- #1

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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...

- Thread starter amibhatta
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- #1

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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...

- #2

- 37

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I can get it reduced to:

[tex]\sin(\alpha x) = x \sin(\alpha)[/tex]

[tex]\sin(\alpha y) = y \sin(\alpha)[/tex]

[tex]x = a + (b/i) \quad \& \quad y= a - (b/i)[/tex]

beyond that, I am a bit stuck. Anyone else?

[tex]\sin(\alpha x) = x \sin(\alpha)[/tex]

[tex]\sin(\alpha y) = y \sin(\alpha)[/tex]

[tex]x = a + (b/i) \quad \& \quad y= a - (b/i)[/tex]

beyond that, I am a bit stuck. Anyone else?

Last edited:

- #3

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Anyway, the solution is:

eq1 - eq2

[tex]sin(\alpha*a)*cosh(\alpha*b)-sinh(\alpha*b)*cos(\alpha*a)=a*sin\alpha-b*sin\alpha[/tex]

Now use the sum difference formula

[tex]sin(u \pm v)=sinucosv \pm cosusinv[/tex]

and you got

[tex]sin(\alpha*a - \alpha*b) = (a-b)sin\alpha[/tex]

[tex]sin((a-b)\alpha)=(a-b)sin\alpha[/tex]

x=a-b

[tex]sin(x\alpha)=xsin\alpha[/tex]

Regards.

- #4

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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.

- #5

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Aaah... I see now...

Could you possibly provide me the original problem?

Regards.

Could you possibly provide me the original problem?

Regards.

- #6

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sinZ = C1*ZAaah... I see now...

Could you possibly provide me the original problem?

Regards.

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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