soopo
Nov1-09, 02:25 PM
1. The problem statement, all variables and given/known data
Find A,B,C in sin^5(x) = Asin(x) + Bsin(3x) + Csin(5x).
3. The attempt at a solution
I get by Euler the double angle identities for sin(3x) and sin(5x).
They are
sin(3x) = s(2x) c(x) + c(2x) s(x)
sin(5x) = s(3x) c(2x) + c(3x) s(2x)
I have the following expression now
sin^5(x) = 1 - cos^5(x) - 5c^4(x)s(x) - 10c^3(x)s^2(x) - 10 c^2(x)s^3(x) - 5 c(x)s^4(x)
where the trigonometric terms are (1/2) s(2x) c^3(x), (1/4) (s(2x))^2 c(x), (1/4) (s2x)^2 s(x), (1/2) s(2x) s^3(x) , respectively.
I get the common term [(1/2)s(2x) = c(x)s(x) by the double angle identity of sine.
So you have (-5/2) s(2x) (c^3(x) + c(x) + s(x) + s^(x) .
By comparing the terms to the given equation, we get
Bsin(3x) = B (s(2x) c(x) + c(2x) s(x) )
Csin(5x) = C ( s(3x) c(2x) + c(3x) s(2x) )
so we have
B c(x) + C c(3x) = (-5/2) ( c^3(x) + c(x) + s(x) + s^3(x) )
which implies that C = - \frac {-5} {2}
The other term is in the form
B c(x) = (-5/2) ( c(x) + s(x) + s^3(x) )
where where s(x) + s^3(x) = s(x) (1 + s^(x) = 2s(x) - c^2(x) s(x) by Pythagoras.
However, I do not see directly how to get the term [itex] c(x) [/tex] to the RHS of the equation.
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My first attempt seems to be useless, since the answer may be found by Fourier series too.
However, I have little experience of them and cannot see to how use them here.
How can you find A, B and C by Fourier series or by other method?
Find A,B,C in sin^5(x) = Asin(x) + Bsin(3x) + Csin(5x).
3. The attempt at a solution
I get by Euler the double angle identities for sin(3x) and sin(5x).
They are
sin(3x) = s(2x) c(x) + c(2x) s(x)
sin(5x) = s(3x) c(2x) + c(3x) s(2x)
I have the following expression now
sin^5(x) = 1 - cos^5(x) - 5c^4(x)s(x) - 10c^3(x)s^2(x) - 10 c^2(x)s^3(x) - 5 c(x)s^4(x)
where the trigonometric terms are (1/2) s(2x) c^3(x), (1/4) (s(2x))^2 c(x), (1/4) (s2x)^2 s(x), (1/2) s(2x) s^3(x) , respectively.
I get the common term [(1/2)s(2x) = c(x)s(x) by the double angle identity of sine.
So you have (-5/2) s(2x) (c^3(x) + c(x) + s(x) + s^(x) .
By comparing the terms to the given equation, we get
Bsin(3x) = B (s(2x) c(x) + c(2x) s(x) )
Csin(5x) = C ( s(3x) c(2x) + c(3x) s(2x) )
so we have
B c(x) + C c(3x) = (-5/2) ( c^3(x) + c(x) + s(x) + s^3(x) )
which implies that C = - \frac {-5} {2}
The other term is in the form
B c(x) = (-5/2) ( c(x) + s(x) + s^3(x) )
where where s(x) + s^3(x) = s(x) (1 + s^(x) = 2s(x) - c^2(x) s(x) by Pythagoras.
However, I do not see directly how to get the term [itex] c(x) [/tex] to the RHS of the equation.
---------------------------------------
My first attempt seems to be useless, since the answer may be found by Fourier series too.
However, I have little experience of them and cannot see to how use them here.
How can you find A, B and C by Fourier series or by other method?