Trig identities

by Gwilim
Tags: identities, trig
 P: 123 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.
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P: 26,148
Hi Gwilim!

(have a square-root: √ and a theta: θ and try using the X2 icon just above the Reply box )
 Quote by Gwilim 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?
Sorry, but yes (with ±s).
If a = cosθ, then 1 - a2 = … ?
P: 123
 Quote by tiny-tim Hi Gwilim! (have a square-root: √ and a theta: θ and try using the X2 icon just above the Reply box ) Sorry for the awful formatting I'm posting with an iPod.
 Sorry, but yes (with ±s). If a = cosθ, then 1 - a2 = … ?
right but I am wondering about that if. What if a is the square root of cos2θ+C for some C>2? wouldn't the equation 1-a2=b2still hold for b = sqrt(sin2θ-C)? And a is no longer equal to cos θ for any θ.

Ah okay so b is a complex number and since the question wanted real solutions I have not found a counterexample as far as the exam is concerned but I still haven't proved that there aren't any.

I suppose we can say that a and b must exist on the interval [-1,1] if we want real solutions. But we can define functions other than sines and cosines with that range.

No no wait since sines and cosines are surjective on that range any number on the interval [-1,1] can be written as the sine or cosine of something, and so a can be written as sinθ at which point we can go ahead and identify b with cosθ.

Alright I guess I'm convinced. Does this mean it's possible to actually define sines and cosines as the functions with that relative property?