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Re: Trig identity for linear combination of sines
*To*: mathgroup at smc.vnet.net
*Subject*: [mg100827] Re: [mg100814] Trig identity for linear combination of sines
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Mon, 15 Jun 2009 05:37:05 -0400 (EDT)
*References*: <200906150120.VAA11516@smc.vnet.net>
On 15 Jun 2009, at 10:20, Chelly wrote:
> Hi:
>
> I have a function which is a linear combination of sines, of the form
>
> y = a*sin(x+q) + b*sin(x+r) + c*sin(x+t)
>
> and I am looking for a trignometric identity that would reduce the
> above to the form
>
> y = A*sin(x+Z)
>
> I would also like to find out how to implement this rule in
> Mathematica.
>
> Thanks
> Chelly
>
I do not know of any such trigonometric identity, and I think it would
have to be pretty complicated. However, you can use Mathematica to
find A and Z.
First consider the functions
ff[x_] := a Sin[x + q] + b Sin[x + r] + c Sin[x + t]
and
gg[x_] := A Sin[x + Z]
where a,b,c, q,r,t, A Z are some real numbers.
Both of these are analytic functions. Suppose, for given a,b,c, q,r,t
we find A and Z such that these two equations hold:
eq1 = ff[0] == gg[0]
a*Sin[q] + b*Sin[r] + c*Sin[t] == A*Sin[Z]
and
eq2 = ff'[0] == gg'[0]
a*Cos[q] + b*Cos[r] + c*Cos[t] == A*Cos[Z]
You can easily convince yourself that in this case the n-th
derivative of ff at 0 must equal the n-th derivative of gg at 0, since
the derivatives of order (n+2) are simply the derivatives of order n,
or - derivatives of order n. So as long as eq1 and eq2 hold the
functions will have equal derivatives of all orders at 0. But since
both ff and gg are analytic functions everywhere in the complex plane
they are determined by their derivatives at 0, and so they must be
equal. So we only need to solve eq1 and eq 2.
Reduce[{eq1, eq2}, {A, Z}]
will now give you a complete solution (which is not unique). You can
get a simpler looking (but complete) answer by using;
Simplify[{A, Z} /. Last[Solve[{eq1, eq2}, {A, Z}]]]
{(Sqrt[a^2 + 2*a*b*Cos[q - r] + 2*a*c*Cos[q - t] + b^2 + 2*b*c*Cos[r
- t] +
c^2]*
Sqrt[(a*Cos[q] + b*Cos[r] + c*Cos[t])^2])/(a*Cos[q] +
b*Cos[r] +
c*Cos[t]),
ArcCos[Sqrt[(a*Cos[q] + b*Cos[r] + c*Cos[t])^2]/
Sqrt[
a^2 + 2*a*b*Cos[q - r] + 2*a*c*Cos[q - t] + b^2 + 2*b*c*Cos[r -
t] +
c^2]]}
Note that some conditions need to be satisfied for these to be valid.
Reduce gives you all the conditions.
Andrzej Kozlowski
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