so, it would not serve as a proof that all solutions to a^2+b^2=1 must take the form a=sin* b=cos*.
I could jump to that conclusion to get the marks in an exam but I would like to see a proof. I'm probably being blind to something very obvious.