Solving simultaneous trigonometric equation

[amibhatta]

how to solve
eqn1 = sin((alpha)*a)*cosh((alpha)*b)=sin(alpha)*a
eqn2 = sinh((alpha)*b)*cos((alpha)*a) = b*sin(alpha)
alpha = 0 to 2pi...

Find a & b...

Plz help me with any method that would help solving this...

 

[jpreed]

I can get it reduced to:\sin(\alpha x) = x \sin(\alpha)

\sin(\alpha y) = y \sin(\alpha)

x = a + (b/i) \quad \& \quad y= a - (b/i)beyond that, I am a bit stuck. Anyone else?

 

[Дьявол]

What is that "h" in front of cos in the first equation and sin in the second equation?

Anyway, the solution is:

eq1 - eq2
sin(\alpha*a)*cosh(\alpha*b)-sinh(\alpha*b)*cos(\alpha*a)=a*sin\alpha-b*sin\alpha

Now use the sum difference formula

sin(u \pm v)=sinucosv \pm cosusinv

and you got

sin(\alpha*a - \alpha*b) = (a-b)sin\alpha

sin((a-b)\alpha)=(a-b)sin\alpha

x=a-b

sin(x\alpha)=xsin\alpha

Regards.

 

[amibhatta]

Hi,

Thank for ur reply,

Reply 1:
I will look into it... thanx...

Reply 2:
'h' means hyperbolic function... and the result u have showed is from where i started to get these equations... these are transcedental equations having many roots so that eqn formed at the end that you got is not solvable(that is i don't know to solve)... if you know any method please reply...

Thank you.

 

[Дьявол]

Aaah... I see now...
Could you possibly provide me the original problem?

Regards.

 

[amibhatta]

[/tex]

Aaah... I see now...
Could you possibly provide me the original problem?

Regards.

sinZ = C1*Z
Z = (\alpha)*(\lambda)
C1 =(+/-)[Sin(\alpha)] / (\alpha)

hence we get,

Sin((\alpha)*(\lambda)) = \lambda *sin(\alpha)

\lambda = a+ib

hence eqn1 and eqn2...