# 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...

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#### 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?

Last edited:

#### 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 in to 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 dont know to solve).... if you know any method plz 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...

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